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^{○}.

(2*l* − *BE*)^{2} = *l*^{2} + *BE*^{2}.

It follows that
tan∠*BEC* = (4)/(3) < √(3) = tan 60^{○}.