Subfield: Geometry


Question: Let $A B C D$ be a convex quadrilateral with $\angle A B C=\angle A D C<90^{\circ}$. The internal angle bisectors of $\angle A B C$ and $\angle A D C$ meet $A C$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $A C$ and let $\omega$ be the circumcircle of triangle $B P D$. Segments $B M$ and $D M$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $X E$ and $Y F$. Prove that $P Q \perp A C$.

Solution:
<img_3447>

Let $\omega_{1}$ be the circumcircle of triangle $A B C$. We first prove that $Y$ lies on $\omega_{1}$. Let $Y^{\prime}$ be the point on ray $M D$ such that $M Y^{\prime} \cdot M D=M A^{2}$. Then triangles $M A Y^{\prime}$ and $M D A$ are oppositely similar. Since $M C^{2}=M A^{2}=M Y^{\prime} \cdot M D$, triangles $M C Y^{\prime}$ and $M D C$ are also oppositely similar. Therefore, using directed angles, we have

$$
\measuredangle A Y^{\prime} C=\measuredangle A Y^{\prime} M+\measuredangle M Y^{\prime} C=\measuredangle M A D+\measuredangle D C M=\measuredangle C D A=\measuredangle A B C
$$

so that $Y^{\prime}$ lies on $\omega_{1}$.

Let $Z$ be the intersection point of lines $B C$ and $A D$. Since $\measuredangle P D Z=\measuredangle P B C=\measuredangle P B Z$, point $Z$ lies on $\omega$. In addition, from $\measuredangle Y^{\prime} B Z=\measuredangle Y^{\prime} B C=\measuredangle Y^{\prime} A C=\measuredangle Y^{\prime} A M=\measuredangle Y^{\prime} D Z$, we also know that $Y^{\prime}$ lies on $\omega$. Note that $\angle A D C$ is acute implies $M A \neq M D$ so $M Y^{\prime} \neq M D$. Therefore, $Y^{\prime}$ is the second intersection of $D M$ and $\omega$. Then $Y^{\prime}=Y$ and hence $Y$ lies on $\omega_{1}$.

Next, by the Angle Bisector Theorem and the similar triangles, we have

$$
\frac{F A}{F C}=\frac{A D}{C D}=\frac{A D}{A M} \cdot \frac{C M}{C D}=\frac{Y A}{Y M} \cdot \frac{Y M}{Y C}=\frac{Y A}{Y C}
$$

Hence, $F Y$ is the internal angle bisector of $\angle A Y C$.

Let $B^{\prime}$ be the second intersection of the internal angle bisector of $\angle C B A$ and $\omega_{1}$. Then $B^{\prime}$ is the midpoint of arc $A C$ not containing $B$. Therefore, $Y B^{\prime}$ is the external angle bisector of $\angle A Y C$, so that $B^{\prime} Y \perp F Y$.



Denote by $l$ the line through $P$ parallel to $A C$. Suppose $l$ meets line $B^{\prime} Y$ at $S$. From

$$
\begin{aligned}
\measuredangle P S Y & =\measuredangle\left(A C, B^{\prime} Y\right)=\measuredangle A C Y+\measuredangle C Y B^{\prime}=\measuredangle A C Y+\measuredangle C A B^{\prime}=\measuredangle A C Y+\measuredangle B^{\prime} C A \\
& =\measuredangle B^{\prime} C Y=\measuredangle B^{\prime} B Y=\measuredangle P B Y,
\end{aligned}
$$

the point $S$ lies on $\omega$. Similarly, the line through $X$ perpendicular to $X E$ also passes through the second intersection of $l$ and $\omega$, which is the point $S$. From $Q Y \perp Y S$ and $Q X \perp X S$, point $Q$ lies on $\omega$ and $Q S$ is a diameter of $\omega$. Therefore, $P Q \perp P S$ so that $P Q \perp A C$.
Denote by $\omega_{1}$ and $\omega_{2}$ the circumcircles of triangles $A B C$ and $A D C$ respectively. Since $\angle A B C=\angle A D C$, we know that $\omega_{1}$ and $\omega_{2}$ are symmetric with respect to the midpoint $M$ of $A C$.

Firstly, we show that $X$ lies on $\omega_{2}$. Let $X_{1}$ be the second intersection of ray $M B$ and $\omega_{2}$ and $X^{\prime}$ be its symmetric point with respect to $M$. Then $X^{\prime}$ lies on $\omega_{1}$ and $X^{\prime} A X_{1} C$ is a parallelogram. Hence, we have

$$
\begin{aligned}
\measuredangle D X_{1} B & =\measuredangle D X_{1} A+\measuredangle A X_{1} B=\measuredangle D C A+\measuredangle A X_{1} X^{\prime}=\measuredangle D C A+\measuredangle C X^{\prime} X_{1} \\
& =\measuredangle D C A+\measuredangle C A B=\angle(C D, A B) .
\end{aligned}
$$

<img_4035>

Also, we have

$$
\measuredangle D P B=\measuredangle P D C+\angle(C D, A B)+\measuredangle A B P=\angle(C D, A B)
$$

These yield $\measuredangle D X_{1} B=\measuredangle D P B$ and hence $X_{1}$ lies on $\omega$. It follows that $X_{1}=X$ and $X$ lies on $\omega_{2}$. Similarly, $Y$ lies on $\omega_{1}$.



