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\title{Application of Algebra and Calculus \\in Euclidean Geometry}
\author{Huy Nguyen, Trung Le}
\date{Le Hong Phong High School - Ho Chi Minh City}

\begin{document}

\maketitle
\tableofcontents

\newpage

\section*{Glossary}
\begin{center}
	\begin{tabular}{p{2cm}|p{14cm}}
		\hline
		\multicolumn{2}{c}{\bf Abbreviations} \\
		\hline \hline
		{\bf MO} & Abbrev. for {\bf Mathematical Olympiad}, the highest-level math contest for high school students in a region - for example, {\bf VMO} (Vietnam), {\bf USAMO} (America), {\bf IMO} (International). \\
		\hline
		{\bf TST} & Abbrev. for {\bf Team Selection Test}, a test given to top performers in a national olympiad to select the nation's representative team for an international olympiad. \\
		\hline
		\multicolumn{2}{c}{\bf Mathematical notions} \\
		\hline \hline
		$a,b,c,r,R \ $ & Unless otherwise specified, in $\triangle ABC$, $a, b, c$ denote the lengths of segments $BC, CA, AB$ respectively. $p$ is its semiperimeter, i.e., $p = \frac{1}{2}(a+b+c)$. $r$ and $R$ denote its inradius and circumradius respectively. \\
		\hline
		$S_{\triangle XYZ}$ & Denote the area of $\triangle XYZ$. \\
		\hline
		$(XYZ)$ & For any three non-collinear points $X, Y, Z$, denote the circumcircle of $\triangle XYZ$. \\
		\hline
		$(I;r)$ and $(I)$ & Denote the circle with center $I$ and radius $r$. When it is clear from the context that there is only one circle with center $I$, the parameter $r$ can be omitted. \\
		\hline
		$\ds{AB}$ & The algebraic length of segment $AB$. If $\vto{AB}$ points in the positive direction, $\ds{AB} = |\vto{AB}|$; otherwise, $\ds{AB} = -|\vto{AB}|$. \\
		\hline
		$(ABCD)$ & For any four collinear points $(A, B, C, D)$, denote the cross-ratio
		$$(ABCD) = \frac{\ds{AC}}{\ds{AD}} : \frac{\ds{BC}}{\ds{BD}}.$$
		When $(ABCD) = -1$, $A$ and $B$ are harmonic conjugates of each other with respect to $C$ and $D$. \\
		\hline
		$(AB,CD)$ & The oriented angle starting at $AB$ and ending at $CD$. \\
		\hline
	\end{tabular}
\end{center}

\newpage

\section{Introduction}
We begin with an analysis of a simple geometry problem, proposed by Dr. Trinh Le, deputy delegate of the Vietnam IMO team.


\setcounter{prob}{-1}
\bpr{}
In \abc consider a point $D$ on segment $BC$. Let $I$ and $J$ be the midpoints of $DB$ and $DC$ respectively. The perpendicular bisectors of $DB$ and $DC$ intersect $AB$ at $M$ and $AC$ at $N$ respectively. Let $O$ be the circumcenter of \abc. Prove that $AMON$ is a cyclic quadrilateral.
\epr
\begin{figure}[h]
	\centering
	\input{geogebra/pr0}
	\caption{Prove that $AMON$ is a cyclic quadrilateral.}
	\label{pr0-prompt}
\end{figure}

The following six solutions were devised by six students from the Ho Chi Minh City VMO team. Let us first take a look at their approaches and analyze the pros and cons of each.

\newpage
\begin{prf}
\addintroproofcontents
\begin{center}
	\input{geogebra/pr0-sol1}
\end{center}
Let $T$ be the midpoint of $BC$. Consider the Cartesian coordinate system $Txy$ where $T$ is the origin, $Tx$ is $BC$ and $Ty$ is the perpendicular bisector of $BC$. \\
Denote the following coordinate points: $B(-1,0), C(1,0), D(d, 0)$ and $A(a,b)$. \\
Let $U$ and $V$ be the midpoints of $AB$ and $AC$ respectively, and let $K$ be the projection of $O$ on $MN$. Furthermore, for any line $l$, denote $\vec{n_l}$ as its normal vector. \\
We can then compute the followings:
\begin{enumerate}

	\item $ (AB): y=\dfrac{b(x+1)}{a+1}, \quad (AC): y=\dfrac{b(x-1)}{a-1}.$

	\item $U \lf \dfrac{a-1}{2}, \dfrac{b}{2} \ri, \ V \lf \dfrac{a+1}{2}, \dfrac{b}{2} \ri, M \lf \dfrac{d-1}{2}, \dfrac{b(d+1)}{2(a+1)} \ri , \ N \lf \dfrac{d+1}{2}, \dfrac{b(d-1)}{2(a-1)} \ri.$

	\item $x_O = x_T=0,\  \vto{OU} \parallel \vto{n_{AB}} \Rightarrow O \lf 0; \dfrac{a^2+b^2-1}{2ab} \ri.$

	\item $(MN): \dfrac{b(a-d)}{a^2-1} \lf x -\dfrac{d-1}{2} \ri  + y - \dfrac{b(d+1)}{2(a+1)}=0.$

	\item $\vto{OK} \parallel \vto{n_{MN}} \Rightarrow \dfrac{x_K}{y_K - \frac{a^2+b^2-1}{2b}}=\dfrac{b(a-d)}{a^2-1}.$
\end{enumerate}

Observe that $y_U = y_v = \frac{b}{2}$, so it turns out the line $UV$ is $(UV): y = \frac{b}{2}$. Furthermore, the point $K$ is uniquely determined from (4) and (5), and we notice that $\lf x_K = \frac{d-a}{2}, y_K = \frac{b}{2} \ri$ satisfies both. Hence $K \lf \frac{d-a}{2}, \frac{b}{2} \ri$ and therefore $K \in UV$. By the converse \href{https://en.wikipedia.org/wiki/Simson_line}{Simson's theorem} $AMON$ is cyclic. (QED)
\end{prf}

\newpage
\begin{prf}
\addintroproofcontents
Let $N' \ne A$ be the intersection point of $(AMO)$ and $AC$. We prove $AN'=AN$, which would imply $N' \equiv N$. Denote $DB = x$, then
\begin{equation}
	BM=\frac{x}{2\cos B}, \ AM=c-\dfrac{x}{2\cos B}, \ AN=b-\dfrac{a-x}{2\cos C}.  \label{am-an}
\end{equation}

\begin{center}
	\input{geogebra/pr0-sol2}
\end{center}

Since $AMON'$ is a cyclic quadrilateral, by \href{https://en.wikipedia.org/wiki/Ptolemy\%27s_theorem}{Ptolemy's theorem}:
\begin{equation}
	AO \cdot MN' =AM \cdot ON' + AN' \cdot OM. \label{ptolemy}
\end{equation}
Further observe that $\angle{OMN'} = \angle{OAC}=90^0-B$ and $\angle{ON'M} =\angle{OAB}=90^0-C.$ Hence, in $\triangle OMN'$, applying the sine Rule, we have 
$$\dfrac{ON'}{OM}=\dfrac{\cos B}{\cos C}.$$
Multiplying both sides of (\ref{ptolemy}) by $\frac{\cos C}{OM}$ we have
$$(AO\cos C) \cdot \frac{MN'}{OM} = (AM \cos C) \cdot \frac{ON'}{OM} + AN'\cos C.$$
which yields
\begin{align*}
	AN'\cos C 
	& = (AO \cos C) \cdot \frac{\sin \angle MON'}{\sin \angle ON'M} - (AM \cos C) \cdot \frac{\cos B}{cos C} \\
	& = \lf \frac{a}{2\sin A} \cdot \cos C \cdot \frac{\sin A}{\cos C} \ri - \lf c - \frac{x}{2\cos B} \ri \cos B \\
	& = \frac{a + x}{2} - c\cos B.
\end{align*}
From (\ref{am-an}) we also have 
$$ AN \cos C=b \cos C - \dfrac{a-x}{2},$$ 
hence $AN' = AN$ would be equivalent to
$$\frac{a + x}{2} - c\cos B = b \cos C - \dfrac{a-x}{2},$$
which can be rewritten as  $a = b\cos C + c\cos B$,
which is true. Thus $AN' = AN$ and therefore $N' \equiv N$. (QED)
\end{prf}

\begin{prf}
\addintroproofcontents
\begin{center}
	\input{geogebra/pr0-sol3}
\end{center}
$(AMN)$ intersects the diameter $AK$ of $(ABC)$ at $O' \ne A$. We will prove that $AK=2AO'$, which implies $O' \equiv O$.\\
Observe that $\triangle O'MN \sim \triangle KBC$ so
$$ \dfrac{OM}{KB}=\dfrac{MN}{BC}=\dfrac{O'N}{KC}=k, $$
which means that 
\begin{equation}
	MN=kBC,\ O'M=kKB, \ O'N=kKC. \label{mn}
\end{equation}
Applying Ptolemy's theorem to cyclic quadrilaterals $AMO'N$ and $ABKC$ we have 
\begin{align}
	AO' \cdot MN & = AM \cdot O'N + AN \cdot O'M. \label{amo'n} \\ 
	AK \cdot BC & = AB \cdot KC + AC \cdot KB. \label{ak}
\end{align}
Dividing both sides of (\ref{amo'n}) by $k$ we have 
\begin{equation}
AO \cdot BC = AM \cdot KC + AN \cdot KB. \label{ao}
\end{equation}

From Solution 2, with $DB = x$, we also have 
$$AM=c-\dfrac{x}{2\cos B}, AN =b-\dfrac{a-x}{2\cos C}.$$
Further note that that $KB=2R\cos C$ and $KC=2R\cos B$. Looking at (\ref{ak}) and (\ref{ao}), to show $AK=2AO'$ we need to prove that
$$AB \cdot KC + AC \cdot KB = 2(AM \cdot KC + AN \cdot KB),$$
which is equivalent to
$$c\cos B + b\cos C= 2\cos B \lf c-\dfrac{x}{2\cos B} \ri+ 2\cos C \lf b-\dfrac{a-x}{2\cos C} \ri,$$
which reduces to $a=b\cos C + c\cos B$, which is true. Hence $O \equiv O' \in (AMN)$ and therefore $AMON$ is cyclic. (QED)
\end{prf}

