Posted by **Knights** on Tuesday, January 15, 2013 at 6:01pm.

Let AB be the diameter of a circle, and let point P be a point on AB. Let CD be a chord parallel to AB. Prove that PA^2 + PB^2 = PC^2 + PD^2

It can be solved using geometry methods (no trig). Anyway, I figured out that PA^2 +PB^2 = 2OP^2 + 2OB^2. However, I cannot find right triangles to help to find PC^2+PD^2. Oh, and I got a hint: "The problem involves sums of squares of lengths. What should you be looking for (or building) in the problem?"

Thanks in advance

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