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Construction of a Similar Triangle, When One Vertex Is Common

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Question 1
In figure, if $∠ B A C = 90^{\circ} a n d A D ⊥ B C$ . Then,

(A) $B D . C D = B C^{2}$
(B) $A B . A C = B C^{2}$
(C) $B D . C D = A D^{2}$
(D) $A B . A C = A D^{2}$

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Solution

$I n Δ A D B a n d Δ A D C,$

$∠ A D B = ∠ A D C = 90^{\circ}$
$∠ D B A = ∠ D A C [e a c h e q u a l t o (90^{\circ} - ∠ C)]$
$∴$ $Δ A D B \sim Δ A D C [b y A A A s i m i l a r i t y c r i t e r i o n]$
$∴ \frac{B D}{A D} = \frac{A D}{C D}$
$\Rightarrow B D . C D = A D^{2}$

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Q. Question 1
In figure, if

∠ B A C = 90^{\circ} a n d A D ⊥ B C

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

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

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

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Construction of a Similar Triangle, When One Vertex Is Common

Standard X Mathematics

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