Let ABC be a triangle whose angles α=∠CAB and β=∠CBA are greater than 45∘. Above the side AB a right isosceles triangle ABR is constructed with AB as the hypotenuse, such that R is inside the triangle ABC. Analogously we construct above the sides BC and AC the right isosceles triangles CBP and ACQ, right at P and in Q, but with these outside the triangle ABC. Prove that CQRP is a parallelogram. geometryparallelogramisosceles right triangle