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# HELP! THANKS!

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In triangle ABC, AC = BC and angle C = 90. Points P and Q are on AB} such that QCP = 45. Show that AP^2 + BQ^2 = PQ^2.

Hint: Rotate the diagram counter-clockwise around C by 90 degrees. Let P go to point P'. What is angle P'CQ?

Feb 25, 2021

$AP=BQ\implies AP=PQ\sqrt{2}$ is desired. Follows trivially from LoC and coordinates as $\angle ACP=\angle BCP=45^{\circ}$. Can't think of anything synthetic...