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In an isosceles triangle ABC,AB=AC,P and Q are points on AC and AB respectively such that CB=BP=PQ=QA.Then prove that angle AQP=900/7

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Please show work, and avoid "questions" that are merely the statement of a problem. – The Chaz 2.0 Aug 10 '13 at 5:39

Hints: Make a drawing, then

$$\alpha:=\angle BCP=\angle ABC\;,\;\;\angle PBC=180^\circ-2\alpha\implies \angle QBP=3\alpha-180^\circ=\angle PQB\implies$$

$$\implies \angle AQP=360^\circ-3\alpha$$

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