This is a JMO Mentoring scheme question, and answer:
Q: In a quadrilateral ABCD, side AB = side BC = diagonal CA = 2 and side AD = 1. D is located so that a circle using diagonal BD as a diameter will pass through A. Prove that BD = sq-root of 5.
A: Angle BAD = 90 degrees because BD is the diameter of the circle through A. Using Pythagoras's rule on triangle BAD we can work out that BD squared= 5.
I want to explain this to my son but the bit I'm struggling with is to find a simple explanation for why Angle BAD = 90 degrees. It's stated as fact in the answer, and with a bit of fiddling around we (or rather DH) did manage to prove it, but is there a named theorem for it that we should have just been able to quote?