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In triangle \( ABC, AC = BC\) and \( \angle ACB = 90^\circ.\)Points P and Q are on \(\overline AB\) such that P is between A and Q and \(\angle QCP = 45^\circ\). Show that \(AP^2 + BQ^2 = PQ^2\)

Hint(s): Rotate the diagram counter-clockwise around C by \(90^\circ\). Let P go to point P'. What is \(\angle P'CQ\)? Show your work.

 Jun 10, 2021
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Hint:          If   AP = BQ     then    PQ = 2 * [tan(22.5º) * AB/2]

 Jun 10, 2021

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