\newpage
\begin{prf}
\addintroproofcontents
\begin{center}
	\input{geogebra/pr0-sol4}
\end{center}
Let $U,V$ be the midpoints of $AB,AC$ respectively and $R,S$ be the projections of $U,V$ onto $BC$ respectively.
It is easy to see that $IR=JS$. Indeed, on axis $BC$, denote the direction of $\vto{BC}$ as the positive direction. We then have 
\begin{align*} 
	& \ds{IR}=\ds{BR}-\ds{BI}=BU \cos (BR, BU) - \dfrac{x}{2} = \dfrac{c\cos B - x}{2}, \\
	& \ds{JS}=\ds{CS}-\ds{CJ}=CV\cos (CS, CA) + \dfrac{a-x}{2}=\dfrac{-b\cos C +a-x}{2}.
\end{align*}
Since $a=b\cos C + c\cos B$, $\ds{IR}=\ds{JS}$ so $IR=JS$.\\
Applying \href{https://en.wikipedia.org/wiki/Thales%27s_theorem}{Thales's theorem} for $UR \parallel MI$ and $VS \parallel NJ$ we have 
$$\dfrac{MU}{IR}=\dfrac{BU}{BR}=\dfrac{1}{\cos B}, \ \dfrac{NV}{SJ}=\dfrac{CN}{CJ}=\dfrac{1}{\cos C}.$$
Hence $$\dfrac{MU}{NV}=\dfrac{IR}{JS} \cdot \dfrac{\cos C}{\cos B}=\dfrac{\cos C}{\cos B} =\dfrac{OU}{OV}.$$
Observe that $\triangle OMU$ and $\triangle ONV$ are directionally similar. Let $f$ be the spiral similarity with center $O$, angle $(OU,OV)$ and ratio $k=\frac{OU}{OV}$, then $f(U)=V, f(M)=N$. \\ 
Hence $(MO,MU) \equiv (NO,NV) \ \pmod{\pi}$, so $AMON$ is a cyclic quadrilateral. (QED)
\end{prf}

\newpage
\begin{prf}
\addintroproofcontents
\begin{center}
	\input{geogebra/pr0-sol5}
\end{center}
Let $E$ be the reflection of $D$ across $MN$. We then have 
$$\angle{MEN}=\angle{MDN}=180^0-\angle{MDB}-\angle{NDC}=180^0-B-C=A.$$
Hence, $AENM$ is a cyclic quadrilateral, which implies $\angle{AME}=\angle{ANE}$. Note that $ME = MD = MB$ and $NE = ND = NC$, so $\triangle MBE$ and $\triangle NCE$ are isosceles. In other words, $\angle{AME} = 2\angle{MBE}$ and $\angle{ANE} = 2\angle{NCE}$. Therefore $\angle{MBE}=\angle{NCE}$, so $AECB$ is cylic and $E \in (ABC)$. \\
Observe that $OM$ and $ON$ are the perpendicular bisectors of $EB$ and $EC$ respectively. Let $X=OM \cap BE, \ Y = ON \cap CE$, then $EXOY$ is a cyclic quadrilateral.\\
Hence $$\angle{MON}=180^0-\angle{XEY}=180^0-A,$$
which means $AMON$ is cyclic. (QED)
\end{prf}

\newpage
\begin{prf}
\addintroproofcontents
\begin{center}
	\input{geogebra/pr0-sol6}
\end{center}
Let $E \ne D$ be the other intersection point of $(DMN)$ and $BC$.\\
We have $\angle{EMN}=\angle{EDN}=C, \ A = \angle{MDN}=\angle{MEN}$ so $\triangle EMN \sim \triangle ACB$.\\
Let $O'$ be the orthocenter of $\triangle EMN$. We observe that $BMO'E, AMO'N, CNO'E$ are cyclic quadrilaterals, hence
$$\angle{O'BE}=\angle{O'ME}=\angle{O'NE}=\angle{O'CE},$$
so $O'C = O'B$. Similarly we have $O'A = O'B \ (=O'C)$ and thus $o'$ is the orthocenter of \abc. In other words, $O' \equiv O$. \\
We also have $\angle{MON}=180^0-\angle{MDN}=180^0-A$, which means $AMON$ is cyclic.
\end{prf}

\begin{remark}
We can observe the relation among the six solutions given above. While all solutions involve the additional construction of some geometric objects (new point, new line, or new circle), the more sophisticated this construction is, the simpler the subsequent proof becomes.
\begin{itemize}
	\item Solution 1 is the most simple approach: by considering the problem in the context of Cartesian coordinates. It is then straightforward to calculate the  coordinates of every point. The proof does apply a small twist, however, by invoking Simson's theorem to turn the problem of cyclicity into that of collinearity, thereby avoiding equations of circles which are usually fairly complicated.

	\item Solution 2 and 3 ``reverse'' the problem by constructing the point $N' \ne A$ as the intersection of $AC$ and $(AMO)$. The motivation here is that, instead of having to deduce cyclicity, we assume cyclicity is already given. What remains is to show that both $N'$ and $N$ share a common unique property, which in turn implies that $N'$ is indeed $N$. This is a recommended approach when one has not figured out how to effectively use all of the information provided by the problem. 

	\item Solution 4 builds upon the realization that all properties of $M$ and $N$ depend only on the location of $D$ on $BC$, i.e., the length of the segment $DB$. Hence we try to convert all computations to those involving only the segments on $BC$. Thanks to this approach, there are much less algebraic work to do.

	\item Solution 5 and 6 are two purely geometric solutions that one needs rich experience in geometry to come up with. In particular, the idea of solution 5 is that since $\angle{MDN}=A$, we create a reflection of $D$ in order to obtain a cyclic quadrilateral.
	In solution 6, we use the following lemma: 
	\begin{lemma}
		Given $\triangle ABC$ with three points $X,Y,Z$ on $BC,CA,AB$ respectively such that $\triangle XYZ \sim \triangle ABC$. It follows that the orthocenter of $\triangle XYZ$ is also the circumcenter of $\triangle ABC$.
	\end{lemma}
\end{itemize}
\end{remark}

While solutions such as 5 and 6 are always desirable, it can be difficult to identify the underlying geometrical insights that in turn lead to such concise solutions, especially for students not strong in geometry\footnote{including us, the authors.}. The goal of this article, then, is to provide a systematic approach to tackling geometry problems using the tools from algebra and calculus. We will show how algebraic techniques along the line of Solution 1 and 2 can be utilized to solve a variety of Olympiad-level problems.

\begin{remark}
The given problem is in fact one in a series of similar geometric properties.
\bpr{}
Let $O, I$ and $H$ be the circumcenter, incenter and orthocenter of \abc respectively. Consider three points $D \in BC, \ M \in CA, \ N \in AB$.
\begin{itemize}
	\item Prove that if $BD = BM$ and $CD = CN$ then $I \in (AMN)$.
	\item Prove that if $MD = MB$ and $ND = NC$ then $O \in (AMN)$.
	\item Prove that if $DM = DB$ and $DN = DC$ then $H \in (AMN)$.
\end{itemize}
\epr
\end{remark}

\section{Background}
We first highlight a number of theorems and solutions that are highly applicable. Note that there are a number of textbook formulas - \href{https://en.wikipedia.org/wiki/Pythagorean_theorem}{Pythagorean's Theorem}, \href{https://en.wikipedia.org/wiki/Thales%27s_theorem}{Thales' theorem}, \href{https://en.wikipedia.org/wiki/Law_of_sines}{Law of sines} and \href{https://en.wikipedia.org/wiki/Law_of_cosines}{Law of cosines} - that are not introduced but also relevant here.
\begin{theme} Useful equatilies. \end{theme}
\begin{itemize}
\item \href{https://en.wikipedia.org/wiki/Ptolemy%27s_theorem}{(Ptolemy)} If $ABCD$ is a cyclic quadrilateral then
$$AC \cdot BD = AB \cdot CD + BC \cdot AD.$$
\item \href{https://en.wikipedia.org/wiki/Stewart%27s_theorem}{(Stewart)} For any collinear points $A, B, C$ and arbitrary point $M$, 
$$MA^2 \cdot \ds{BC} + MB^2 \cdot \ds{CA}+MC^2 \cdot \ds{AB}=- \ds{AB} \cdot \ds{BC} \cdot \ds{CA}.$$
\item \href{https://www.cut-the-knot.org/pythagoras/Carnot.shtml}{(Carnot)} For any four arbitrary points $A,B,C,D$, $$AD \bot BC \Leftrightarrow AB^2-AC^2=DB^2-DC^2.$$
\item \href{https://en.wikipedia.org/wiki/Heron%27s_formula}{(Heron)} The area of \abc with side lengths $a, b$ and $c$ and semiperimeter $p = \frac{a+b+c}{2}$ is
$$S_{\triangle ABC} = \sqrt{p(p-a)(p-b)(p-c)}.$$
\end{itemize}

\begin{theme} Triangle lengths. \end{theme}
Given \abc $(AB < AC)$ with altitude $AD$ and bisector $AF$. Let $E$ be the tangent point of \abc's incircle and $BC$. We then see that
$$BD=\dfrac{a^2+c^2-b^2}{2a}, \ BE = \dfrac{a+c-b}{2}, \ BF=\dfrac{ac}{b+c},$$
which in turn yields
$$DE=BE-BD=\dfrac{(b-c)(b+c-a)}{2a}.$$
Furthermore, from Stewart's theorem, we have
$$AD=\sqrt{AB \cdot AC - DB \cdot DC }= \dfrac{2\sqrt{bcp(p-a)}}{b+c}=\dfrac{2bc}{b+c} \cdot \cos \dfrac{A}{2}.$$

\begin{theme} Trigonometric expressions. \end{theme}
In any $\triangle ABC$, we have
$$\begin{aligned} 
	& \cos \dfrac{A}{2}=\sqrt{\dfrac{p(p-a)}{bc}} , \ \tan \dfrac{A}{2} =\sqrt{\dfrac{(p-b)(p-c)}{p(p-a)}},\\ 
	& \cot A =\dfrac{b^2+c^2-a^2}{4S} , \ \cot \dfrac{A}{2}=\dfrac{b+c-a}{2r}.
\end{aligned}$$

\begin{theme} Concurrency and collinearity. \end{theme}
In \abc consider points $D \in BC, E \in CA, F \in AB$. Let $L$ be another point on $BC$.
\begin{itemize}
	\item \href{https://en.wikipedia.org/wiki/Ceva%27s_theorem}{(Ceva)} $AD, BE, CF$ are concurrent if and only if
	$$\dfrac{\ds{DB}}{\ds{DC}} \cdot \dfrac{\ds{EA}}{\ds{EB}} \cdot \dfrac{\ds{FC}}{\ds{FA}} =-1.$$

