Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it

9. (20 points) From point $D$ , the midpoint of the base $B C$ of isosceles triangle $A B C$ , a perpendicular $D E$ is dropped to the lateral side $A C$ . The circumcircle of triangle $A B D$ intersects line $B E$ at points $B$ and $F$ . Prove that line $A F$ passes through the midpoint of segment $D E$ .

Solution

Solution: In the right triangle $A C D: \angle A D E=90^{\circ}-\angle C D E=\angle A C D=\angle A B D$ . It follows that $D E$ is a tangent to the circumcircle of triangle $A B D$ . Then $G D^{2}=G F \cdot G A$ , where $G$ is the intersection point of lines $D E$ and $A F$ . $\angle A F B=\angle A D B=90^{\circ}$ - as inscribed angles subtending the same arc. In the right triangle $A E G$ , segment $E F$ is the altitude. $\frac{G F}{G E}=\cos \angle A G E=\frac{G E}{A G} \Rightarrow G E^{2}=G F \cdot G A=G D^{2}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.