Question 475918: O is any point in triangle PQR. Parallelograms QORX, ROQY and POQZ are drawn. Prove that triangle RQP ≅ triangle XYZ.
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! Here I will have to assume you typo-ed and meant ROPY instead of ROQY. This is mainly because of labeling techniques (one usually labels vertices of a polygon in order).
At first this looks like a very scary geometry problem, but the solution is quite straightforward. We take advantage of all the parallel lines here and say that
XR = OQ = PZ
QZ = OP = YR
PY = OR = QX
Furthermore,
XR || OQ || PZ
QZ || OP || YR
PY || OR || QX
simply because we established parallelograms. Now we can say that QRYZ, RPXZ, and PQXY are all parallelograms, because each quadrilateral has two parallel lines that are equal in length. Because they are parallelograms, we have QR = ZY, RP = ZX, and PQ = XY, so by SSS congruency, triangles RPQ and XZY are congruent (note that I said RPQ and XZY instead of RPQ and XYZ since we want to compare corresponding sides).
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