	\item \href{https://en.wikipedia.org/wiki/Menelaus%27s_theorem}{(Menelaus)} $L, E, F$ are collinear if and only if
	$$\dfrac{\ds{LB}}{\ds{LC}} \cdot \dfrac{\ds{EA}}{\ds{EB}} \cdot \dfrac{\ds{FC}}{\ds{FA}}= 1.$$
\end{itemize}
\begin{center}
	\input{geogebra/harmonic}
\end{center}
Combining the two theorems also shows that
$$\dfrac{\ds{DB}}{\ds{DC}} : \dfrac{\ds{LB}}{\ds{LC}} = -1.$$
In other words, if $AD, BE, CF$ are concurrent and $L, E, F$ are collinear then $(DLBC)= -1 =(LDBC)$.

\begin{theme} Oriented angles. \end{theme}
Oriented angles are considered in mod $\pi$ and have the following properties:
\begin{itemize}
	\item $ (AB,CD) \equiv (AB,DC) \equiv (BA,CD)$.
	\item $a \parallel b$ if and only if $(a,b) \equiv 0$.
	\item $a \bot b$ if and only if $(a,b) \equiv \frac{\pi}{2}$.
	\item $(a,b) \equiv -(b,a)$.
	\item $(a,b) \equiv (a,c)+(c,b)$.
	\item $A,B,C,D$ are cyclic if and only if $(AB,AD) \equiv (CB,CD)$.
	\item $AT$ is tangent to $(ABC)$ if and only if $(AT,AB) \equiv (CA,CB)$.
\end{itemize}

\begin{theme} Cartesian coordinates. \end{theme}
Besides textbook formulas on vectors and line equations, we introduce the expressions of several useful geometric constructs in the context of Cartesian coordinates. \\
Given \abc with centroid $G$, incenter $I$, orthocenter $H$, circumcenter $O$ and excenter $J$ opposite to $A$, we have:
\begin{itemize}
	\item $a\vto{IA}+b\vto{IB}+c\vto{IC}=\vto{0}$, so
	$$I \lf \dfrac{ax_A+bx_B+cx_C}{a+b+c}, \dfrac{ay_A+by_B+cy_C}{a+b+c} \ri.$$

	\item $\vto{GA}+\vto{GB}+\vto{GC}=\vto{0}$, so
	$$G \lf \dfrac{x_A+x_B+x_C}{3}, \dfrac{y_A+y_B+y_C}{3} \ri.$$

	\item $\vto{AH} \cd \vto{BC}={0}$ and $\vto{BH} \cd \vto{AC}={0}$, so
	\begin{align*} 
		x_H(x_C-x_B)+y_H(y_C-y_B)=x_A(x_C-x_B)+y_A(y_C-y_B), \\ 
		x_H(x_C-x_A)+y_H(y_C-y_A)=x_B(x_C-x_A)+y_B(y_C-y_A).
	\end{align*}

	\item $OA^2 = OB^2 = OC^2$, so
	\begin{align*} 
		& -2x_Ox_A+x_A^2-2y_Oy_A+y_A^2 \\ 
		= & -2x_Ox_B+x_B^2-2y_Oy_B+y_B^2 \\ 
		= & -2x_Ox_C+x_C^2-2y_Oy_C+y_C^2. 
	\end{align*}

	\item $(\vto{AJ},\vto{AB})=(\vto{AJ},\vto{AC})$ and $(\vto{BJ},\vto{BC})=(\vto{BJ},\vto{AB})$, so
	$$\dfrac{\vto{AJ}\cd \vto{AB}}{c}=\dfrac{\vto{AJ} \cd \vto{AC}}{b}, \ \dfrac{\vto{BJ} \cd \vto{BC}}{a}=\dfrac{\vto{BJ} \cd \vto{AB}}{c},$$
	which yields
	$$J \lf \dfrac{-ax_A+bx_B+cx_C}{-a+b+c}, \dfrac{-ay_A+by_B+cy_C}{-a+b+c} \ri.$$
\end{itemize}

\section{Example problems}
\bpr{} \label{strong-lemma}
Consider \abc with incircle $(I)$ touching $BC$ at $D$. Let $DE$ be a diameter of $(I)$. The excircle to vertex $A$ touches $BC$ at $L$. Prove that $A, E, L$ are collinear.
\epr
\begin{proof}
We know from the definition of $L$ that $CL = BD = \frac{a+c-b}{2}$. To prove that $A, E, L$ are collinear, let $L'$ be the intersection of $AE$ and $BC$. It remains to show that $CL' = CL$, so $L' \equiv L$. To make use of $DE \bot BC$ let's create an altitude $AH$. In this case, $\triangle AHL$ has $DE \parallel AH$ - since we already know $DH, DE$ and $AH$, Thales would give us $CL'$, as desired.
\begin{figure}[h]
	\centering
	\input{geogebra/pr2}
	\caption{Prove that $A, E, L$ are collinear.}
\end{figure}

Let $L'$ be the intersection of $AE$ and $BC$. Draw an altitude $AH$ of \abc. WLOG assume $AB \le AC$, so $H$ lies between $B$ and $D$. \\
Let $BL' = x$. As $DE \parallel AH$ we have
$$\dfrac{DL'}{HL'}=\dfrac{ED}{AH}=\dfrac{2r}{AH}=\dfrac{a}{p}=\dfrac{2a}{a+b+c},$$
which leads to 
$$\dfrac{x-BD}{x-BH}= \frac{DL'}{HL'} = \dfrac{2a}{a+b+c}.$$
Having 
$$BH=\frac{a^2+c^2-b^2}{2a}, \ BD=\frac{a+c-b}{2}$$
then gives us 
$$x=\dfrac{-2a \cd BH+(a+b+c) \cd BD}{b+c-a} =\dfrac{b+a-c}{2},$$
so 
$$CL'=a-x=\frac{a+c-b}{2} = CL.$$
In other words, $L' \equiv L$ and therefore $A, E, L$ are collinear.
\end{proof}

\begin{remark}
This example serves as a strong lemma for many Olympiad problems. Let's take a look at one of them next.
\end{remark}

\bpr{\href{https://artofproblemsolving.com/wiki/index.php?title=2001_USAMO_Problems}{\bf USAMO 2001}}
The incircle $(I)$ of \abc touches $CA$ and $CB$ at $E_1$ and $D_1$ respectively. Consider $E_2 \in CA$ and $D_2 \in CB$ such that $CE_2 = AE_1$ and $CD_2 = BD_1$. $AD_2$ intersects $BE_2$ at $P$. Let $Q$ be the intersection of $(I)$ and $AD_2$ that is closer to $A$. Prove that $AQ = D_2P$.
\epr
\begin{proof}
Construct an altitude $AH$ of \abc. As $BD_2 = CD_1 = \frac{a+b-c}{2}$, $D_1Q$ is in fact a diameter of $(I)$, so $D_1Q \parallel AH$, from which we can calculate $AQ$. \\
To make use of the intersecion $P$, we can apply Menelaus to the transversal line $BPE_2$ of $\triangle CAD_2$, which would give us $PD_2$.

\begin{figure}[h]
	\centering
	\input{geogebra/pr3}
	\label{pr3-prompt}
	\caption{Prove that $AQ = DP_2$.}
\end{figure}

Construct an altitude $AH$ of \abc. According to Problem \ref{strong-lemma}, $D_1Q$ is the diameter of $(I)$, so $AH \parallel D_1Q$. This would imply that
$$\dfrac{QD_2}{AD_2}=\dfrac{QD_1}{AH} \Rightarrow \dfrac{AD_2 - QD_2}{AD_2} = \dfrac{AH - QD_1}{AH}.$$
In other words,
\begin{equation} \label{aq}
	\dfrac{AQ}{AD_2}=\dfrac{AH-QD_1}{AH}=\dfrac{ \frac{2S}{a}-\frac{2S}{p} }{\frac{2S}{a}} = \dfrac{p-a}{p}=\dfrac{b+c-a}{b+c+a}.
\end{equation}
According to Menelaus, since $BPE_2$ is a transversal line of $\triangle CAD_2$,
\begin{align*} 
	& \dfrac{PD_2}{PA} \cd \dfrac{BC}{BD_2} \cd \dfrac{E_2A}{E_2C}=1 \\ 
	& \dfrac{PD_2}{PA} \cd \dfrac{2a}{a+b-c} \cd \dfrac{a+b-c}{b+c-a} =1 \\ 
	& \dfrac{PD_2}{PA}=\dfrac{b+c-a}{2a} \\ 
	& \dfrac{PD_2}{AD_2}=\dfrac{b+c-a}{b+c-a+2a}=\dfrac{b+c-a}{b+c+a}. \numberthis\label{d_2p}
\end{align*} 
From (\ref{aq}) and (\ref{d_2p}) we have $AQ = D_2P$.
\end{proof}

\begin{remark}
	Many more applications of the lemma introduced in Problem \ref{strong-lemma} can be found in [1]. For now we will move on to a different application of Menelaus' theorem.
\end{remark}

\bpr{\href{https://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems}{\bf USAMO 2008}}
In an acute, scalene triangle $\triangle ABC$, let $M, N, P$ be the midpoints of $BC, CA, AB$ respectively. The perpendicular bisectors of $AB$ and $AC$ intersect ray $AM$ at $D$ and $E$ respectively. $BD$ intersects $CE$ at $F$. Prove that $APFN$ is a cyclic quadrilateral.
\epr
\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/pr4}
	\label{pr4-prompt}
	\caption{Prove that $APFN$ is a cylic quadrilateral.}
\end{figure}
To make use of the intersection $F$, we can apply Menelaus to the transversal line $DEM$ or $FEC$. Note that using $DEM$ would involve the ratio $\frac{MB}{MC} = 1$, since $M$ is the midpoint of $BC$; this would make the calculations simpler. \\
It's easy to see that $DA = DB$ and $FA = FC$. Let $O$ be the circumcenter of $\triangle ABC$, then $O$ lies on both $EN$ and $DP$. \\
$\triangle BFC$ has transversal line $DEM$ so by Menelaus:
$$1 = \dfrac{DF}{DB} \cd \dfrac{EC}{EF} \cd \dfrac{MB}{MC} = \frac{DF}{DB} \cd \frac{EC}{EF},$$
so 
$$\dfrac{DF}{EF}=\dfrac{DB}{EC}=\dfrac{AD}{AE}.$$
Hence $FA$ is the external bisector at $F$ of $\triangle DEF$. Observe that $\angle APO = \angle ANO = 90\degr$ so $APON$ is cyclic. To show that $APFN$ is also cyclic, we will prove that $\angle AFO = 90\degr$ as well, or, in other words, $FO$ is the internal bisector at $F$ of $\triangle DEF$. \\
Indeed, $\triangle ADB$ has $DA = DB$ and median $DP$ so $DP$ is also the angular bisector of $\angle ADB$. In other words, $DO$ is the external bisector at $D$ of $\triangle DEF$. Similarly, $EO$ is the external bisector at $F$ of $\triangle DEF$. Hence $O$ is the excenter relative to $F$ of $\triangle DEF$. It then follows that $FO$ is the internal bisector at $F$ of $\triangle DEF$.
\end{proof}

