Question 106
Let $\triangle ABC$ be an acute and non-isosceles triangle. Its angle bisectors $AL_1$ and $BL_2$ intersect at the point $I$. Points $D$ and $E$ are chosen on the segments $AL_1$ and $BL_2$ in such a way that $\angle DBC = \frac{1}{2}\angle A$ and $\angle EAC = \frac{1}{2}\angle B$. The lines $AE$ and $BD$ intersect at a point $P$. Let $K$ be the point that is symmetric to the point $I$ with respect to the line $DE$. Prove that the lines $KP$ and $DE$ intersect at a point on the circumcircle of $\triangle ABC$.
(Danylo Khilko)
(Danylo Khilko)