Asked by Knights
                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
Answers
                                                    There are no human answers yet.
                                            
                
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.