Track / Stage 5 / 244 of 400 #844 of 1964

Problem 844

AIME late

Geometry Difficulty 5.6 Prove it

2.88*. Points $A, B, C$ and $D$ lie on a circle with center $O$ . Lines $A B$ and $C D$ intersect at point $E$ , and the circumcircles of triangles $A E C$ and $B E D$ intersect at points $E$ and $P$ . Prove that:

a) points $A, D, P$ and $O$ lie on the same circle;

b) $\angle E P O=90^{\circ}$

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

2.88. a) Since $\angle(A P, P D)=\angle(A P, P E)+\angle(P E, P D)=\angle(A C, C D)+$ $+\angle(A B, B D)=\angle(A O, O D)$ , points $A, P, D$ and $O$ lie on the same circle.

6) It is clear that $\angle(E P, P O)=\angle(E P, P A)+\angle(P A, P O)=\angle(D C, C A)+$ $+\angle(D A, D O)=90^{\circ}$ , since the arcs on which these angles subtend make up half of the circle.

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