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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 4.6 AIME Prove it Soviet Union

Problem:

In the acute-angled triangle $ABC$ , the altitudes $BD$ and $CE$ are drawn. Let $F$ and $G$ be the points of the line $ED$ such that $BF$ and $CG$ are perpendicular to $ED$ . Prove that $EF = DG$ .

Solution

Solution:

$\angle BDC = \angle BEC = 90^{\circ}$ , so $BCDE$ is cyclic, so $\angle BDE = \angle BCE = 90^{\circ} - \angle B$ . Hence $\angle DGC = 90^{\circ} - \angle CDG = \angle BDE = 90^{\circ} - B$ . So $DG = CD \sin DGC = BC \sin CBD \sin DGC = BC \cos C \cos B$ . Similarly $EF$ .

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