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Converse of Pythagoras' Theorem

In a triangle...

Question

In a triangle, $A B C$ , angle $A = 90^{o}, C A = A B$ and $D$ is a point on $A B$ produced. Prove that ${D C}^{2} - {B D}^{2} = 2 A B . A D$ .

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Solution

Given $△ A B C, ∠ A = 90^{\circ}, C A = A B$
In $△ A C D, C D^{2} = A C^{2} + A D^{2} = A B^{2} + A D^{2}$
$C D^{2} - B D^{2} = A B^{2} + A D^{2} - (A B - B D)^{2} = A B^{2} + A D^{2} - (A B^{2} + A D^{2} - 2 A B A D) = 2 A B . A D$

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△ A B C

∠ A = 90^{0}, C A = A B

D

A B

D C^{2} - B D^{2} = 2 A B . A D

In a triangle ABC, angle A=90°, CA=AB and D is a point on AB produced. Prove that

DC² - BD²= 2AB × AD

△ A B C

∠ A = 90^{o},

{D C}^{2} = {B D}^{2} = 2 A B \times A D

∠

90^{\circ},

D C^{2} - B D^{2} =

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△ A B C

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A B = B R, B C = C P, C A = A F

A (△ P E R) = 7 A (△ A B C)

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