Question

Let $ABC$ be a triangle with incenter $I$ and incircle $\omega.$ Line $AI$ intersects $\omega$ at $J$ and $K,$ with $K$ lying on segment $AJ.$ Let $\omega$ be tangent to $BC$ at $D,$ and let $E$ be the antipode of $D$ on $\omega.$ Let $L$ be the foot of the altitude from $A$ to $BC.$ Let $P$ be a point such that $LDEP$ is a rectangle, let $Q$ be a point on $\omega$ with $AB \perp KQ,$ and let $R$ be a point with $\angle CAR=90^\circ$ and $AI=RI.$ Let $X$ be a point on $AL$ such that $\angle LXK+\angle BAQ=90^\circ,$ and let $Y$ be a point on the circumcircle of $APJ$ with $AI=YI.$ Prove that the circumcircles of $ALR$ and $AXY$ intersect again at a point lying on line $AI.$

Answer 1

Interesting development.

Answer 2

i posted this in the wrong subjeck im too lazey to find my open question s and close them

Answer 3

no way you have original poster. i love this problem. it is so difficult. i will try my best.

Answer 4

im late but ill tey to help