ABC is an isosceles triangle with vertex angle
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Copyright © 1996-2018 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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