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Help with triangles.

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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