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.
figure fig_Problem_8_1..jpg
Solution 1. Let E such that BCE is equilateral. Then BEE is isoscel (EBE’ = ∠EEB = 75 o). Also AEB = △EEB (EB = BC = BA, BE = BE, ABE = ∠EBE = 75 o. It follows that ABE is isoscel, therefore AE = AB = AD. In the same way, DE = DC = AD.Then ABE is equilateral.
figure Problem 8..jpg
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.
figure fig_Problem_8_3..jpg