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Geometry Difficulty 5.4 AIME, harder Prove it United States

Problem:

In convex quadrilateral $A B C D$ we have $A B=15$ , $B C=16$ , $C D=12$ , $D A=25$ , and $B D=20$ . Let $M$ and $\gamma$ denote the circumcenter and circumcircle of $\triangle A B D$ . Line $C B$ meets $\gamma$ again at $F$ , line $A F$ meets $M C$ at $G$ , and line $G D$ meets $\gamma$ again at $E$ . Determine the area of pentagon $A B C D E$ .

Solution

Solution:

Note that $\angle A D B = \angle D C B = 90^{\circ}$ and $B C \parallel A D$ . Now by Pascal's theorem on $D D E B F A$ implies that $B, M, E$ are collinear. So $[A D E] = [A B D] = 150$ and $[B C D] = 96$ , so the total area is $396$ .

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