ABC is an isosceles triangle with vertex angle BAC = 20° and AB = AC. Point E is on AB such that AE = BC. Find the measure of AEC.

Solution

Copyright © 1996-2009 Alexander Bogomolny

The solution is by Vladimir Dubrovsky and, besides [Honsberger], it appeared in the Russian Kvant and subsequently in the now defunct Quantum magazine (May-June 1994).

Imagine a circle with center A and radius AB = AC. B is away from C by an arc subtending an angle of 20°. Step twice in this direction by the same angular measure getting points U and V on the circle. Naturally,

and

BAC = BAU = UAV = 20°

But this means that CAV = 60° and AC = AV making ΔCAV equilateral, so that CV = AC.

Inscribed CVU is subtended by an arc of 40° and is, therefore, equal 20°:

CVU = 20°.

In addition, UV = BU = BC. Consider now triangles AEC and VUC. They are equal by SAS: AE = UV, AC = CV, and CVU = 20° = CAE. But in ΔVUC, VCU = 10° (as an inscribed angle subtended by a 20° arc). It follows that ACE = 10° and AEC = 180° - 10° - 20° = 150°.

Reference

1. R. Honsberger, Mathematical Chestnuts from Around the World, MAA, 2001, Ch 2.

Copyright © 1996-2009 Alexander Bogomolny