\bpr{\href{https://kinhnghiemhoctoan.files.wordpress.com/2017/08/vmo2006a.pdf}{\bf VMO 2006, Board A}}
Let $ABCD$ be a convex quadrilateral. Consider a moving point $M$ on $AB$ such that $M \ne A$ and $M \ne B$. Let $N \ne M$ be the other intersection of $(MAC)$ and $(MBD)$. Prove that $N$ is always on a fixed circle and $MN$ always goes through a fixed point.
\epr
\begin{proof}
\begin{figure}[h!]
	\centering
	\input{geogebra/pr5}
	\label{pr5-prompt}
	\caption{Prove that $APFN$ is a cylic quadrilateral.}
\end{figure}
Let $I$ be the intersection of $AC$ and $BD$, then
\begin{align*}
	(CI,CN) & \equiv (CA,CN) \equiv (MA,MN) \\
	& \equiv (MB,MN) \equiv (DB,DN)\\
	& \equiv (DI,DN).
\end{align*}
So $C, I, D, N$ are cyclic or, in other words, $N$ is always on the fixed circle $(CDI)$. \\
Let $t$ be the line that goes through $I$ and is parallel to $AB$. $t$ intersects $MN$ at $K$. \\
Observe $(MA, MN) \equiv (KI, KN)$ because $AM \parallel KI$, while $(MA, MN) \equiv (CI, CN)$ as well because $A, M, C, N$ are cyclic. Hence $(KI, KN) \equiv (CI, CN)$ so $C, I, K, N$ are cyclic. It follows that $K$ is the other intersection of $t$ and $(CDI)$ besides $I$. \\
As both $t$ and $(CDI)$ are fixed, $K$ is a fixed point, and this is the point that $MN$ always goes through.
\end{proof}

\begin{remark}
	This problem is an example of when oriented angles work best. Due to the positioning of $M$, it's very hard to prove that $C, I, D, N$ are cyclic with normal angles; both cases of $\angle ICN = \angle IDN$ or $\angle ICN = 180\degr - \angle IDN$ can happen under different circumstances. With oriented angle, we can simply express this condition as $(CI, CN) \equiv (DI, DN)$.
\end{remark}

\bpr{\href{https://diendantoanhoc.net/topic/117115-vietnam-tst-2014/}{\bf Vietnam TST 2014}}
Prove that:
\begin{enumerate}[(a)]
	\item In \abc with altitude $AD$, consider a point $P$ on $AD$. $PB$ intersects $AC$ at $E$ and $PC$ intersects $AB$ at $F$. If $AEDF$ is a cyclic quadrilateral, prove that
	$$\frac{PA}{PD}=(\tan B+\tan C) \cdot \cot \frac{A}{2}.$$

	\item Consider \abc with orthocenter $H$. Let $P$ be a moving point on $AH$. The line perpendicular to $AC$ at $C$ intersects $BP$ at $M$. The line perpendicular to $AB$ at $B$ intersects $CP$ at $N$. Let $K$ be the projection of $A$ on $MN$. Prove that $\angle BKC + \angle MAN$ is constant.
\end{enumerate}
\epr
\begin{proof}
We first state, without proof, a common application of the harmonic series:
\begin{lemma} \label{vmo-tst-2014-lemma}
	Consider \abc with altitude $AD$ and a point $P$ on $AD$. $BP$ intersects $AC$ at $E$ and $CP$ intersects $AB$ at $F$. Then $DA$ is the angular bisector of $\angle DHF$.
\end{lemma}
Using this lemma, we see that $\angle AEF = \angle ADF = \angle ADE = \angle AFE$, so $\triangle AEF$ is isosceles at $A$. Now consider two cases:
\begin{enumerate}[(a)]
	\item
		\begin{figure}[h]
			\centering
			\input{geogebra/pr6}
			\label{pr6-prompt-a}
			\caption{Prove that $\frac{PA}{PD} = (\tan B + \tan C) \cdot \cot \frac{A}{2}$.}
		\end{figure}
		\begin{itemize}
			\item If $AB = AC$ then $AD$ is the angular bisector of $A$. It then follows that $DF \bot AB, DE \bot AC$. Hence 
			$$EC=\dfrac{DC^2}{AC}=\dfrac{a^2}{4b}, \ EA=AC-EC=\dfrac{4b^2-a^2}{4b}.$$
			Applying Menelaus to the transversal line $EPB$ of $\triangle ADC$ yields
			$$\dfrac{PA}{PD}=\dfrac{EA}{EC} \cd \dfrac{BC}{BD}=\dfrac{4b^2-a^2}{2a^2}.$$
			Note that $\angle B = \angle C = 90\degr - \frac{\angle A}{2}$, so we also have
			$$(\tan B + \tan C) \cd \cot \dfrac{A}{2} = 2 \cot^2 \dfrac{A}{2}=\dfrac{(b+c-a)^2}{2r^2}=\dfrac{(2b-a)^2}{r^2},$$
			Using algebraic manipulations and noting that \abc has $b = c, r = \frac{S}{p}$, we see that
			\begin{align*}
				\frac{(2b - a)^2}{r^2} & = (2b - a)^2 \cdot \frac{p^2}{S^2} \\
				& = (2b - a)^2 \cdot \frac{p^2}{p(p-a)(p-b)(p-c)} \\
				& = (2b - a)^2 \cdot \frac{4(2b + a)^2}{a^2(2b + a)(2b - a)} \\
				& = \frac{4b^2-a^2}{2a^2}.
			\end{align*}
			Hence $\frac{PA}{PD}=(\tan B+\tan C)\cot \frac{A}{2}.$

			\item If $AB \ne AC$ then $DB \ne DC$. Applying Lemma \ref{vmo-tst-2014-lemma}, we see that $DA$ is the angular bisector of $\angle EDF$. Furthermore, $AFDE$ is cyclic, so $AE = AF = x$. \\
			We now see that
			$$DB=\dfrac{a^2+c^2-b^2}{2a}, \ DC=\dfrac{a^2+b^2-c^2}{2a}, AD=\dfrac{2S}{a}, \ \cot \dfrac{A}{2}=\dfrac{b+c-a}{2r},$$
			which implies
			\begin{align*} 
				(\tan B + \tan C) \cdot \cot \frac{A}{2} & = AD \left ( \dfrac{1}{DB}+\dfrac{1}{DC} \right) \cdot \cot \frac{A}{2} \\
				& =\dfrac{2S}{a} \left ( \dfrac{2a}{a^2+c^2-b^2}+\dfrac{2a}{a^2+b^2-c^2} \right) \cdot \frac{b + c - a}{2r} \\
				& = \dfrac{2a^2(b+c-a)(b+c+a)}{(a^2+c^2-b^2)(a^2+b^2-c^2)}. \numberthis \label{tanB-tanC}
			\end{align*}
			As $AD, BE, CF$ are collinear, according to Ceva,
			$$\b1 = \frac{DB}{DC} \cdot \frac{EC}{EA} \cdot \frac{FA}{FB} = \dfrac{DB}{DC} \cdot \dfrac{x}{c-x} \cdot \dfrac{b-x}{x}.$$
			which yields
			$$x=\dfrac{b \cdot DB - c \cdot DC}{DB-DC}, \ b-x=\dfrac{DC(c-b)}{DB-DC}.$$
			Applying Menelaus to the transversal line $EPB$ of $\triangle ADC$:
			\begin{equation}
				\dfrac{PA}{PD} = \dfrac{EA}{EC} \cd \dfrac{BC}{BD}= \frac{x}{b-x} \cdot \frac{a}{BD} = \frac{b \cdot DB - c \cdot DC}{DC(c - b)} \cdot \frac{a}{DB}. \label{pa-pd} 
			\end{equation}
			Substituting $DB = \frac{a^2 + c^2 - b^2}{2a}$ and $DC = \frac{a^2 + b^2 - c^2}{2a}$ into (\ref{pa-pd}) and simplifying the expression, we get
			\begin{equation}
				\frac{PA}{PD} = \dfrac{2a^2(b+c-a)(b+c+a)}{(a^2+c^2-b^2)(a^2+b^2-c^2)} = (\tan B + \tan C) \cdot \cot \frac{A}{2}.
			\end{equation}
		\end{itemize}

	\item Let $I$ be the intersection of $NB$ and $MC$, then $I$ lies on $(O)$. \\
	Note that $AKBN$ and $AKCM$ are cyclic, so 
	$$\angle NAM + \angle BKC = \angle KBI + \angle KCI + \angle BKC = 360\degr - \angle BIC,$$
	which is constant.
\end{enumerate}
\end{proof}

\newpage

\bpr{\href{https://artofproblemsolving.com/community/c4737_2008_vietnam_national_olympiad}{VMO 2008}}
Consider \abc with median $AD$ and a line $d$ perpendicular to $AD$. Let $M$ be a mobile point on $d$ and let $E, F$ be the midpoints of $MB, MC$ respectively. The line passing through $E$ and perpendicular to $d$ intersects $AB$ at $P$. The line passing through $F$ and perpendicular to $d$ intersects $AC$ at $Q$. Let $d'$ be the line passing through $M$ and perpendicular to $PQ$. Prove that $d'$ always goes through a fixed point.
\epr
\begin{proof}
Since this problem only concerns perpendicular and parallel lines, a natural solution would be to consider it the context of a two-dimensional coordinate $Oxy$. To simplify subsequent calculations we will pick $D$ as the origin.
\begin{figure}[h]
	\centering
	\input{geogebra/pr7}
	\label{pr7-prompt}
	\caption{Prove that $d'$ always goes through a fixed point.}
\end{figure}

