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’ = ∠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.
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