Competition problem — Geometry — Maths Olympiad Prep

Library / /360 of 520

Geometry Difficulty 3.4 AMC 10/12 Prove it

Points $A$ , $B$ , $C$ , $D$ , $E$ lie on a circle $\omega$ and point $P$ lies outside the circle. The given points are such that (i) lines $PB$ and $PD$ are tangent to $\omega$ , (ii) $P$ , $A$ , $C$ are collinear, and (iii) $\overline{DE} \parallel \overline{AC}$ . Prove that $\overline{BE}$ bisects $\overline{AC}$ .

Solution

Since $\overline{DE} \parallel \overline{AC}$ , we have that $AD=CE\rightarrow \angle ABD=\angle CBE$ . But it is well known that $ABCD$ is a harmonic quadrilateral, thus $\overline{BD}$ is a symmedian of triangle $ABC$ , from which it follows that $\overline{BE}$ is a median of $\triangle ABC$ . $\blacksquare$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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