Question
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
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