Consider the Cartesian coordinates $Dxy$ with origin $D$ and $Dy \equiv DA$. As $Dx \bot Dy$ and $d \bot DA$, we have $Dx \parallel d$. \\
Now consider the following point coordinates: $A(0, a), B(b, c), C(-b, -c)$, where $a, b \ne 0$. We can then calculate
$$\vto{AB} = (b,c-a), \ \vto{AC} = (-b,-c-a),$$
which yields
$$\begin{aligned} 
	& (AB):(a-c)x+by-ab=0, \\ 
	& (AC):(a+c)x-by+ab=0.
\end{aligned}$$
Note that $d$ is fixed and $d \parallel Dx$ so $y_M$ is fixed. Let $y_M = h$, then
$$E \lf \dfrac{b+x_M}{2}, \dfrac{h+c}{2} \ri , \ F \lf \dfrac{x_M-b}{2}, \dfrac{h-c}{2} \ri.$$
Let $d_1, d_2$ be the lines passing through $E, F$ respectively and perpendicular to $d$. It follows that
$$(d_1):x=\dfrac{x_M+b}{2}, \ (d_2):x=\dfrac{x_M-b}{2},$$
so
$$\begin{aligned} 
	P = d_1 \cap AB = \lf \dfrac{x_M+b}{2}, a- \dfrac{(a-c)(x_M+b)}{2b} \ri, \\
	Q = d_2 \cap AC = \lf \dfrac{x_M-b}{2}, a+\dfrac{(a+c)(x_M-b)}{2b} \ri.
\end{aligned}$$
As a result, $\vto{PQ}= \lf -b,\dfrac{ax_M-bc}{b} \ri$. We can now calculate
$$(d'): -b(x-x_M)+\dfrac{ax_M-bc}{b} \cdot (y-h)=0,$$
which can be rewritten as
$$b^2 \lf x-\dfrac{bc}{a} \ri - (ax_M-bc) \lf y -h+\dfrac{b^2}{a} \ri = 0.$$
Hence $d'$ always passes through a fixed point with coordinates $\lf \frac{bc}{a}, h-\frac{b^2}{a} \ri$.
\end{proof}

\begin{remark}
This was a very difficult problem in VMO 2008, with 81.2\% participants getting a 0/7 score on it. A number of purely geometrical solutions to this problem can be viewed \href{https://artofproblemsolving.com/community/c6h185640p1020457}{here}. The fixed point in question is the intersection of $HH'$ and $KK'$, where $H, K$ are the projections of $B, C$ on $d$ respectively and $H', K'$ are the reflections of $H$ about $AB$ and $K$ about $AC$ respectively. An analytical solution, in our case, turns out to be much simpler.
\end{remark}

\section{Practice Exercises}

\bex{\href{https://artofproblemsolving.com/community/c3667_2002_china_girls_math_olympiad}{China Girls MO 2002}}
An acute $\triangle ABC$ has three heights $ AD, BE$ and $ CF$ respectively. Prove that the perimeter of triangle $ DEF$ is not over half of the perimeter of triangle $ ABC.$
\eex

\bex{\href{https://artofproblemsolving.com/community/c5503_2000_italy_tst}{Italy TST 2000}}
Let $ABC$ be an isosceles right triangle and $M$ be the midpoint of its hypotenuse $AB$. Points $D$ and $E$ are taken on the legs $AC$ and $BC$ respectively such that $AD=2DC$ and $BE=2EC$. Lines $AE$ and $DM$ intersect at $F$. Show that $FC$ bisects $\angle DFE$.
\eex

\bex{\href{http://mathworld.wolfram.com/MixtilinearIncircles.html}{Mixtinilear Incircle}}
Consider \abc with circumcircle $(O)$. A circle $\omega$ touches $(O)$ and also touches $AB, AC$ at $D, E$ respectively. Let $J$ be the incenter of \abc. Prove that $D, J, E$ are collinear.
\eex

\bex{\href{https://artofproblemsolving.com/community/c4634_2004_usa_team_selection_test}{USA TST 2004}}
Let $ABC$ be a triangle. Choose a point $D$ in its interior. Let $\omega_1$ be a circle passing through $B$ and $D$ and $\omega_2$ be a circle passing through $C$ and $D$ so that the other point of intersection of the two circles lies on $AD$. Let $\omega_1$ and $\omega_2$ intersect side $BC$ at $E$ and $F$, respectively. Denote by $X$ the intersection of $DF$, $AB$ and $Y$ the intersection of $DE, AC$. Show that $XY \parallel BC$.
\eex

\bex{\href{http://forum.mathscope.org/showthread.php?t=36749}{Ho Chi Minh City Regional Team Selection 2012}}
\abc has circumcircle $(O)$ and heights $AM, BN$. Let $D$ be a point on the arc $\hat{BC}$ that does not contain $A$ ($D \ne B, D \ne C$). $DA$ intersects $BN$ at $Q$; $DB$ intersects $AM$ at $P$. Let $I$ be the midpoint of $PQ$. Prove that $M, N, I$ are collinear.
\eex

\bex{\href{https://artofproblemsolving.com/community/c3840_2013_imo}{IMO 2013}}
Let the excircle of $\triangle ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of $(A_1B_1C_1)$ lies on the circumcircle $(ABC)$. Prove that $\triangle ABC$ is right-angled.
\eex

\bex{\href{https://artofproblemsolving.com/community/c3956_2005_imo_shortlist}{IMO Shortlist 2005}}
Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$. Let $H$ be the orthocenter of $\triangle ABC$, and let $M$ be the midpoint of the side $BC$. Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$, $H$, $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$.
\eex


\bex{\href{https://artofproblemsolving.com/community/c3962_2011_imo_shortlist}{IMO Shortlist 2011}}
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with $AI$ and $BI$, respectively. The chord $DE$ meets $AC$ at a point $F$, and $BC$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $AD$ and the line through $G$ parallel to $BE$. Suppose that the tangents to $\omega$ at $A$ and $B$ meet at a point $K$. Prove that the three lines $AE,BD$ and $KP$ are either parallel or concurrent.
\eex

\bex{\href{https://artofproblemsolving.com/community/c5169_2009_allrussian_olympiad}{All-Russian MO 2009}}
Let be given a parallelogram $ ABCD$ and two points $ A_1$, $ C_1$ on its sides $ AB$, $ BC$, respectively. Lines $ AC_1$ and $ CA_1$ meet at $ P$. Assume that the circumcircles of triangles $ AA_1P$ and $ CC_1P$ intersect at the second point $ Q$ inside triangle $ ACD$. Prove that $ \angle PDA = \angle QBA$.
\eex

\bex{\href{https://artofproblemsolving.com/community/c4079_2007_balkan_mo}{Balkan MO 2007}}
Let $ABCD$ a convex quadrilateral with $AB=BC=CD$, with $AC$ not equal to $BD$ and $E$ be the intersection point of it's diagonals. Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120\degr$.
\eex

\bex{\href{https://artofproblemsolving.com/community/c4964_2008_china_team_selection_test}{China TST 2008}}
Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE = CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF = FG$.
\eex

\bex{\href{https://artofproblemsolving.com/community/c4085_2013_balkan_mo}{Balkan MO 2013}}
In $\triangle ABC$, the excircle $\omega_a$ opposite $A$ touches $AB$ at $P$ and $AC$ at $Q$, while the excircle $\omega_b$ opposite $B$ touches $BA$ at $M$ and $BC$ at $N$. Let $K$ be the projection of $C$ onto $MN$ and let $L$ be the projection of $C$ onto $PQ$. Show that the quadrilateral $MKLP$ is cyclic.
\eex

\bex{\href{https://artofproblemsolving.com/community/c3956_2005_imo_shortlist}{IMO Shortlist 2005}}
Given $\triangle ABC$ satisfying $AC+BC=3\cdot AB$. The incircle of $\triangle ABC$ has center $I$ and touches the sides $BC$ and $CA$ at the points $D$ and $E$, respectively. Let $K$ and $L$ be the reflections of the points $D$ and $E$ with respect to $I$. Prove that the points $A$, $B$, $K$, $L$ lie on one circle.
\eex

\section{Solution to Practice Exercises}

\bsol{China Girls MO 2002}
An acute triangle $ ABC$ has three heights $ AD, BE$ and $ CF$ respectively. Prove that the perimeter of triangle $ DEF$ is not over half of the perimeter of triangle $ ABC.$
\esol

\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice1}
	\label{practice1-fig}
	\caption{Prove that $DE + EF + FD \le \frac{1}{2}(AB + BC + CA)$.}
\end{figure}
From $\triangle AEF \sim \triangle ABC \sim \triangle DBF \sim \triangle DEC$ we get
\begin{align*} 
	EF= BC\cd \cos A= a \cd \cos A \\
 	DE= AB \cd \cos C= c \cd \cos C \\
 	FD= AC \cd \cos C= b \cd \cos B 
\end{align*}
We need to prove that
$$EF + DE + FD \le \frac{1}{2}( AB + BC+ CA),$$
which is equivalent to
\begin{align*}
&\  a \cd \cos A + b \cd \cos B+ c \cd \cos C \le \dfrac{1}{2}( a+b+c)\\
\Leftrightarrow &\ \sum a \cd \frac{b^2+c^2-a^2}{2bc} \le \dfrac{1}{2}(a+b+c) \\
\Leftrightarrow &\  \sum a^2(b^2+c^2-a^2) \le  abc(a+b+c) \\
\Leftrightarrow &\  2(a^2b^2+b^2c^2+c^2a^2) \le a^4+b^4+c^4+abc(a+b+c).
\end{align*}
According to \href{https://artofproblemsolving.com/wiki/index.php?title=Schur%27s_Inequality}{Schur's Inequality}, we have
$$a^4 + b^4 + c^4 + abc(a + b + c) \ge \sum (a^3b + ab^3).$$
while based on AM-GM we have
$$\sum (a^3b + ab^3) \ge \sum 2 \sqrt{a^3b \cdot ab^3} = 2 \sum a^2b^2.$$
Hence $2(a^2b^2+b^2c^2+c^2a^2) \le a^4+b^4+c^4+abc(a+b+c)$.
\end{proof}

\bsol{Italy TST 2000}
Let $ABC$ be an isosceles right triangle and $M$ be the midpoint of its hypotenuse $AB$. Points $D$ and $E$ are taken on the legs $AC$ and $BC$ respectively such that $AD=2DC$ and $BE=2EC$. Lines $AE$ and $DM$ intersect at $F$. Show that $FC$ bisects $\angle DFE$.
\esol

