Subfield: Geometry


Question: Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.

Solution:
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Let $\Gamma$ be the circle with centre $E$ passing through $B$ and $C$. Since $E D \perp A C$, the point $F$ symmetric to $C$ with respect to $D$ lies on $\Gamma$. From $\angle D C I=\angle I C B=\angle C B I$, the line $D C$ is a tangent to the circumcircle of triangle $I B C$. Let $J$ be the symmetric point of $I$ with respect to $D$. Using directed lengths, from

$$
D C \cdot D F=-D C^{2}=-D I \cdot D B=D J \cdot D B
$$

the point $J$ also lies on $\Gamma$. Let $I^{\prime}$ be the reflection of $I$ in $A C$. Since $I J$ and $C F$ bisect each other, $C J F I$ is a parallelogram. From $\angle F I^{\prime} C=\angle C I F=\angle F J C$, we find that $I^{\prime}$ lies on $\Gamma$. This gives $E I^{\prime}=E B$.

Note that $A C$ is the internal angle bisector of $\angle B D I^{\prime}$. This shows $D E$ is the external angle bisector of $\angle B D I^{\prime}$ as $D E \perp A C$. Together with $E I^{\prime}=E B$, it is well-known that $E$ lies on the circumcircle of triangle $B D I^{\prime}$.
Let $I^{\prime}$ be the reflection of $I$ in $A C$ and let $S$ be the intersection of $I^{\prime} C$ and $A I$. Using directed angles, we let $\theta=\measuredangle A C I=\measuredangle I C B=\measuredangle C B I$. We have

$$
\measuredangle I^{\prime} S E=\measuredangle I^{\prime} C A+\measuredangle C A I=\theta+\left(\frac{\pi}{2}+2 \theta\right)=3 \theta+\frac{\pi}{2}
$$

and

$$
\measuredangle I^{\prime} D E=\measuredangle I^{\prime} D C+\frac{\pi}{2}=\measuredangle C D I+\frac{\pi}{2}=\measuredangle D C B+\measuredangle C B D+\frac{\pi}{2}=3 \theta+\frac{\pi}{2} .
$$

This shows $I^{\prime}, D, E, S$ are concyclic.

Next, we find $\measuredangle I^{\prime} S B=2 \measuredangle I^{\prime} S E=6 \theta$ and $\measuredangle I^{\prime} D B=2 \measuredangle C D I=6 \theta$. Therefore, $I^{\prime}, D, B, S$ are concyclic so that $I^{\prime}, D, E, B, S$ lie on the same circle. The result then follows.



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Let $I^{\prime}$ be the reflection of $I$ in $A C$, and let $D^{\prime}$ be the second intersection of $A I$ and the circumcircle of triangle $A B D$. Since $A D^{\prime}$ bisects $\angle B A D$, point $D^{\prime}$ is the midpoint of the arc $B D$ and $D D^{\prime}=B D^{\prime}=C D^{\prime}$. Obviously, $A, E, D^{\prime}$ lie on $A I$ in this order.

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We find that $\angle E D^{\prime} D=\angle A D^{\prime} D=\angle A B D=\angle I B C=\angle I C B$. Next, since $D^{\prime}$ is the circumcentre of triangle $B C D$, we have $\angle E D D^{\prime}=90^{\circ}-\angle D^{\prime} D C=\angle C B D=\angle I B C$. The two relations show that triangles $E D^{\prime} D$ and $I C B$ are similar. Therefore, we have

$$
\frac{B C}{C I^{\prime}}=\frac{B C}{C I}=\frac{D D^{\prime}}{D^{\prime} E}=\frac{B D^{\prime}}{D^{\prime} E}
$$

Also, we get

$$
\angle B C I^{\prime}=\angle B C A+\angle A C I^{\prime}=\angle B C A+\angle I C A=\angle B C A+\angle D B C=\angle B D A=\angle B D^{\prime} E .
$$

These show triangles $B C I^{\prime}$ and $B D^{\prime} E$ are similar, and hence triangles $B C D^{\prime}$ and $B I^{\prime} E$ are similar. As $B C D^{\prime}$ is isosceles, we obtain $B E=I^{\prime} E$.

