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Singapore Mathematical Olympiad (SMO)

Geometry Difficulty 5.4 AIME, harder Prove it Singapore

In the triangle $ABC$ with $AC > AB$ , $D$ is the foot of the perpendicular from $A$ onto $BC$ and $E$ is the foot of the perpendicular from $D$ onto $AC$ . Let $F$ be the point on the line $DE$ such that $EF \cdot DC = BD \cdot DE$ . Prove that $AF$ is perpendicular to $BF$ .

Solution

Since we are supposed to prove $\angle AFB = 90^\circ$ , it means that the 4 points $A$ , $B$ , $D$ , $F$ are concyclic. Note that $AC > AB$ implies that $\angle B > \angle C$ . If $TD$ is the tangent to the circumcircle of the triangle $ABD$ with $B$ and $T$ lying on opposite sides of the line $AD$ , then $\angle ADT = \angle B > \angle C = \angle ADE$ so that $\omega$ intersects the interior of $DE$ at $F$ . Therefore $F$ can only be in the interior of $DE$ .

Now observe that the triangles $ADE$ and $DCE$ are similar so that $AD/AE = DC/DE$ . By the given condition, this can be written as $AD/AE = BD/EF$ . This means the triangles $ABD$ and $AFE$ are similar. Thus $\angle ABD = \angle AFE$ . This shows that $A$ , $B$ , $D$ , $F$ are concyclic. Therefore $\angle AFB = \angle ADB = 90^\circ$ .

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