Track / Stage 5 / 296 of 400 #896 of 2000

Problem 896

AIME late

Geometry Difficulty 5.8 Prove it

There is a circle and a line passing through its center. From a randomly chosen point $P$ outside the circle, only using a straightedge, a perpendicular must be constructed to the line!

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

Let the two intersection points of the circle and the given line be $A$ and $B$ . The lines $P A$ and $P B$ intersect the circle at two new points: $C$ and $D$ , respectively. The extensions of segments $C B$ and $A D$ meet at a point $E$ , which, when connected to $P$ , gives the desired perpendicular. Indeed, according to Thales' theorem, $C B$ and $D B$ are two altitudes of triangle $A P E$ , and the third altitude is $A B$ , which is perpendicular to the corresponding side $P E$ .

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