Next, we prove that $Q$ lies on $\omega$. Suppose the perpendicular bisector of $A C$ meet $\omega_{1}$ at $B^{\prime}$ and $M_{1}$ and meet $\omega_{2}$ at $D^{\prime}$ and $M_{2}$, so that $B, M_{1}$ and $D^{\prime}$ lie on the same side of $A C$. Note that $B^{\prime}$ lies on the angle bisector of $\angle A B C$ and similarly $D^{\prime}$ lies on $D P$.

If we denote the area of $W_{1} W_{2} W_{3}$ by $\left[W_{1} W_{2} W_{3}\right]$, then

$$
\frac{B A \cdot X^{\prime} A}{B C \cdot X^{\prime} C}=\frac{\frac{1}{2} B A \cdot X^{\prime} A \sin \angle B A X^{\prime}}{\frac{1}{2} B C \cdot X^{\prime} C \sin \angle B C X^{\prime}}=\frac{\left[B A X^{\prime}\right]}{\left[B C X^{\prime}\right]}=\frac{M A}{M C}=1
$$

As $B E$ is the angle bisector of $\angle A B C$, we have

$$
\frac{E A}{E C}=\frac{B A}{B C}=\frac{X^{\prime} C}{X^{\prime} A}=\frac{X A}{X C}
$$

Therefore, $X E$ is the angle bisector of $\angle A X C$, so that $M_{2}$ lies on the line joining $X, E, Q$. Analogously, $M_{1}, F, Q, Y$ are collinear. Thus,

$$
\begin{aligned}
\measuredangle X Q Y & =\measuredangle M_{2} Q M_{1}=\measuredangle Q M_{2} M_{1}+\measuredangle M_{2} M_{1} Q=\measuredangle X M_{2} D^{\prime}+\measuredangle B^{\prime} M_{1} Y \\
& =\measuredangle X D D^{\prime}+\measuredangle B^{\prime} B Y=\measuredangle X D P+\measuredangle P B Y=\measuredangle X B P+\measuredangle P B Y=\measuredangle X B Y,
\end{aligned}
$$

which implies $Q$ lies on $\omega$.

Finally, as $M_{1}$ and $M_{2}$ are symmetric with respect to $M$, the quadrilateral $X^{\prime} M_{2} X M_{1}$ is a parallelogram. Consequently,

$$
\measuredangle X Q P=\measuredangle X B P=\measuredangle X^{\prime} B B^{\prime}=\measuredangle X^{\prime} M_{1} B^{\prime}=\measuredangle X M_{2} M_{1}
$$

This shows $Q P / / M_{2} M_{1}$. As $M_{2} M_{1} \perp A C$, we get $Q P \perp A C$.
We first state two results which will be needed in our proof.

- Claim 1. In $\triangle X^{\prime} Y^{\prime} Z^{\prime}$ with $X^{\prime} Y^{\prime} \neq X^{\prime} Z^{\prime}$, let $N^{\prime}$ be the midpoint of $Y^{\prime} Z^{\prime}$ and $W^{\prime}$ be the foot of internal angle bisector from $X^{\prime}$. Then $\tan ^{2} \measuredangle W^{\prime} X^{\prime} Z^{\prime}=\tan \measuredangle N^{\prime} X^{\prime} W^{\prime} \tan \measuredangle Z^{\prime} W^{\prime} X^{\prime}$.

Proof.

<img_3585>

Without loss of generality, assume $X^{\prime} Y^{\prime}>X^{\prime} Z^{\prime}$. Then $W^{\prime}$ lies between $N^{\prime}$ and $Z^{\prime}$. The signs of both sides agree so it suffices to establish the relation for ordinary angles. Let $\angle W^{\prime} X^{\prime} Z^{\prime}=\alpha, \angle N^{\prime} X^{\prime} W^{\prime}=\beta$ and $\angle Z^{\prime} W^{\prime} X^{\prime}=\gamma$. We have

$$
\frac{\sin (\gamma-\alpha)}{\sin (\alpha-\beta)}=\frac{N^{\prime} X^{\prime}}{N^{\prime} Y^{\prime}}=\frac{N^{\prime} X^{\prime}}{N^{\prime} Z^{\prime}}=\frac{\sin (\gamma+\alpha)}{\sin (\alpha+\beta)}
$$



This implies

$$
\frac{\tan \gamma-\tan \alpha}{\tan \gamma+\tan \alpha}=\frac{\sin \gamma \cos \alpha-\cos \gamma \sin \alpha}{\sin \gamma \cos \alpha+\cos \gamma \sin \alpha}=\frac{\sin \alpha \cos \beta-\cos \alpha \sin \beta}{\sin \alpha \cos \beta+\cos \alpha \sin \beta}=\frac{\tan \alpha-\tan \beta}{\tan \alpha+\tan \beta}
$$

Expanding and simplifying, we get the desired result $\tan ^{2} \alpha=\tan \beta \tan \gamma$.

