Subfield: Geometry


Question: Let $A B C$ be a triangle with $A B=A C$, and let $D$ be the midpoint of $A C$. The angle bisector of $\angle B A C$ intersects the circle through $D, B$, and $C$ in a point $E$ inside the triangle $A B C$. The line $B D$ intersects the circle through $A, E$, and $B$ in two points $B$ and $F$. The lines $A F$ and $B E$ meet at a point $I$, and the lines $C I$ and $B D$ meet at a point $K$. Show that $I$ is the incenter of triangle $K A B$.

Solution:
Let $D^{\prime}$ be the midpoint of the segment $A B$, and let $M$ be the midpoint of $B C$. By symmetry at line $A M$, the point $D^{\prime}$ has to lie on the circle $B C D$. Since the $\operatorname{arcs} D^{\prime} E$ and $E D$ of that circle are equal, we have $\angle A B I=\angle D^{\prime} B E=\angle E B D=I B K$, so $I$ lies on the angle bisector of $\angle A B K$. For this reason it suffices to prove in the sequel that the ray $A I$ bisects the angle $\angle B A K$.

From

$$
\angle D F A=180^{\circ}-\angle B F A=180^{\circ}-\angle B E A=\angle M E B=\frac{1}{2} \angle C E B=\frac{1}{2} \angle C D B
$$

we derive $\angle D F A=\angle D A F$ so the triangle $A F D$ is isosceles with $A D=D F$.

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Applying Menelaus's theorem to the triangle $A D F$ with respect to the line $C I K$, and applying the angle bisector theorem to the triangle $A B F$, we infer

$$
1=\frac{A C}{C D} \cdot \frac{D K}{K F} \cdot \frac{F I}{I A}=2 \cdot \frac{D K}{K F} \cdot \frac{B F}{A B}=2 \cdot \frac{D K}{K F} \cdot \frac{B F}{2 \cdot A D}=\frac{D K}{K F} \cdot \frac{B F}{A D}
$$

and therefore

$$
\frac{B D}{A D}=\frac{B F+F D}{A D}=\frac{B F}{A D}+1=\frac{K F}{D K}+1=\frac{D F}{D K}=\frac{A D}{D K}
$$



It follows that the triangles $A D K$ and $B D A$ are similar, hence $\angle D A K=\angle A B D$. Then

$$
\angle I A B=\angle A F D-\angle A B D=\angle D A F-\angle D A K=\angle K A I
$$

shows that the point $K$ is indeed lying on the angle bisector of $\angle B A K$.
It can be shown in the same way as in the first solution that $I$ lies on the angle bisector of $\angle A B K$. Here we restrict ourselves to proving that $K I$ bisects $\angle A K B$.

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Denote the circumcircle of triangle $B C D$ and its center by $\omega_{1}$ and by $O_{1}$, respectively. Since the quadrilateral $A B F E$ is cyclic, we have $\angle D F E=\angle B A E=\angle D A E$. By the same reason, we have $\angle E A F=\angle E B F=\angle A B E=\angle A F E$. Therefore $\angle D A F=\angle D F A$, and hence $D F=D A=D C$. So triangle $A F C$ is inscribed in a circle $\omega_{2}$ with center $D$.

Denote the circumcircle of triangle $A B D$ by $\omega_{3}$, and let its center be $O_{3}$. Since the arcs $B E$ and $E C$ of circle $\omega_{1}$ are equal, and the triangles $A D E$ and $F D E$ are congruent, we have $\angle A O_{1} B=2 \angle B D E=\angle B D A$, so $O_{1}$ lies on $\omega_{3}$. Hence $\angle O_{3} O_{1} D=\angle O_{3} D O_{1}$.

The line $B D$ is the radical axis of $\omega_{1}$ and $\omega_{3}$. Point $C$ belongs to the radical axis of $\omega_{1}$ and $\omega_{2}$, and $I$ also belongs to it since $A I \cdot I F=B I \cdot I E$. Hence $K=B D \cap C I$ is the radical center of $\omega_{1}$, $\omega_{2}$, and $\omega_{3}$, and $A K$ is the radical axis of $\omega_{2}$ and $\omega_{3}$. Now, the radical axes $A K, B K$ and $I K$ are perpendicular to the central lines $O_{3} D, O_{3} O_{1}$ and $O_{1} D$, respectively. By $\angle O_{3} O_{1} D=\angle O_{3} D O_{1}$, we get that $K I$ is the angle bisector of $\angle A K B$.
Again, let $M$ be the midpoint of $B C$. As in the previous solutions, we can deduce $\angle A B I=\angle I B K$. We show that the point $I$ lies on the angle bisector of $\angle K A B$.

Let $G$ be the intersection point of the circles $A F C$ and $B C D$, different from $C$. The lines



$C G, A F$, and $B E$ are the radical axes of the three circles $A G F C, C D B$, and $A B F E$, so $I=A F \cap B E$ is the radical center of the three circles and $C G$ also passes through $I$.

<img_3981>

The angle between line $D E$ and the tangent to the circle $B C D$ at $E$ is equal to $\angle E B D=$ $\angle E A F=\angle A B E=\angle A F E$. As the tangent at $E$ is perpendicular to $A M$, the line $D E$ is perpendicular to $A F$. The triangle $A F E$ is isosceles, so $D E$ is the perpendicular bisector of $A F$ and thus $A D=D F$. Hence, the point $D$ is the center of the circle $A F C$, and this circle passes through $M$ as well since $\angle A M C=90^{\circ}$.

Let $B^{\prime}$ be the reflection of $B$ in the point $D$, so $A B C B^{\prime}$ is a parallelogram. Since $D C=D G$ we have $\angle G C D=\angle D B C=\angle K B^{\prime} A$. Hence, the quadrilateral $A K C B^{\prime}$ is cyclic and thus $\angle C A K=\angle C B^{\prime} K=\angle A B D=2 \angle M A I$. Then

$$
\angle I A B=\angle M A B-\angle M A I=\frac{1}{2} \angle C A B-\frac{1}{2} \angle C A K=\frac{1}{2} \angle K A B
$$

and therefore $A I$ is the angle bisector of $\angle K A B$.