- ACB -90- and CD - AB- Prove that BC 2- AC 2- BD - AD

ACD ABC So, AC/AB = AD/AC or, AC^2 = AB.AD (1) Similarly, Δ BCD Δ B ...

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Relation between Inradius and Perimeter of Triangle

∠ ACB =90∘ an...

Question

$∠ A C B$ = $90^{\circ}$ and $C D ⊥ A B$ . Prove that $\frac{{B C}^{2}}{{A C}^{2}}$ = $\frac{B D}{A D}$

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∠ A C B = 90^{\circ}, C D ⊥ A B

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

In the given figure, $∠ A C B = 90^{o} a n d C D ⊥ A B$ .

Prove that $\frac{B C^{2}}{A C^{2}} = \frac{B D}{A D}$ .

A B C

C

C D

A B

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

∠ C = 90^{\circ}

C D ⊥ A B

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

∠ A C B = 90^{0}

C D ⊥ A B

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

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