Let ${ABC}$ be a triangle inscribed in a circle. Point ${P}$ is the center of the arc ${BAC}$. The circle with the diameter ${CP}$ intersects the angle bisector of angle ${\angle BAC}$ at points ${K, L}$ ${(|AK| <|AL|)}$. Point ${M}$ is the reflection of ${L}$ with respect to line ${BC}$. Prove that the circumcircle of the triangle ${BKM}$ passes through the center of the segment ${BC}$ .
Problem
Source: Czech-Polish-Slovak Match 2013 day 2 P3
Tags: geometry, circumcircle, reflection, arc midpoint
timon92
14.05.2019 20:00
This problem was proposed by Burii.
Euler365
28.09.2019 13:20
Good problem Let $D$ be the midpoint of $BC$. Then $D$ lies on circle with diameter $CP$. Let $E$ be the midpoint of arc $BC$ not containing $A$. Then $E$ lies on angle bisector of $\angle BAC$ and also on $PD$. $\therefore ED.EP = EK.EL = EB^2 = EC^2$. $\therefore \angle EKB = \angle EBL = 180^{\circ}-\angle BLE-\angle BEL = 180^{\circ}-\angle BLE-C$ Also $\angle DKL = \angle CKE-\angle CKD = \angle LCE-\left(90^{\circ}-\frac{A}{2}\right) = 180^{\circ}-\angle LEC-\angle CLE-\left(90^{\circ}-\frac{A}{2}\right)$. But $\angle CEL = B$ and $\angle CLE = \angle CLD + \angle DLE =\left (90^{\circ}-\frac{A}{2}\right) + \angle DLE$. This gives us that $\angle DKL = C - \angle DLE$ $\therefore \angle BKD = \angle BKE + \angle DKL = 180^{\circ} - (\angle DLE + \angle BLE) = 180^{\circ}-\angle BLD = 180^{\circ}-\angle BMD$ $\therefore$ $BKMD$ is cyclic as desired. Q.E.D.
Tafi_ak
26.06.2022 12:05
It is sufficient to show $K,L$ are isogonal conjugate wrt $\triangle ABC$ but notice that $K,L$ both lie on $A$ angle bisector and $\triangle ACK\sim\triangle DLC$ where $D$ is the midpoint of $BC$ because $\angle CAK=\angle DLC=\angle C/2$ and $\angle AKD=\angle LDC$.