P and Q are points nontheatrical sides CA and CB, respectively of triangle ABC, right angled at C. Prove that AQ^2+BP^2 = AB^2+ PQ^2
Given: Triangle ABC is right-angled at C, and P and Q are points on non-hypotenuse sides CA and CB, respectively.
To prove: AQ^2 + BP^2 = AB^2 + PQ^2
Proof:
1. By the Pythagorean theorem, we have:
AB^2 = AC^2 + BC^2
2. Similarly, by the Pythagorean theorem in triangles APC and BQC, we have:
AP^2 = AC^2 + PC^2,
BQ^2 = BC^2 + CQ^2
3. Adding the above two equations, we get:
AP^2 + BQ^2 = AC^2 + PC^2 + BC^2 + CQ^2
4. Since P and Q are points on sides CA and CB respectively, we can write:
PC = CQ = PQ
5. Substituting PC = PQ and CQ = PQ in equation 3, we get:
AP^2 + BQ^2 = AC^2 + BC^2 + 2(PQ)^2
6. Now, substitute AB^2 = AC^2 + BC^2 (from equation 1) in equation 5, we get:
AP^2 + BQ^2 = AB^2 + 2(PQ)^2
7. Rearranging the terms, we get:
AP^2 + BQ^2 = AB^2 + PQ^2
8. Therefore, AQ^2 + BP^2 = AB^2 + PQ^2
Hence, proved.