Maths Olympiad Prep

Track / Stage 5 / 7 of 400 #607 of 1964

Problem 607

AIME late
Geometry Difficulty 5.0 Find the answer

1. Inside square ABCDA B C D, a point EE is chosen so that triangle DECD E C is equilateral. Find the measure of AEB\angle A E B.

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

Answer: 150150^{\circ}

Solution: Since DEC\triangle D E C is an equilateral triangle, then DE=CE|D E|=|C E| each angle has a measure of 6060^{\circ}. This implies that ADE\angle A D E has measure of 3030^{\circ}. Since AD=DE|A D|=|D E|, then ADE\triangle A D E is an isosceles triangle. Thus, DAE=DEA=75\angle D A E=\angle D E A=75^{\circ}. The same argument on triangle BECB E C will give us CBE=\angle C B E= CEB=75\angle C E B=75^{\circ}. Thus, AEB=150\angle A E B=150^{\circ}.

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