Problem 5

Let ABCD be a square. The midpoint O of the segment AD is the center of the circle with the diameter AD. If E ∈ AB and EC is tangent to the circle O, prove that ∠BEC < 60○.
figure fig_Problem_6.jpg
Solution. We have CT = CD = l, where l is the square side, ET = AE = l − BE, therefore EC = 2l − BE. Then
(2l − BE)2 = l2 + BE2.
It follows that BE = (3)/(4)l, therefore
tan∠BEC = (4)/(3) < √(3) = tan 60○.