Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.7 AIME, harder Prove it Ukraine

A convex pentagon $ABCDE$ is inscribed into the circle $\omega$ . The diagonal $AD$ is a diameter of that circle. The diagonals $BE$ and $AC$ are perpendicular to each other. The diagonals $CE$ and $AD$ meet at point $P$ . Prove that the area of the triangle $APE$ is equal to the sum of the areas of the triangles $ABC$ and $CDP$ .

Solution

The problem will be solved if we prove that the area of triangle $ACE$ is equal to the area of quadrilateral $ABCD$ .

Let $K$ be the intersection point of segments $AC$ and $BE$ . Since $S(ACE) = \frac{1}{2} AC \cdot KE$ , $S(ABCD) = \frac{1}{2} AC \cdot BK + \frac{1}{2} AC \cdot CD$ , it remains to show that $KE = BK + CD$ .

Let point $H$ on diagonal $BE$ be symmetric to point $B$ with respect to line $AC$ . Then, as is known, $H$ is the orthocenter of triangle $ACE$ , and therefore $CH \perp AE$ , $DE \perp AE$ , $CH \parallel DE$ , $CD \perp AC$ , $EH \perp AC$ , $EH \parallel CD$ . Thus, quadrilateral $CDEH$ is a parallelogram. Hence, $CD = EH$ , and we have $KE = KH + HE = BK + CD$ .

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