ABC is an isosceles triangle with vertex angle

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Copyright © 1996-2017 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,

AC = AB = AU = AV

and

∠BAC = ∠BAU = ∠UAV = 20°

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

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:

### Reference

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

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