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Pythagoras Theorem

In triangle A...

Question

In triangle ABC, CD perpendicular to AB,CA=2AD and BD=3AC. Prove that angle BCA $= 90^{\circ}$

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Solution

Denote AD = x, AC = 2x, BD = 3x.
As $∠ A D C = 90^{\circ}$ you can use pythagoras theorem to show $C D = x \sqrt{3}$ and $B C = 2 \sqrt{(3 x)}$ .
Now $(A C)^{2} + (B C)^{2}$
$= 4 x^{2} + 12 x^{2} = 16 x^{2} = (A B)^{2}$
Hence $∠ B C A = 90^{\circ}$

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