\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice2}
	\label{practice2-fig}
	\caption{Prove that $FC$ bisects $\angle DFE$.}
\end{figure}
Let $AC = BC = x$ we have
$$AM=\dfrac{x\sqrt{2}}{2}, \ AD=BE=\dfrac{2x}{3}, \ CD=CE=\dfrac{x}{3}, DE^2=\dfrac{2x^2}{9},$$
which yields
$$ME^2=MB^2+BE^2-2MB \cd BE \cd \cos \dfrac{\pi}{4} = \dfrac{5x^2}{18}.$$
Hence $AM^2-AD^2=ME^2-DE^2$. According to Carnot's Theorem, $AE \bot DM$. \\
Hence $DCEF$ is cyclic, and we also have $CD = CE$, so $\angle DFC = \angle DEC = \angle CDE = \angle CEF$. In other words, $FE$ bisects $\angle DFE$.
\end{proof}


\bsol{Mixtilinear Incircle}
Consider \abc with circumcircle $(O)$. A circle $\omega$ touches $(O)$ and also touches $AB, AC$ at $D, E$ respectively. Let $J$ be the incenter of \abc. Prove that $D, J, E$ are collinear.
\esol

\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice3}
	\label{practice3-fig}
	\caption{Prove that $E, D, F$ are collinear.}
\end{figure}
Denote $I$ as the center of $\omega$ and let $ID=IE=x$, we have 
$$IA=\dfrac{x}{\sin \frac{A}{2}}, \ JA=\dfrac{r}{\sin \frac{A}{2}}, \ OI=R-x, \ OJ=\sqrt{R^2-2Rr}.$$
According to Stewart's Theorem,
$$OA^2 \cd IJ + OI^2 \cd JA = OJ^2 \cd AI + IA \cd AJ \cd JI,$$
which is equivalent to
$$R^2(x-r) + (R-x)^2r = (R^2-2Rr)x+\dfrac{rx(x-r)}{\sin^2 \frac{A}{2}},$$
which implies that $\sin^2 \frac{A}{2}=\frac{x - r}{x}$, so
\begin{align*}
IJ \cdot IA & = (AI - AJ) \cdot AI = AI^2 - AI \cdot AJ \\
& = \frac{1}{\sin^2 \frac{A}{2}}(x^2 - rx) = \frac{x}{x - r} \cdot (x^2 - rx) \\
& = x^2 = IE^2.
\end{align*}
Hence $JE \bot AI$. By similar arguments, $JD \bot AI$. Thus $D, J, E$ are collinear.
\end{proof}


\bsol{USA TST 2004}
Let $ABC$ be a triangle. Choose a point $D$ in its interior. Let $\omega_1$ be a circle passing through $B$ and $D$ and $\omega_2$ be a circle passing through $C$ and $D$ so that the other point of intersection of the two circles lies on $AD$. Let $\omega_1$ and $\omega_2$ intersect side $BC$ at $E$ and $F$, respectively. Denote by $X$ the intersection of $DF$, $AB$ and $Y$ the intersection of $DE, AC$. Show that $XY \parallel BC$.
\esol

\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice4}
	\label{practice4-fig}
	\caption{Prove that $XY \parallel BC$.}
\end{figure}
Let $S$ be the intersection of $AD$ and $BC$ and $R$ be the other intersection of $\omega_1$ and $\omega_2$ besides $D$, then $R$ lies on $AD$. \\
Applying Menelaus to a transversal line $XDF$ $\triangle ABS$ and line $ACS$ of $\triangle EDY$ we have
$$\dfrac{\ds{AX}}{\ds{XB}}=\dfrac{\ds{AD}}{\ds{SD}} \cd \frac{\ds{SF}}{\ds{BF}} \text{ and } \frac{\ds{AY}}{\ds{CY}}=\frac{\ds{AD}}{\ds{SD}} \cd \frac{\ds{SE}}{\ds{CE}}.$$
To prove that $XY \parallel BC$ we need $\frac{\ds{AX}}{\ds{BX}}=\frac{\ds{AY}}{\ds{CY}}$, or, in other words, $\frac{\ds{SF}}{\ds{BF}}=\frac{\ds{SE}}{\ds{CE}}$.\\
Observe that $\ds{SF} \cd \ds{SC} = \ds{SR} \cd \ds{SD} = \ds{SE} \cd \ds{SB}$, so
$$\ds{SF} \cd \ds{CE}=\ds {SF} (\ds{SE}-\ds{SC}) \text{ and } \ds{SE} \cd \ds{BF} = \ds{SE} (\ds{SF}-\ds{SB}).$$
We note that $\triangle SDC \sim \triangle SFR$ and $\triangle SDB \sim \triangle SER$ so
$$\ds{SF} \cd \ds{SC} = \ds{SD} \cd \ds{SR} = \ds{SE} \cdot \ds{SB},$$
which in turn means $\ds{SF} \cd \ds{CE} = \ds{SE} \cd \ds{BF}$. Thus $XY \parallel BC$.
\end{proof}

\newpage
\bsol{Ho Chi Minh City Regional Team Selection 2012}
\abc has circumcircle $(O)$ and heights $AM, BN$. Let $D$ be a point on the arc $\hat{BC}$ that does not contain $A$ ($D \ne B, D \ne C$). $DA$ intersects $BN$ at $Q$; $DB$ intersects $AM$ at $P$. Let $I$ be the midpoint of $PQ$. Prove that $M, N, I$ are collinear.
\esol

\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice5}
	\label{practice5-fig}
	\caption{Prove that $M, N, I$ are collinear.}
\end{figure}
Let $\angle CAD = \angle CDB = x$. Consider the following ratios:
\begin{align}
& \dfrac{NQ}{NH} = \dfrac{S_{ANQ}}{S_{ANH}} = \dfrac{AN \cdot AQ \cdot \sin x}{AN \cdot AH \cdot \sin \angle{HAC}} = \dfrac{AQ\sin x}{AH\cos C}, \label{nq-nh} \\ &
\dfrac{MP}{MH} = \dfrac{S_{BMP}}{S_{BMH}} = \dfrac{BM \cdot BP \cdot \sin \angle{HBC}}{BM.BH. \sin x} = \dfrac{BP \sin x}{BH \cos C}. \numberthis \label{mp-mh}
\end{align}
Observe that $\angle{AQH} + \angle{BPH} = 180\degr$, so $DPHQ$ is cyclic. Therefore
$$\frac{\sin \angle{AHQ}}{\sin \angle{AQH}} = \frac{\sin \angle{BHP}}{\sin \angle{BPH}}.$$
It then follows that $\frac{AQ}{AH} = \frac{BP}{BH}$, and, according to (\ref{nq-nh}) and (\ref{mp-mh}), $\frac{NQ}{NH} = \frac{MP}{MH}$. \\
Now consider three points $M, I, N$ and $\triangle HPQ$, we see that
$$\frac{IP}{IQ} \cdot \frac{NQ}{NH} \cdot \frac{MH}{MP} = \frac{NQ}{NH} \cdot \frac{MH}{MP} = 1,$$
According to the converse Menelaus' theorem, $M, N, I$ are collinear.
\end{proof}

\bsol{IMO 2013}
Let the excircle of $\triangle ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of $(A_1B_1C_1)$ lies on the circumcircle $(ABC)$. Prove that $\triangle ABC$ is right-angled.
\esol

\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice6}
	\label{practice6-fig}
	\caption{Prove that $\triangle ABC$ is right-angled.}
\end{figure}

We will extend the problem and prove both directions: {\it $\triangle ABC$ is right-angled if and only if the circumcentre of $(A_1B_1C_1)$ lies on the circumcircle of $\triangle ABC$.}
\begin{itemize}
	\item {\bf If $\triangle ABC$ is right-angled}, WLOG assume $\angle BAC = 90\degr$. Let $O$ be the midpoint of $BC$ and $M$ be the midpoint of the arc $\widehat{BC}$ that contains $A$ in the circumcircle $(ABC)$. As 
	$$MC = MB,\ CB_1 = BC_1 \text{ and } \angle MCB_1 = \angle MBC_1,$$
	we get $\triangle MCB_1 = \triangle MBC_1$ and therefore $MB_1 = MC_1$. We now show that $MA_1 = MC_1$. \\
	Note that 
	$$BA_1 = \frac{a+b-c}{2}, \ OM = \frac{a}{2}, \ OA_1 = \frac{c-b}{2}, MA_1^2 = \frac{a^2 + (b-c)^2}{4}.$$
	Let $H$ be the projection of $M$ on $AB$. $\triangle MBC$ is a right isosceles triangle so $MB = MC = \frac{a}{\sqrt 2}$. By Carnot's Theorem, 
	$$MB^2 - MA^2 = BH^2 - AH^2 = (BH + AH)(BH - AH),$$
	which would give us $MA$:
	\begin{align*}
		\frac{a^2}{2} - MA^2 & = c(c - \sqrt 2 \cdot MA) \\
		\frac{b^2 + c^2}{2} - MA^2 & = c^2 - \sqrt 2 \cdot c \cdot MA \\
		MA & = \frac{\sqrt 2}{2} (c-b).
	\end{align*}
	Applying the law of cosine on $\triangle MAC_1$:
	$$\begin{aligned}
		MC_1^2 & = MA^2+AC_1^2 - 2 \cdot MA \cdot AC_1 \cdot \cos 45\degr \\ 
		& = \frac{(b-c)^2}{2} + \frac{(a+c-b)^2}{4}-2 \cd \frac{\sqrt{2}}{2} \cd \frac{a+c-b}{2}\cd \frac{c-b}{\sqrt{2}} \\
		& = \frac{a^2 + (b-c)^2}{4}.
	\end{aligned}$$
	Hence $MA_1^2 = MC_1^2$, so $MA_1 = MC_1 = MB_1$, which means $M$ is the circumcenter of $(A_1B_1C_1)$.

