SOLUTION: In right triangle ABC, AC = BC and angle C = 90. Let P and Q be points on hypotenuse AB, as shown below, such that angle PCQ=45. Show that AP^2 + BQ^2 = PQ^2. I think the first ste

Algebra ->  Points-lines-and-rays -> SOLUTION: In right triangle ABC, AC = BC and angle C = 90. Let P and Q be points on hypotenuse AB, as shown below, such that angle PCQ=45. Show that AP^2 + BQ^2 = PQ^2. I think the first ste      Log On


   

Question 1153587: In right triangle ABC, AC = BC and angle C = 90. Let P and Q be points on hypotenuse AB, as shown below, such that angle PCQ=45. Show that AP^2 + BQ^2 = PQ^2. I think the first step is to rotate it 90 degrees counter clockwise around C, let P go to P’ and find angle P’CQ
Answer by foofy(2) About Me  (Show Source):
You can put this solution on YOUR website!
Perform the rotation. Then angle PCP' = 90 degrees, so PP' = CP*sqrt(2). Then PQ = PP'/2 and CQ = CP = CP', so angle B'PQ = angle BPQ = 45.
Since CA = BC, angle CBQ = 45 and angle CAP = 45. Therefore, angle ACP + angle BCQ = 45. But BQ = QP' and P'Q' = B'Q', so angle CQ'P' = angle C'Q'P', which means angle CQ'B = 90.
Finally, PQ = P'Q' = (AP + BQ)/2 = 1/2 and angle PCQ = 135, so angle CAP = 45 - angle Q'CB. So by the Pythagorean Theorem on right triangle ACQ, AP^2 + BC^2 = PC^2, which means AP^2 + BQ^2 = PQ^2.