Above the side AB we construct a right-angled isosceles triangle ABR with AB as hypotenuse, such that R lies

*inside*the triangle ABC.

Analogously we erect above the sides BC and AC right-angled isosceles triangles CBP and ACQ, but with their (right-angled) vertices P and Q

*outside*of the triangle ABC.

Show that CQRP is a parallelogram.

[OEMO Beginners 2001 Q4]

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