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Macedonian Mathematical Competitions

Geometry Difficulty 5.5 AIME, harder Prove it North Macedonia

Four points $A$ , $B$ , $C$ and $D$ in the plane are given such that $\overline{AB} = \overline{AC}$ and $\overline{AD} = \overline{BD}$ . Let $E$ be a point from the plane $AC$ , such that $A$ lies between $E$ and $C$ (see picture).

If $\alpha = \angle BAE$ and $\beta = \angle ADB$ and $\alpha + \beta = 200^\circ$ , find $\varphi = \angle CBD$ .

Solution

Since $ABC$ is an isosceles triangle with base $BC$ , and $\alpha = \angle BAE$ , then $\angle ACB = \angle CBA = \frac{180^\circ - \angle BAC}{2} = \frac{\alpha}{2}$ .

Since $ABD$ is isosceles triangle with base $AB$ , then $\angle DBA = \angle BAD = \frac{180^\circ - \beta}{2}$ .

From there, we get
$\varphi = \angle CBD = \angle CBA - \angle DBA = \frac{\alpha}{2} - \frac{180^\circ - \beta}{2} = \frac{\alpha + \beta}{2} - 90^\circ = 10^\circ.$

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