- Claim 2. Let $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be a quadrilateral inscribed in circle $\Gamma$. Let diagonals $A^{\prime} C^{\prime}$ and $B^{\prime} D^{\prime}$ meet at $E^{\prime}$, and $F^{\prime}$ be the intersection of lines $A^{\prime} B^{\prime}$ and $C^{\prime} D^{\prime}$. Let $M^{\prime}$ be the midpoint of $E^{\prime} F^{\prime}$. Then the power of $M^{\prime}$ with respect to $\Gamma$ is equal to $\left(M^{\prime} E^{\prime}\right)^{2}$.

Proof.

<img_3656>

Let $O^{\prime}$ be the centre of $\Gamma$ and let $\Gamma^{\prime}$ be the circle with centre $M^{\prime}$ passing through $E^{\prime}$. Let $F_{1}$ be the inversion image of $F^{\prime}$ with respect to $\Gamma$. It is well-known that $E^{\prime}$ lies on the polar of $F^{\prime}$ with respect to $\Gamma$. This shows $E^{\prime} F_{1} \perp O^{\prime} F^{\prime}$ and hence $F_{1}$ lies on $\Gamma^{\prime}$. It follows that the inversion image of $\Gamma^{\prime}$ with respect to $\Gamma$ is $\Gamma^{\prime}$ itself. This shows $\Gamma^{\prime}$ is orthogonal to $\Gamma$, and thus the power of $M^{\prime}$ with respect to $\Gamma$ is the square of radius of $\Gamma^{\prime}$, which is $\left(M^{\prime} E^{\prime}\right)^{2}$.

We return to the main problem. Let $Z$ be the intersection of lines $A D$ and $B C$, and $W$ be the intersection of lines $A B$ and $C D$. Since $\measuredangle P D Z=\measuredangle P B C=\measuredangle P B Z$, point $Z$ lies on $\omega$. Similarly, $W$ lies on $\omega$. Applying Claim 2 to the cyclic quadrilateral $Z B D W$, we know that the power of $M$ with respect to $\omega$ is $M A^{2}$. Hence, $M X \cdot M B=M A^{2}$.

Suppose the line through $B$ perpendicular to $B E$ meets line $A C$ at $T$. Then $B E$ and $B T$ are the angle bisectors of $\angle C B A$. This shows $(T, E ; A, C)$ is harmonic. Thus, we have $M E \cdot M T=M A^{2}=M X \cdot M B$. It follows that $E, T, B, X$ are concyclic.



<img_3636>

The result is trivial for the special case $A D=C D$ since $P, Q$ lie on the perpendicular bisector of $A C$ in that case. Similarly, the case $A B=C B$ is trivial. It remains to consider the general cases where we can apply Claim 1 in the latter part of the proof.

Let the projections from $P$ and $Q$ to $A C$ be $P^{\prime}$ and $Q^{\prime}$ respectively. Then $P Q \perp A C$ if and only if $P^{\prime}=Q^{\prime}$ if and only if $\frac{E P^{\prime}}{F P^{\prime}}=\frac{E Q^{\prime}}{F Q^{\prime}}$ in terms of directed lengths. Note that

$$
\frac{E P^{\prime}}{F P^{\prime}}=\frac{\tan \measuredangle E F P}{\tan \measuredangle F E P}=\frac{\tan \measuredangle A F D}{\tan \measuredangle A E B}
$$

Next, we have $\frac{E Q^{\prime}}{F Q^{\prime}}=\frac{\tan \measuredangle E F Q}{\tan \measuredangle F E Q}$ where $\measuredangle F E Q=\measuredangle T E X=\measuredangle T B X=\frac{\pi}{2}+\measuredangle E B M$ and by symmetry $\measuredangle E F Q=\frac{\pi}{2}+\measuredangle F D M$. Combining all these, it suffices to show

$$
\frac{\tan \measuredangle A F D}{\tan \measuredangle A E B}=\frac{\tan \measuredangle M B E}{\tan \measuredangle M D F}
$$

We now apply Claim 1 twice to get

$$
\tan \measuredangle A F D \tan \measuredangle M D F=\tan ^{2} \measuredangle F D C=\tan ^{2} \measuredangle E B A=\tan \measuredangle M B E \tan \measuredangle A E B .
$$

The result then follows.