, 2016

Let $AM$ meet $\Gamma$ again at $Y$ and $XY$ meet $BC$ at $D'$. It suffices to show $D' = D$. We shall apply the following fact.

- Claim. For any cyclic quadrilateral $PQRS$ whose diagonals meet at $T$, we have
$$\frac{QT}{TS} = \frac{PQ \cdot QR}{PS \cdot SR}$$
Proof. We use $[W_1 W_2 W_3]$ to denote the area of $W_1 W_2 W_3$. Then
$$\frac{QT}{TS} = \frac{[PQR]}{[PSR]} = \frac{\frac{1}{2} PQ \cdot QR \sin \angle PQR}{\frac{1}{2} PS \cdot SR \sin \angle PSR} = \frac{PQ \cdot QR}{PS \cdot SR}$$
Applying the Claim to $ABYC$ and $XBYC$ respectively, we have $1 = \frac{BM}{MC} = \frac{AB \cdot BY}{AC \cdot CY}$ and $\frac{BD'}{D'C} = \frac{XB \cdot BY}{XC \cdot CY}$. These combine to give
$$\begin{equation*} \frac{BD'}{CD'} = \frac{XB}{XC} \cdot \frac{BY}{CY} = \frac{XB}{XC} \cdot \frac{AC}{AB} \tag{1} \end{equation*}$$
Next, we use directed angles to find that $\measuredangle XBF = \measuredangle XBA = \measuredangle XCA = \measuredangle XCE$ and $\measuredangle XFB = \measuredangle XFA = \measuredangle XEA = \measuredangle XEC$. This shows triangles $XBF$ and $XCE$ are directly similar. In particular, we have
$$\begin{equation*} \frac{XB}{XC} = \frac{BF}{CE} \tag{2} \end{equation*}$$
In the following, we give two ways to continue the proof.

- Method 1. Here is a geometrical method. As $\angle FIB = \angle AIB - 90^{\circ} = \frac{1}{2} \angle ACB = \angle ICB$ and $\angle FBI = \angle IBC$, the triangles $FBI$ and $IBC$ are similar. Analogously, triangles $EIC$ and $IBC$ are also similar. Hence, we get
$$\begin{equation*} \frac{FB}{IB} = \frac{BI}{BC} \quad \text{ and } \quad \frac{EC}{IC} = \frac{IC}{BC} \tag{3} \end{equation*}$$

Next, construct a line parallel to $BC$ and tangent to the incircle. Suppose it meets sides $AB$ and $AC$ at $B_1$ and $C_1$ respectively. Let the incircle touch $AB$ and $AC$ at $B_2$ and $C_2$ respectively. By homothety, the line $B_1I$ is parallel to the external angle bisector of $\angle ABC$, and hence $\angle B_1IB = 90^{\circ}$. Since $\angle BB_2I = 90^{\circ}$, we get $BB_2 \cdot BB_1 = BI^2$, and similarly $CC_2 \cdot CC_1 = CI^2$. Hence,
$$\begin{equation*} \frac{BI^2}{CI^2} = \frac{BB_2 \cdot BB_1}{CC_2 \cdot CC_1} = \frac{BB_1}{CC_1} \cdot \frac{BD}{CD} = \frac{AB}{AC} \cdot \frac{BD}{CD} . \tag{4} \end{equation*}$$
Combining (1), (2), (3) and (4), we conclude
$$\frac{BD'}{CD'} = \frac{XB}{XC} \cdot \frac{AC}{AB} = \frac{BF}{CE} \cdot \frac{AC}{AB} = \frac{BI^2}{CI^2} \cdot \frac{AC}{AB} = \frac{BD}{CD}$$
so that $D' = D$. The result then follows.

- Method 2. We continue the proof of Solution 1 using trigonometry. Let $\beta = \frac{1}{2} \angle ABC$ and $\gamma = \frac{1}{2} \angle ACB$. Observe that $\angle FIB = \angle AIB - 90^{\circ} = \gamma$. Hence, $\frac{BF}{FI} = \frac{\sin \angle FIB}{\sin \angle IBF} = \frac{\sin \gamma}{\sin \beta}$. Similarly, $\frac{CE}{EI} = \frac{\sin \beta}{\sin \gamma}$. As $FI = EI$, we get
$$\frac{BD'}{CD'} = \frac{AC}{AB} \cdot \left(\frac{\sin \gamma}{\sin \beta}\right)^2 = \frac{\sin 2\beta}{\sin 2\gamma} \cdot \left(\frac{\sin \gamma}{\sin \beta}\right)^2 = \frac{\tan \gamma}{\tan \beta} = \frac{ID / CD}{ID / BD} = \frac{BD}{CD} .$$
This shows $D' = D$ and the result follows.

Let $\omega_A$ be the $A$-mixtilinear incircle of triangle $ABC$. From the properties of mixtilinear incircles, $\omega_A$ touches sides $AB$ and $AC$ at $F$ and $E$ respectively. Suppose $\omega_A$ is tangent to $\Gamma$ at $T$. Let $AM$ meet $\Gamma$ again at $Y$, and let $D_1, T_1$ be the reflections of $D$ and $T$ with respect to the perpendicular bisector of $BC$ respectively. It is well-known that $\angle BAT = \angle D_1AC$ so that $A, D_1, T_1$ are collinear.

We then show that $X, M, T_1$ are collinear. Let $R$ be the radical centre of $\omega_A, \Gamma$ and the circumcircle of triangle $AEF$. Then $R$ lies on $AX, EF$ and the tangent at $T$ to $\Gamma$. Let $AT$ meet $\omega_A$ again at $S$ and meet $EF$ at $P$. Obviously, $SFTE$ is a harmonic quadrilateral. Projecting from $T$, the pencil ($R, P ; F, E$) is harmonic. We further project the pencil onto $\Gamma$ from $A$, so that $XBTC$ is a harmonic quadrilateral. As $TT_1 \parallel BC$, the projection from $T_1$ onto $BC$ maps $T$ to a point at infinity, and hence maps $X$ to the midpoint of $BC$, which is $M$. This shows $X, M, T_1$ are collinear.
We have two ways to finish the proof.

- Method 1. Note that both $AY$ and $XT_1$ are chords of $\Gamma$ passing through the midpoint $M$ of the chord $BC$. By the Butterfly Theorem, $XY$ and $AT_1$ cut $BC$ at a pair of symmetric points with respect to $M$, and hence $X, D, Y$ are collinear. The proof is thus complete.

- Method 2. Here, we finish the proof without using the Butterfly Theorem. As $DTT_1D_1$ is an isosceles trapezoid, we have
$$\measuredangle YTD = \measuredangle YTT_1 + \measuredangle T_1TD = \measuredangle YAT_1 + \measuredangle AD_1D = \measuredangle YMD$$
so that $D, T, Y, M$ are concyclic. As $X, M, T_1$ are collinear, we have
$$\measuredangle AYD = \measuredangle MTD = \measuredangle D_1T_1M = \measuredangle AT_1X = \measuredangle AYX$$
This shows $X, D, Y$ are collinear.