Track / Stage 5 / 292 of 400 #892 of 2000

Problem 892

AIME late

Geometry Difficulty 5.7 Prove it

184. Line $l$ is perpendicular to segment $A B$ and passes through $B$ . A circle with center on $l$ passes through $A$ and intersects $l$ at points $C$ and $D$ . Tangents to the circle at points $A$ and $C$ intersect at $N$ . Prove that line $D N$ bisects segment $A B$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

184. Let $M$ be the intersection point of $N D$ and $A B$ , and $P$ be the intersection point of the tangents to the circle at points $A$ and $D$ .

Since the lines $N C, A B$ and $P D$ are parallel, from the similarity of the corresponding triangles we get:

$\begin{gathered} |A M|=|D P| \cdot \frac{|A N|}{|N P|} \\ \frac{|M B|}{|N C|}=\frac{|M D|}{|N D|}=\frac{|A P|}{|N P|},|M B|=|N C| \frac{|A P|}{|N P|} \end{gathered}$

but $|D P|=|A P|,|N C|=|A N|$ . Therefore, the right-hand sides of expressions (1) and (2) are equal, i.e., $|A M|=|M B|$ .

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