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In the given figure, \angle A C B=90^{\circ} CD ⊥ AB Prove that \frac{B C^{2}}{A C^{2}}=\frac{B D}{A D}
CBSE | Class 10 | Exercise 4D | Maths | RS Aggarwal | Triangles
In the given figure, \angle A C B=90^{\circ} CD ⊥ AB Prove that \frac{B C^{2}}{A C^{2}}=\frac{B D}{A D}

 

 

 

Answer:

Given,

∠???????????? = 90o

???????? ⊥ ????B

In ∆ ACB and ∆ CDB

∠???????????? = ∠???????????? = 90o

∠???????????? = ∠????????????

By AA similarity-criterion,

∆ ACB ~ ∆CDB

When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.

\begin{aligned} \therefore \frac{B C}{B D} &=\frac{A B}{B C} \\ B C^{2} &=B D \cdot A B \end{aligned} ………(1)

In ∆ ACB and ∆ ADC

∠???????????? = ∠???????????? = 90o

∠???????????? = ∠????????????

By AA similarity-criterion,

∆ ACB ~ ∆ADC

When two triangles are similar, then the ratios of their corresponding sides are proportional.

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

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

Dividing (2) by (1),

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

i

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