As $D E$ is the external angle bisector of $\angle B D I^{\prime}$ and $E I^{\prime}=E B$, we know that $E$ lies on the circumcircle of triangle $B D I^{\prime}$.
Let $A I$ and $B I$ meet the circumcircle of triangle $A B C$ again at $A^{\prime}$ and $B^{\prime}$ respectively, and let $E^{\prime}$ be the reflection of $E$ in $A C$. From

$$
\begin{aligned}
\angle B^{\prime} A E^{\prime} & =\angle B^{\prime} A D-\angle E^{\prime} A D=\frac{\angle A B C}{2}-\frac{\angle B A C}{2}=90^{\circ}-\angle B A C-\frac{\angle A B C}{2} \\
& =90^{\circ}-\angle B^{\prime} D A=\angle B^{\prime} D E^{\prime},
\end{aligned}
$$

points $B^{\prime}, A, D, E^{\prime}$ are concyclic. Then

$$
\angle D B^{\prime} E^{\prime}=\angle D A E^{\prime}=\frac{\angle B A C}{2}=\angle B A A^{\prime}=\angle D B^{\prime} A^{\prime}
$$

and hence $B^{\prime}, E^{\prime}, A^{\prime}$ are collinear. It is well-known that $A^{\prime} B^{\prime}$ is the perpendicular bisector of $C I$, so that $C E^{\prime}=I E^{\prime}$. Let $I^{\prime}$ be the reflection of $I$ in $A C$. This implies $B E=C E=I^{\prime} E$. As $D E$ is the external angle bisector of $\angle B D I^{\prime}$ and $E I^{\prime}=E B$, we know that $E$ lies on the circumcircle of triangle $B D I^{\prime}$.

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Let $F$ be the intersection of $C I$ and $A B$. Clearly, $F$ and $D$ are symmetric with respect to $A I$. Let $O$ be the circumcentre of triangle $B I F$, and let $I^{\prime}$ be the reflection of $I$ in $A C$.

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From $\angle B F O=90^{\circ}-\angle F I B=\frac{1}{2} \angle B A C=\angle B A I$, we get $E I / / F O$. Also, from the relation $\angle O I B=90^{\circ}-\angle B F I=90^{\circ}-\angle C D I=\angle I^{\prime} I D$, we know that $O, I, I^{\prime}$ are collinear.

Note that $E D / / O I$ since both are perpendicular to $A C$. Then $\angle F E I=\angle D E I=\angle O I E$. Together with $E I / / F O$, the quadrilateral $E F O I$ is an isosceles trapezoid. Therefore, we find that $\angle D I E=\angle F I E=\angle O E I$ so $O E / / I D$. Then $D E O I$ is a parallelogram. Hence, we have $D I^{\prime}=D I=E O$, which shows $D E O I^{\prime}$ is an isosceles trapezoid. In addition, $E D=O I=O B$ and $O E / / B D$ imply $E O B D$ is another isosceles trapezoid. In particular, both $D E O I^{\prime}$ and $E O B D$ are cyclic. This shows $B, D, E, I^{\prime}$ are concyclic.
Let $I^{\prime}$ be the reflection of $I$ in $A C$. Denote by $T$ and $M$ the projections from $I$ to sides $A B$ and $B C$ respectively. Since $B I$ is the perpendicular bisector of $T M$, we have

$$
D T=D M\tag{1}
$$

Since $\angle A D E=\angle A T I=90^{\circ}$ and $\angle D A E=\angle T A I$, we have $\triangle A D E \sim \triangle A T I$. This shows $\frac{A D}{A E}=\frac{A T}{A I}=\frac{A T}{A I^{\prime}}$. Together with $\angle D A T=2 \angle D A E=\angle E A I^{\prime}$, this yields $\triangle D A T \sim \triangle E A I^{\prime}$. In particular, we have

$$
\frac{D T}{E I^{\prime}}=\frac{A T}{A I^{\prime}}=\frac{A T}{A I}
\tag{2}
$$

Obviously, the right-angled triangles $A M B$ and $A T I$ are similar. Then we get

$$
\frac{A M}{A B}=\frac{A T}{A I}\tag{3}
$$

Next, from $\triangle A M B \sim \triangle A T I \sim \triangle A D E$, we have $\frac{A M}{A B}=\frac{A D}{A E}$ so that $\triangle A D M \sim \triangle A E B$. It follows that

$$
\frac{D M}{E B}=\frac{A M}{A B}\tag{4}
$$

Combining (1), (2), (3) and (4), we get $E B=E I^{\prime}$. As $D E$ is the external angle bisector of $\angle B D I^{\prime}$, we know that $E$ lies on the circumcircle of triangle $B D I^{\prime}$.

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