Ukrainian Mathematical Olympiad

Question

A convex pentagon ABCDE is inscribed into the circle . 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.

Accepted Answer

The problem will be solved if we prove that the area of triangle

A C E

is equal to the area of quadrilateral

A B C D

.

Let

K

be the intersection point of segments

A C

and

B E

. Since

S (A C E) = \frac{1}{2} A C \cdot K E

,

S (A B C D) = \frac{1}{2} A C \cdot B K + \frac{1}{2} A C \cdot C D

, it remains to show that

K E = B K + C D

.

Let point

H

on diagonal

B E

be symmetric to point

B

with respect to line

A C

. Then, as is known,

H

is the orthocenter of triangle

A C E

, and therefore

C H ⊥ A E

,

D E ⊥ A E

,

C H ∥ D E

,

C D ⊥ A C

,

E H ⊥ A C

,

E H ∥ C D

. Thus, quadrilateral

C D E H

is a parallelogram. Hence,

C D = E H

, and we have

K E = K H + H E = B K + C D

.