Subfield: Geometry


Question: Let $A B C$ be a triangle with incenter $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with the lines $A I$ and $B I$, respectively. The chord $D E$ meets $A C$ at a point $F$, and $B C$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $A D$ and the line through $G$ parallel to $B E$. Suppose that the tangents to $\omega$ at $A$ and at $B$ meet at a point $K$. Prove that the three lines $A E, B D$, and $K P$ are either parallel or concurrent.

Solution:
Since

$$
\angle I A F=\angle D A C=\angle B A D=\angle B E D=\angle I E F
$$

the quadrilateral $A I F E$ is cyclic. Denote its circumcircle by $\omega_{1}$. Similarly, the quadrilateral $B D G I$ is cyclic; denote its circumcircle by $\omega_{2}$.

The line $A E$ is the radical axis of $\omega$ and $\omega_{1}$, and the line $B D$ is the radical axis of $\omega$ and $\omega_{2}$. Let $t$ be the radical axis of $\omega_{1}$ and $\omega_{2}$. These three lines meet at the radical center of the three circles, or they are parallel to each other. We will show that $t$ is in fact the line $P K$.

Let $L$ be the second intersection point of $\omega_{1}$ and $\omega_{2}$, so $t=I L$. (If the two circles are tangent to each other then $L=I$ and $t$ is the common tangent.)

<img_3513>

Let the line $t$ meet the circumcircles of the triangles $A B L$ and $F G L$ at $K^{\prime} \neq L$ and $P^{\prime} \neq L$, respectively. Using oriented angles we have

$$
\angle\left(A B, B K^{\prime}\right)=\angle\left(A L, L K^{\prime}\right)=\angle(A L, L I)=\angle(A E, E I)=\angle(A E, E B)=\angle(A B, B K),
$$



so $B K^{\prime} \| B K$. Similarly we have $A K^{\prime} \| A K$, and therefore $K^{\prime}=K$. Next, we have

$$
\angle\left(P^{\prime} F, F G\right)=\angle\left(P^{\prime} L, L G\right)=\angle(I L, L G)=\angle(I D, D G)=\angle(A D, D E)=\angle(P F, F G),
$$

hence $P^{\prime} F \| P F$ and similarly $P^{\prime} G \| P G$. Therefore $P^{\prime}=P$. This means that $t$ passes through $K$ and $P$, which finishes the proof.
Let $M$ be the intersection point of the tangents to $\omega$ at $D$ and $E$, and let the lines $A E$ and $B D$ meet at $T$; if $A E$ and $B D$ are parallel, then let $T$ be their common ideal point. It is well-known that the points $K$ and $M$ lie on the line $T I$ (as a consequence of PASCAL's theorem, applied to the inscribed degenerate hexagons $A A D B B E$ and $A D D B E E$ ).

The lines $A D$ and $B E$ are the angle bisectors of the angles $\angle C A B$ and $\angle A B C$, respectively, so $D$ and $E$ are the midpoints of the arcs $B C$ and $C A$ of the circle $\omega$, respectively. Hence, $D M$ is parallel to $B C$ and $E M$ is parallel to $A C$.

Apply PASCAL's theorem to the degenerate hexagon $C A D D E B$. By the theorem, the points $C A \cap D E=F, A D \cap E B=I$ and the common ideal point of lines $D M$ and $B C$ are collinear, therefore $F I$ is parallel to $B C$ and $D M$. Analogously, the line $G I$ is parallel to $A C$ and $E M$.

<img_3833>

Now consider the homothety with scale factor $-\frac{F G}{E D}$ which takes $E$ to $G$ and $D$ to $F$. Since the triangles $E D M$ and $G F I$ have parallel sides, the homothety takes $M$ to $I$. Similarly, since the triangles $D E I$ and $F G P$ have parallel sides, the homothety takes $I$ to $P$. Hence, the points $M, I, P$ and the homothety center $H$ must lie on the same line. Therefore, the point $P$ also lies on the line $T K I M$.