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Berkeley Math Circle Monthly Contest 6

Geometry Difficulty 4.7 AIME Prove it United States

Problem:

Let $ABC$ be a triangle with $\angle ABC = 90^{\circ}$ . Points $D$ and $E$ on $AC$ and $BC$ respectively satisfy $BD \perp AC$ and $DE \perp BC$ . The circumcircle of $\triangle CDE$ intersects $AE$ at two points, $E$ and $F$ . Prove that $BF \perp AE$ .

Solution

Solution:

By Power of a Point,
$AF \cdot AE = AD \cdot AC$
because $\triangle ABC$ is right,
$AD \cdot AC = AB^2$
Combining,
$AF \cdot AE = AB^2.$
On the other hand, if $F'$ is the foot of the altitude from $B$ to $AE$ , then
$AF' \cdot AE = AB^2$
Consequently $AF' = AF$ and $F' = F$ .

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