Competition problem — Geometry — Maths Olympiad Prep

Library / /1 of 18

Geometry Difficulty 4.3 AIME Prove it Austria

Let $ABCD$ be a rhombus with $\angle BAD < 90^\circ$ . The circle passing through $D$ with center $A$ intersects the line $CD$ a second time in point $E$ . Let $S$ be the intersection of the lines $BE$ and $AC$ .
Prove that the points $A, S, D$ and $E$ lie on a circle.

Solution

By the inscribed angle theorem, it is enough to show that $\angle SED = \angle SAD$ .
Since $ABCD$ is a rhombus, we have
$\angle SAD = \frac{1}{2} \angle BAD.$
Since $ABCE$ is an isosceles trapezoid, we have by symmetry that
$\angle SED = \angle ECS = \frac{1}{2} \angle DCB = \frac{1}{2} \angle BAD,$
which finishes the proof.

(Theresia Eisenkölbl) □

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: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.