In $Δ$ ABC, B is a right angle, and BD is perpendicular to AC and DE is perpendicular to BC. Then AD.DC =

B D^{2}

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B E . B C

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A B . B C

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B D . D C

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Solution

The correct options are
A $B D^{2}$
B $B E . B C$
$I n Δ A B D a n d Δ B D C ∠ A D B = ∠ B D C = 90^{\circ} ∠ A = ∠ D B C (S i n c e i n Δ B D C, ∠ D B C = 90^{\circ} - ∠ C = ∠ A a n d I n Δ A B C ∠ A = 90^{\circ} - ∠ C) T h e r e f o r e, Δ A D B \sim Δ B D C b y A A s i m i l a r i t y \frac{A D}{D B} = \frac{B D}{D C} = \frac{A B}{B C} \Rightarrow B D^{2} = A D . D C S i m i l a r l y Δ B D C \sim Δ B E D ∠ D B C = ∠ D B E (C o m m o n a n g l e) ∠ B D C = ∠ D E B = 90^{\circ} \frac{B D}{B E} = \frac{D C}{E D} = \frac{B C}{B D} \Rightarrow B D^{2} = B E . B C \Rightarrow B D^{2} = A D . D C = B E . B C$

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