Track / Stage 5 / 272 of 400 #872 of 1964

Problem 872

AIME late

Geometry Difficulty 5.7 Prove it

Example 2 In the circumcircle of $\triangle A B C$ , take points $K$ and $L$ on the arcs $\overparen{A B}$ (not containing point $C$ ) and $\overparen{B C}$ (not containing point $A$ ), respectively, such that the line $K L$ is parallel to the line $A C$ . Prove: The distances from the incenter of $\triangle A B K$ and the incenter of $\triangle C B L$ to the midpoint of $\overparen{A C}$ (containing point $B$ ) are equal.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

If $A B=B C$ , then the conclusion is obviously true.
Assume $A B<B C$ as shown in Figure 2. Let $I_{1}$ and $I_{2}$ represent the incenters of $\triangle A K B$ and $\triangle C L B$ , respectively. Connect $B I_{1}$ and $B I_{2}$ and extend them to intersect the circumcircle of $\triangle A B C$ at points $P$ and $Q$ , respectively. Let the midpoint of $\overparen{A B C}$ be $R$ .

Since $K L / / A C$ , we have $\overparen{A K}=\overparen{C L}$ . As $P$ and $Q$ are the midpoints of $\overparen{A K}$ and $\overparen{C L}$ , respectively, it follows that
$R P=R Q, P A=Q C .$

Thus, by property 3 of the incenter, we have
$P I_{1}=P A=Q C=Q I_{2} \text {. }$

Also, $\angle I_{1} P R=\angle I_{2} Q R$ , so
$\triangle I_{1} P R \cong \triangle I_{2} Q R \text {. }$

Therefore, $R I_{1}=R I_{2}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.