	\item {\bf If the circumcenter of $(A_1B_1C_1)$ lies on the circumcircle $(ABC)$}, WLOG assume $M$, the circumcenter of $(A_1B_1C_1)$, lies on the arc $\widehat{BC}$ that contains $A$ of $(ABC)$. \\
	We then have $MB_1 = MC_1$, and also note that from the property of excircle, $BC_1 = CB_1$. It follows that $\triangle MB_1C = \triangle MC_1B$, so $MB = MC$, or, in other words, $M$ is the midpoint of arc $\widehat{BC}$. \\
	Let $H$ and $K$ be the projections of $M$ on $AB$ and $AC$ respectively. As $AB > AC$, note that $H$ lies inside the segment $AB$ and $K$ lies outside the segment $AC$. It's easy to see that $\triangle MHB = \triangle MKC$, so $MH = MK$ and therefore $AH = AK$. In other words,
	$$\begin{aligned}
		AB - AM \cdot \sin \lf \frac{180\degr - {A}}{2} \ri & = AC + AM \cdot \sin \lf \frac{180\degr - {A}}{2} \ri, \\
		AB-AC & = 2AM \cos \frac{A}{2},\\
		\sin C - \sin B  & = 2 \sin \lf \dfrac{C-B}{2} \ri \cos \dfrac{A}{2},
	\end{aligned}$$
	which means $\tan \frac{A}{2} = 1$ and therefore $\angle BAC = 90\degr$. Thus, \abc is right-angled.
\end{itemize}
\end{proof}

\bsol{IMO Shortlist 2005}
Let $\triangle ABC$ be an acute-angled triangle with $AB \ne AC$. Let $H$ be the orthocenter of $\triangle ABC$, and let $M$ be the midpoint of the side $BC$. Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$, $H$, $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$.
\esol
\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice7}
	\label{practice7-fig}
	\caption{Prove that $HM \bot AK$.}
\end{figure}
Let $O$ and $O'$ be the center of the circumcircles $(ABC)$ and $(ADE)$ respectively. Let $I$ and $P$ be the midpoints of $AH$ and the arc $\widehat{BC}$ not containing $A$ of $(ABC)$ respectively. $AP$ intersects $HM$ at $N$ and $(ADE)$ at $N'$. We will show that $N \equiv N'$. \\
Draw the diameter $AA'$ of $(ABC)$. As $BHCA'$ is a parallelogram, $H, M, A'$ are collinear and $OM$ is the midsegment of $\triangle AHA'$. Now observe that 
$$AH = 2OM = 2R\cos A \text{ and } MP=OP-OM=R(1-\cos A).$$
Furthermore, since $AH \parallel MP$,
\begin{equation}
	\dfrac{AN}{AP}=\dfrac{AH}{AH+MP}=\dfrac{2 \cos A}{1+\cos A}. \label{an-ap}
\end{equation}
Note that 
$$\sin \angle{AHD}=\sin \lf B+\dfrac{A}{2} \ri,\ \sin \angle{ADE}=\cos \dfrac{A}{2}.$$
Applying the sine rule in $\triangle ADH$ we get
\begin{equation}
	AD=\dfrac{1}{\cos \frac{A}{2}} \cd AH \sin \lf B+\dfrac{A}{2} \ri. \label{ad}
\end{equation}
As $\triangle ADE$ is isosceles and $AP$ is the angular bisector of $A$, $AN'$ is the diameter of $(ADE)$. Hence $AD = AN' \cos \frac{A}{2}$. Combining this with (\ref{ad}) we get
\begin{equation}
	AN'=\dfrac{1}{\cos^2 \frac{A}{2}} \cd AH \sin \lf B+\dfrac{A}{2} \ri = \dfrac{4R \cos A}{1 + \cos A} \cdot \sin \lf B + \frac{A}{2} \ri. \label{an'}
\end{equation}
On the other hand, 
$$AP=2R\sin \angle{ABP}=2R \sin \lf B+\dfrac{A}{2} \ri,$$
so (\ref{an-ap}) yields
$$AN = \frac{2 \cos A}{1 + \cos A} \cdot AP = \dfrac{4R \cos A}{1 + \cos A} \cdot \sin \lf B + \frac{A}{2} \ri.$$
Combining this with (\ref{an'}) we get $AN = AN'$; in other words, $N \equiv N'$. Hence $H, N, A'$ are collinear. Also, $O, O', I'$ are collinear and lie on the midsegment of $\triangle AHA'$. \\
As a result, $HM \parallel OI$, while we also have $OI \bot AN$, so $HM \bot AN$.
\end{proof}

\begin{remark}
	A similar problem has appreared in Vietnam TST 2006: {\it Consider \abc with orthocenter $H$. The exterior angular bisector of $\angle BHC$ intersects $AB$ and $AC$ at $D$ and $E$ respectively. The angular bisector of $\angle BAC$ interesects $(ADE)$ at $K \ne A$. Prove that $HK$ goes through the midpoint of $BC$.}
\end{remark}

\bsol{IMO Shortlist 2011}
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with $AI$ and $BI$, respectively. The chord $DE$ meets $AC$ at a point $F$, and $BC$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $AD$ and the line through $G$ parallel to $BE$. Suppose that the tangents to $\omega$ at $A$ and $B$ meet at a point $K$. Prove that the three lines $AE,BD$ and $KP$ are either parallel or concurrent.
\esol
\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice8}
	\label{practice8-fig}
	\caption{Prove that $AE,BD$ and $KP$ are either parallel or concurrent.}
\end{figure}
As $\angle{IAF}=\angle{DAC}=\angle{BAD}=\angle{DEB}=\angle{FEI}$, we see that $AIFE$ is cyclic. Similarly, $BDGI$ is cyclic. Denote $\omega_1$ as the circle passing through $A, I, F, E$ and $\omega_2$ as the circle passing through $B, D, G, I$. \\
If $\omega_1$ touches $\omega_2$ then denote $d$ as the two circles' common tangent line. Otherwise, let $L$ be the other intersection point of $\omega_1$ and $\omega_2$ besides $I$ and denote $d$ as the line $IL$. \\
$d$ intersects $(ABL)$ and $(FGL)$ at $K',P' \ne L$ respectively. Note that
$$(AB,BK') \equiv (AL,LK') \equiv (AL,LI) \equiv (AE,EI) \equiv (AE,EB) \equiv (AB,AB).$$
Therefore $BK' \equiv BK$. Similarly, $AK' \equiv AK$. It follows that $K \equiv K'$. On the other hand,
$$(P'F,FG) \equiv (P'L,LG) \equiv (IL,LG) \equiv (ID,DG) \equiv (AD,DE) \equiv (PF,FG).$$
Hence $P'F \equiv PF$ and similarly $P'G \equiv PG$. As a result, $P' \equiv P$. Thus $d$ passes through $K$ and $P$. Finally, as $KP, AE, BD$ are [...] of $\omega_1$ and $\omega_2$, $\omega$ and $\omega_1$, $\omega$ and $\omega_2$ respectively, they are either parallel or concurrent.
\end{proof}

\bsol{All-Russian MO 2009}
Consider a parallelogram $ ABCD$ and two points $ A_1$, $ C_1$ on its sides $ AB$, $ BC$, respectively. Lines $ AC_1$ and $ CA_1$ meet at $ P$. Assume that the circumcircles of triangles $ AA_1P$ and $ CC_1P$ intersect at the second point $ Q$ inside triangle $ ACD$. Prove that $ \angle PDA = \angle QBA$.
\esol

\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice9}
	\label{practice9-fig}
	\caption{Prove that $ \angle PDA = \angle QBA$.}
\end{figure}
Let $E$ be the intersection of $AD$ and $CP$, then
$$\dfrac{\sin \angle{ABP}}{\sin \angle{CBQ}} = \dfrac{AA_1}{CC_1}, \ \dfrac{\sin \angle{ADP}}{\sin \angle{CDP}} \cd \dfrac{EP}{PC}=\dfrac{CD}{ED}.$$
In addition,
$$\dfrac{EP}{PC}=\dfrac{EA}{CC_1}, \ \dfrac{AD}{ED}=1-\dfrac{EA}{ED}=1-\dfrac{AA_1}{CD}=\dfrac{CD-AA_1}{CD},$$
so
\begin{align*} 
& \ ED=\dfrac{CD \cd AD}{CD-AA_1}=\dfrac{CD \cd AD}{AB-AA_1}=\dfrac{CD \cd AD}{BA_1} \\ 
\Rightarrow & \ \dfrac{\sin \angle{ADP}}{\sin \angle{CDP}}=\dfrac{EA}{CC_1} \cd \dfrac{CD}{\frac{CD \cd AD}{BA_1}} = \dfrac{EA}{BC} \cd \dfrac{BA_1}{CC_1}  
\\ \Rightarrow & \ \dfrac{\sin \angle{ADP}}{\sin \angle{CDP}}= \dfrac{AA_1}{BA_1} \cd \dfrac{BA_1}{CC_1} = \dfrac{AA_1}{CC_1}=\dfrac{\sin \angle{ABQ}}{\sin \angle{CBQ}} \numberthis\label{sin}.
\end{align*}
Let $\angle{ABC}=\angle{ADC}=\alpha$ and consider 
$$f: (0,\pi) \to \mathbb{R}, \ f(x)=\frac{\sin (\alpha - x)}{\sin x}.$$ Note that $f$ is a decreasing function on $(0, \pi)$, and from (\ref{sin}) we see that $f(\angle{PDA}) = f(\angle QBA)$, so $\angle{PDA}=\angle{QBA}$.
\end{proof}


\bsol{Balkan MO 2007}
Let $ABCD$ a convex quadrilateral with $AB=BC=CD$, with $AC$ not equal to $BD$ and $E$ be the intersection point of it's diagonals. Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120$.
\esol
\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice10}
	\label{practice10-fig}
	\caption{Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120$.}
\end{figure}
We will prove both directions:
\begin{itemize}
	\item {\bf If $AE = ED$}, then $\triangle EAD$ is isoceles at $E$ so $\angle EAD = \angle EDA = \beta$. Applying the sine rule in $\triangle ABD$ and $\triangle ACD$ we get
	$$\frac{AB}{\sin \beta}=\frac{AD}{\sin \angle{ABD}}, \ \frac{CD}{\sin \beta}=\frac{AD}{\sin \angle{ACD}}.$$
	Also, $AB = CD$ so $\sin \angle ABD = \sin \angle ACD$, or $\angle ABD + \angle ACD = 180\degr$. \\
	Let $\angle ACD = x$ and denote $M$ as the intersection of $AB$ and $CD$. We now have
	$$\angle{MBD}=\angle{BAD}+\angle{BDC}=\angle{BAC}+2\beta,$$
	which implies $x - 2\beta = \angle BAC = \angle CBA$. By similar arguments, 
	$$\angle CDB = 180\degr - x - 2\beta.$$
	On the other hand, $\angle BEA = \angle EBC + \angle BCE$, so $\angle DBC = 4\beta - x$. Combining this with $\angle CBD = \angle CDB$, we get the equation
	$$180\degr - x - 2\beta = 4\beta - x,$$
	which yields $x = 30\degr$, from which it's easy to deduce that $\angle BAD+\angle ADC = 120$.

