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Problem 8

In the square ABCD we choose the point E such that each of the angles EBC and ECB is 15°. Prove that ADE is equilateral.

**Solution 1**. Let *E*’ such that △*BCE*’ is equilateral. Then △*BEE*’ is isoscel (∠*EBE*’ = ∠*E*’*EB* = 75^{ o}). Also △*AEB* = △*E*’*EB* (*E*’*B* = *BC* = *BA*, *BE* = *BE*, ∠*ABE* = ∠*E*’*BE* = 75^{ o}. It follows that △*ABE* is isoscel, therefore *AE* = *AB* = *AD*. In the same way, *DE* = *DC* = *AD*.Then △*ABE* is equilateral.

**Solution **2. Suppose that the triangle △*ABE* is not equilateral. Let’s consider *E*^{′} such that the triangle △*ABE*^{′} is equilateral. Then ∠*CBE*^{′} = ∠*BCE*^{′} = 15^{ o}, therefore *E* = *E*^{′}. Then the triangle △*ABE* should be equilateral.