A
                               / \
                              /   \
                             /     \
                            /       \
                           /         \
                          /           \
                         /             \
                        /               \
                       /                 \
                      /                   \
                     /                     \
                    /                       \
                   D---                   ---E
                  /    ---             ---    \
                 /        ---       ---        \
                /            --- ---            \
               /              -----              \
              /            ---     ---            \
             /          ---           ---          \
            /        ---                 ---        \
           /      ---                       ---      \
          /    ---                             ---    \
         /  ---                                   ---  \
        / --                                         -- \
       B-------------------------------------------------C

     This was presented to me as a hard problem in elementary, Euclidean geometry. It sounds easy, but I didn't find it so.

     My high-school math teacher Len Shakt was a math-problem-solving whiz. Every year he got all the problem on the high-school math-context exams in half the alloted time and he got them all right. He was on his college "integrating" team where they would get some problem like "the integral of tangent to the fourth" and they would race to get the answer. When I mentioned this little exercise he showed me his proof. Before I spill the beans, I figure I'll present the problem.

     Prove a triangle with equal angle bisectors is isosceles.

     In more mathematical lingo, let BE and CD be angle bisectors of   ⃤ ABC and BE=CD.

     Prove (∡ABC)=(∡ACB).

     (Don't worry, I'll post Mr. Shakt's proof later.)