Problem 19
Let ABCD be a square. P and Q are taken on CD and BC respectively such that ∠PAQ = ∠QAB. Show that:
AP = BQ + PD
Here’s a simulation that demonstrates the problem.
Try dragging the point P. Is AP always equal to BQ + PD? How can we prove the same?
Solution
Here’s a hint: Rotate the square about A by 90°. Let P and Q land at P’ and Q’ respectively. Also, let C and D land at C’ and D’ respectively. Then,
BQ + PD = BQ’ + PD = PQ’
Can we prove that PQ’ = AP? For that, can we prove that ∠APQ’ = ∠PAQ’? We haven’t used the angle bisector yet. How does that come into play here?
Well, that’s all from my side. I suggest you proceed on your own from here.
Were you able to solve it using the above hints? Or, did you solve using any other method? Or, do you need some more help? You can reach out at the Telegram group.
Problem Source: CutTheKnot
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