Problem 20: Segments of a Square’s Diagonal

Problem 20

Let ABCD be a square. P and Q are taken on CD and BC respectively such that ∠PAQ = 45°. The diagonal BD intersects AP at E and AQ at F. Show that:

EF2 = DE2 + BF2

Here’s a simulation that demonstrates the problem.

Try dragging the point P. Is EF2 always equal to DE2 + BF2? How can we prove the same?

Solution

To solve this problem, I’ll give you just one hint. Try folding the square along AP and AQ. Let D and B land at D’ and B’ respectively after folding.

Here are some questions for you.

  1. Would D’ and B’ coincide after folding? Why or why not?
  2. Would AD’ and AB’ coincide after folding? Why or why not?
  3. What can you say about the measures of ∠ED’F and ∠FB’E?

Okay, another hint. We need to prove that EF2 = DE2 + BF2. Generally, in such cases where squares of lengths are involved, Pythagoras’ theorem comes into play. A right triangle might be lurking around. Can you see it?

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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