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Ukrainian National Mathematical Olympiad

Geometry Difficulty 5.0 AIME, harder Prove it Ukraine

In a convex pentagon $ABCDE$ the following conditions are satisfied: $AB \parallel CD$ , $BC \parallel DE$ and $\angle BAE = \angle AED$ . Prove that $AB + BC = CD + DE$ .

Fig. 16

Solution

Let the rays $AB$ and $DE$ intersect at the point $O$ (fig. 16). Then $BCDO$ is a parallelogram, and $\Delta AOE$ is isosceles, since $\angle OAE = \angle AEO$ . So,
$AB + BC = (OB - OA) + BC = CD - OE + OD = CD + DE.$

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