Mathematics
Hamilton Mathematical Olympiad
2023
4

4

In the parallelogram ABCD, a line through A meets BD at P, CD at Q and BC extended at R. Prove the PQ/PR = (PD/PB)2.

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Answer
SOLUTIOn

∠PAB = ∠PQD by alternate angles.

∠PBA = ∠PDQ by alternate angles.

Therefore, trianglesPAB and PQD are similar because they have the same angles.

So PA/PQ= PB/PD.

∠PAD = ∠PRB by alternate angles.

∠PDA = ∠PBR by alternate angles.

Therefore, triangles PAD and PRB are similar because they have the same angles.

So PA/PD= PR/PB.

These equations give us PA = PR × PD / PB = PB × PQ / PD.

This rearranges to give PQ/PR = (PD/PB)2

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