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China Western Mathematical Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it China

The circles $O_1$ and $O_2$ meet at points $A$ and $B$ . The line $DC$ passes through $O_1$ , intersects the circle $O_1$ at $D$ and is a tangent to the circle $O_2$ at $C$ . Also, $CA$ is a tangent to the circle $O_1$ at $A$ . The secant $AE$ of the circle $O_1$ is perpendicular to $DC$ . $AF$ is perpendicular to $DE$ and meets $DE$ at $F$ .

Prove that $BD$ bisects the line segment $AF$ . (posed by Bian Hongping)

Solution

Let $AE$ intersect $DC$ at point $H$ ,

and $AF$ intersect $BD$ at point $G$ . Join $AB$ , $BC$ , $BH$ , $BE$ , $CE$ and $GH$ . By symmetry, $CE$ is also a tangent line of the circle $O_1$ and $H$ is the midpoint of $AE$ .

Since $AF \perp DE$ , we have
$\angle AGB = \frac{\pi}{2} - \angle BDE.$

By $AE \perp DC$ ,
$\angle AHB = \frac{\pi}{2} - \angle BHC.$

By ①, ② and ③, $\angle AGB = \angle AHB$ . Therefore $A$ , $G$ , $H$ , $B$ lie on the same circle, and $\angle AHG = \angle ABG = \angle AED$ . Thus $GH \parallel DE$ . Since $H$ is the midpoint of $AE$ , $G$ is the midpoint of $AF$ .

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