	\item {\bf If $\angle BAD+\angle ADC = 120\degr$}, let $\angle EAD = \alpha$ and $\angle EDA = \beta$. It follows that $\alpha + \beta = 60 \degr$. Applying the sine rule in $\triangle ABD$ and $\triangle ACD$ we get
	$$\frac{\sin\alpha}{\sin\beta}=\frac{\sin \angle{ACD}}{\sin \angle{ABD}}.$$
	Let $\angle CDB = \angle CBD = x$, we can calculate 
	$$\angle ACD = 180\degr-\alpha-\beta-x = 120\degr - x.$$
	Now note that $\angle{AEB}=\angle{EBC}+\angle{ECB}$, so 
	$$\alpha + \beta - x = \angle ACB = \angle CAB,$$
	which implies
	$$\angle ABD = 180\degr - 2\alpha - 2\beta + x = 60\degr + x.$$
	Finally, we see that
	$$\frac{\sin\alpha}{\sin\beta}=\frac{\sin \angle{ACD}}{\sin \angle{ABD}}=\frac{60\degr+x}{120\degr-x}=1,$$
	which means $\sin \alpha = \sin \beta$, so $\alpha = \beta$. Consequently, $AE = DE$.
\end{itemize}
\end{proof}

\bsol{China TST 2008}
Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE = CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF = FG$.
\esol
\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice11}
	\label{practice11-fig}
	\caption{Prove that $CF = FG$.}
\end{figure}

Let $I$ be the incenter of \abc and $M$ be the midpoint of $BC$. Construct the heights $AH$ of \abc and $CT$ of $\triangle CEF$. \\
To prove that $CF = FG$, we will show that $MF \parallel BD$ or, equivalently, $\frac{EM}{EB} = \frac{EF}{ED}$. Note that
$$\dfrac{EM}{EB}=\dfrac{BE-BM}{EB}=\dfrac{\frac{a+c-b}{2}-\frac{a}{2}}{\frac{a+c-b}{2}}=\frac{c-b}{a+c-b}.$$
On the other hand,
\begin{align*}
AE^2 & \ = AH^2+HE^2 =\frac{4S_{ABC}^2}{a^2}+ \lf \frac{a+b-c}{2} - \frac{a^2+b^2-c^2}{2a} \ri^2 \\ 
& = \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4a^2} + \frac{(c-b)^2(b+c-a)^2}{4a^2}. \numberthis\label{ae^2}
\end{align*}
$(I)$ touches $AB$ at $K$, then 
\begin{equation}
	AD \cdot AE = AK^2= \lf \dfrac{b+c-a}{2} \ri^2. \label{ad.ae}
\end{equation}
As $ED \cdot EA = AE^2 - AD \cdot AE$, from (\ref{ae^2}) and (\ref{ad.ae}) we get
\begin{align*}
	ED \cd EA & = \dfrac{4S^2}{a^2} + \dfrac{(c-b)^2(b+c-a)^2}{4a^2} - \lf \dfrac{b+c-a}{2} \ri^2 \\ 
	& = \dfrac{4S^2}{a^2}+\dfrac{(b+c-a)^2(c-b-a)(c-b+a)}{4a^2} \\ 
	& = \dfrac{(a+b-c)(b+c-a)(c+a-b)}{2a}. \numberthis\label{ed.ea}
\end{align*}
As $ATHC$ is cyclic,
\begin{align*}
	EF \cd EA & = 2 EI \cd EA =2 \cd EH \cd EC  \\
	& =2 \cd \lf \dfrac{a^2+c^2-b^2}{2a}-\dfrac{a+c-b}{2} \ri \cd \dfrac{a+b-c}{2} \\
	& = \dfrac{(c-b)(b+c-a)(a+b-c)}{2a}. \numberthis\label{ef.ea}
\end{align*}
From (\ref{ed.ea}) and (\ref{ef.ea}), we get
$$\dfrac{EF}{ED}=\dfrac{c-b}{a+c-b}=\dfrac{EM}{EB},$$
which yields $CF = FG$.
\end{proof}

\bsol{Balkan MO 2013}
In $\triangle ABC$, the excircle $\omega_a$ opposite $A$ touches $AB$ at $P$ and $AC$ at $Q$, while the excircle $\omega_b$ opposite $B$ touches $BA$ at $M$ and $BC$ at $N$. Let $K$ be the projection of $C$ onto $MN$ and let $L$ be the projection of $C$ onto $PQ$. Show that the quadrilateral $MKLP$ is cyclic.
\esol
\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice12}
	\label{practice12-fig}
	\caption{Prove that $MKLP$ is cyclic.}
\end{figure}
Let $I_b, I_c$ be the excenters opposite $B$ and $C$ in \abc respectively. Let $\omega_c$ be the excircle opposite to $C$. $\omega_c$ touches $BC$ at $T$. $S$ is the intersection of $MN$ and $PQ$. \\
Let $X, Y$ be the projections of $M$ on $PQ$ and $P$ on $MN$ respectively. $AX$ intersects $PY$ at $H$. $HC$ intersects $MN$ at $S_1$ and $PQ$ at $S_2$. \\
Observe that
$$CK=\frac{b+c-a}{2} \cd \cos \dfrac{B}{2}, \ CL=\dfrac{a+c-b}{2} \cd \cos \dfrac{A}{2}.$$
On the other hand, $HA \cd HX = HP \cd HY$ so 
$$\dfrac{YH}{XH}=\dfrac{MH}{PH} = \dfrac{\sin \frac{B}{2}}{\sin \frac{A}{2}}.$$
In addition,
$$\tan \dfrac{A}{2} = \dfrac{b+c-a}{2TI_c}, \ \tan \dfrac{B}{2}=\dfrac{c+a-b}{2TI_c},$$
which yields
$$(a+c-b) \cd \tan \dfrac{B}{2} = (b+c-a) \cd \tan \dfrac{A}{2},$$
which implies that
$$\dfrac{CK}{CL} = \dfrac{(b+c-a) \cd \cos \frac{B}{2}}{(a+c-b) \cos \frac{A}{2}} =  \dfrac{\sin \frac{B}{2}}{\sin \frac{A}{2}} = \dfrac{YH}{XH}.$$
Now, as $CK \parallel YH$ and $CL \parallel XH$,
$$\dfrac{S_1C}{S_1H}=\dfrac{CK}{YH} = \dfrac{CL}{XH} = \dfrac{S_2C}{S_2H},$$
so $S_1 \equiv S_2$ and therefore $H, C, S$ are collinear. \\
As $\angle{SKL}=90^0-\angle{CKL}$ and $\angle{SYX}=90^0-\angle{HYX}$, $\angle{SKL}=\angle{HYX}$, so $KL \parallel XY$. It follows that $\angle{YXP}=\angle{KLP}$, while we also have $\angle{YXP}=180^{\circ}-\angle{PMY}$. Thus $\angle KLP = 180^{\circ}-\angle{PMY}$ and so $MPLK$ is cyclic.
\end{proof}

\newpage
\bsol{IMO Shortlist 2005}
Given $\triangle ABC$ satisfying $AB + BC = 3CA$. The incircle of $\triangle ABC$ has center $I$ and touches the sides $AB$ and $BC$ at the points $D$ and $E$, respectively. Let $K$ and $L$ be the reflections of the points $D$ and $E$ with respect to $I$. Prove that the points $A$, $C$, $K$, $L$ lie on one circle.
\esol
\begin{proof}
\begin{figure}[h]
	\centering
	\input{geogebra/practice13}
	\label{practice13-fig}
	\caption{Prove that $A, C, K, L$ lie on one circle.}
\end{figure}
We have $c + a = 3b$. Note that if $b = c$ then $a = 2b = b + c$, which contradicts the triangle inequality. Hence $b \ne c$, and WLOG assume $b < c$. \\
$AL$ intersects $BC$ at $F$. Assume the positive direction is that of vector $\vto{BC}$, then
$$\ds{EF}=\ds{CF}-\ds{CE}=-\ds{BE}-\ds{CE}=\dfrac{a+c-b}{2} - \dfrac{a+b-c}{2} = c-b,$$
so $EF = |\ds{EF}| = c - b$. Now $\triangle LFE$ and $\triangle CLE$ have
$$\tan \angle{LFC}=\dfrac{LE}{EF}=\dfrac{2r}{c-b}, \ \tan \angle{LCE}=\dfrac{LE}{CE}=\dfrac{4r}{a+c-b},$$
so
$$\begin{aligned} 
\tan \angle{ALC} & =\tan (\angle{LFC}+\angle{LCE}) = \dfrac{\tan \angle{LFC}+\tan \angle{LCE}}{1- \tan \angle {LFC} \cd \tan \angle {LCE}} \\
 & = \lf \frac{2r}{c-b} + \frac{4r}{a+b-c} \ri : \lf 1 - \frac{8r^2}{(c-b)(a+b-c)} \ri \\ 
 & = \frac{2r(a+c-b)}{(c-b)(a+b-c)-8r^2}.
 \end{aligned}$$
As $a+c=3b$, we can compute the inradius $r$ of \abc from
$$r^2=\dfrac{S^2}{p^2}=\dfrac{(p-a)(p-b)(p-c)}{p}=\dfrac{(b+c-a)(a+b-c)}{8}.$$
It then follows that
\begin{align*}
\tan \angle{ALC} & =\dfrac{2r(a+c-b)}{(c-b)(a+b-c)-(b+c-a)(a+b-c)} \\
& =\dfrac{2r(a+c-b)}{(a+b-c)(a-2b)}=\dfrac{2br}{(2b-c)(b-c)}.
\end{align*}
By similar arguments,
$$\tan \angle{AKC}=\dfrac{2br}{(2b-a)(b-a)}.$$
From $a + c = 3b$ it is easy to see that $(2b-c)(b-c)=(2b-a)(b-a)$. Thus $\tan \angle{ALC}= \tan \angle{AKC}$. In other words, $\angle ALC = \angle AKC$, so $ALKC$ is cyclic.
\end{proof}
\end{document}
