In the given figure- - ACB - 90o-and-CD - AB- Prove that BC2-AC2 - BD-AD-

On ABC and ACD ∠ ACB = ∠ CDA, ∠ CDA = ∠ CAB ΔABC and ΔACD are similar AC/AB= AD/AC AC^2= AB × AD Similarly, ΔBCD and ΔBAC are similar. BC/BA = BD/BC BC^2 ...

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Standard X

Mathematics

Relation between Areas and Sides of Similar Triangles

In the given ...

Question

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

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

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Solution

On $△ A B C$ and $△ A C D$

$∠ A C B = ∠ C D A,$
$∠ C D A = ∠ C A B$

ΔABC and ΔACD are similar

$\frac{A C}{A B} = \frac{A D}{A C}$

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

Similarly, ΔBCD and ΔBAC are similar.

$\frac{B C}{B A} = \frac{B D}{B C}$

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

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

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

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In the given figure, ∠ACB=90o and CD⊥AB. Prove that BC2AC2=BDAD.

On △ABC and △ACD ∠ACB=∠CDA, ∠CDA=∠CAB ΔABC and ΔACD are similar ACAB=ADAC AC2=AB×AD Similarly, ΔBCD and ΔBAC are similar. BCBA=BDBC BC2=BA×BD BC2AC2=AB×BDAB×AD ∴BC2AC2=BDAD

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

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