Subfield: Geometry


Question: Let $A B C$ be a triangle with circumcircle $\Gamma$ and incentre $I$. Let $M$ be the midpoint of side $B C$. Denote by $D$ the foot of perpendicular from $I$ to side $B C$. The line through $I$ perpendicular to $A I$ meets sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose the circumcircle of triangle $A E F$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $X D$ and $A M$ meet on $\Gamma$.

Solution:
Let $A M$ meet $\Gamma$ again at $Y$ and $X Y$ meet $B C$ at $D^{\prime}$. It suffices to show $D^{\prime}=D$. We shall apply the following fact.

- Claim. For any cyclic quadrilateral $P Q R S$ whose diagonals meet at $T$, we have

$$
\frac{Q T}{T S}=\frac{P Q \cdot Q R}{P S \cdot S R}
$$

Proof. We use $\left[W_{1} W_{2} W_{3}\right]$ to denote the area of $W_{1} W_{2} W_{3}$. Then

$$
\frac{Q T}{T S}=\frac{[P Q R]}{[P S R]}=\frac{\frac{1}{2} P Q \cdot Q R \sin \angle P Q R}{\frac{1}{2} P S \cdot S R \sin \angle P S R}=\frac{P Q \cdot Q R}{P S \cdot S R}
$$

Applying the Claim to $A B Y C$ and $X B Y C$ respectively, we have $1=\frac{B M}{M C}=\frac{A B \cdot B Y}{A C \cdot C Y}$ and $\frac{B D^{\prime}}{D^{\prime} C}=\frac{X B \cdot B Y}{X C \cdot C Y}$. These combine to give

$$
\frac{B D^{\prime}}{C D^{\prime}}=\frac{X B}{X C} \cdot \frac{B Y}{C Y}=\frac{X B}{X C} \cdot \frac{A C}{A B}
\tag{1}
$$

Next, we use directed angles to find that $\measuredangle X B F=\measuredangle X B A=\measuredangle X C A=\measuredangle X C E$ and $\measuredangle X F B=\measuredangle X F A=\measuredangle X E A=\measuredangle X E C$. This shows triangles $X B F$ and $X C E$ are directly similar. In particular, we have

$$
\frac{X B}{X C}=\frac{B F}{C E}
\tag{2}
$$

In the following, we give two ways to continue the proof.

- Method 1. Here is a geometrical method. As $\angle F I B=\angle A I B-90^{\circ}=\frac{1}{2} \angle A C B=\angle I C B$ and $\angle F B I=\angle I B C$, the triangles $F B I$ and $I B C$ are similar. Analogously, triangles $E I C$ and $I B C$ are also similar. Hence, we get

$$
\frac{F B}{I B}=\frac{B I}{B C} \quad \text { and } \quad \frac{E C}{I C}=\frac{I C}{B C}
\tag{3}
$$



<img_3668>

Next, construct a line parallel to $B C$ and tangent to the incircle. Suppose it meets sides $A B$ and $A C$ at $B_{1}$ and $C_{1}$ respectively. Let the incircle touch $A B$ and $A C$ at $B_{2}$ and $C_{2}$ respectively. By homothety, the line $B_{1} I$ is parallel to the external angle bisector of $\angle A B C$, and hence $\angle B_{1} I B=90^{\circ}$. Since $\angle B B_{2} I=90^{\circ}$, we get $B B_{2} \cdot B B_{1}=B I^{2}$, and similarly $C C_{2} \cdot C C_{1}=C I^{2}$. Hence,

$$
\frac{B I^{2}}{C I^{2}}=\frac{B B_{2} \cdot B B_{1}}{C C_{2} \cdot C C_{1}}=\frac{B B_{1}}{C C_{1}} \cdot \frac{B D}{C D}=\frac{A B}{A C} \cdot \frac{B D}{C D}
\tag{4}
$$

Combining (1), (2), (3) and (4), we conclude

$$
\frac{B D^{\prime}}{C D^{\prime}}=\frac{X B}{X C} \cdot \frac{A C}{A B}=\frac{B F}{C E} \cdot \frac{A C}{A B}=\frac{B I^{2}}{C I^{2}} \cdot \frac{A C}{A B}=\frac{B D}{C D}
$$

so that $D^{\prime}=D$. The result then follows.

- Method 2. Let $\beta=\frac{1}{2} \angle A B C$ and $\gamma=\frac{1}{2} \angle A C B$. Observe that $\angle F I B=\angle A I B-90^{\circ}=\gamma$. Hence, $\frac{B F}{F I}=\frac{\sin \angle F I B}{\sin \angle I B F}=\frac{\sin \gamma}{\sin \beta}$. Similarly, $\frac{C E}{E I}=\frac{\sin \beta}{\sin \gamma}$. As $F I=E I$, we get

$$
\frac{B F}{C E}=\frac{B F}{F I} \cdot \frac{E I}{C E}=\left(\frac{\sin \gamma}{\sin \beta}\right)^{2}
\tag{5}
$$



Together with (1) and (2), we find that

$$
\frac{B D^{\prime}}{C D^{\prime}}=\frac{A C}{A B} \cdot\left(\frac{\sin \gamma}{\sin \beta}\right)^{2}=\frac{\sin 2 \beta}{\sin 2 \gamma} \cdot\left(\frac{\sin \gamma}{\sin \beta}\right)^{2}=\frac{\tan \gamma}{\tan \beta}=\frac{I D / C D}{I D / B D}=\frac{B D}{C D}
$$

This shows $D^{\prime}=D$ and the result follows.
Let $\omega_{A}$ be the $A$-mixtilinear incircle of triangle $A B C$. From the properties of mixtilinear incircles, $\omega_{A}$ touches sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose $\omega_{A}$ is tangent to $\Gamma$ at $T$. Let $A M$ 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 $B C$ respectively. It is well-known that $\angle B A T=\angle D_{1} A C$ so that $A, D_{1}, T_{1}$ are collinear.

<img_3166>

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 $A E F$. Then $R$ lies on $A X, E F$ and the tangent at $T$ to $\Gamma$. Let $A T$ meet $\omega_{A}$ again at $S$ and meet $E F$ at $P$. Obviously, $S F T E$ 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 $X B T C$ is a harmonic quadrilateral. As $T T_{1} / / B C$, the projection from $T_{1}$ onto $B C$ maps $T$ to a point at infinity, and hence maps $X$ to the midpoint of $B C$, which is $M$. This shows $X, M, T_{1}$ are collinear.

We have two ways to finish the proof.

- Method 1. Note that both $A Y$ and $X T_{1}$ are chords of $\Gamma$ passing through the midpoint $M$ of the chord $B C$. By the Butterfly Theorem, $X Y$ and $A T_{1}$ cut $B C$ 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 $D_{1}$ is an isosceles trapezoid, we have

$$
\measuredangle Y T D=\measuredangle Y T T_{1}+\measuredangle T_{1} T D=\measuredangle Y A T_{1}+\measuredangle A D_{1} D=\measuredangle Y M D
$$

so that $D, T, Y, M$ are concyclic. As $X, M, T_{1}$ are collinear, we have

$$
\measuredangle A Y D=\measuredangle M T D=\measuredangle D_{1} T_{1} M=\measuredangle A T_{1} X=\measuredangle A Y X
$$

This shows $X, D, Y$ are collinear.