If in Fig, B A C=90^ and A D B C, then - Vidyadip

MCQ

If in Fig, $\angle B A C=90^{\circ}$ and $A D \perp B C$, then

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

✓

Answer

(C)$B D \times C D=A D^2$

In triangles $A B D$ and $A B C$, we find that
$
\angle A D B=\angle B A C\qquad [Each equal to 90^{\circ} ]
$
and, $\ \qquad angle A B D=\angle A B C \qquad$ [Common]
So, by using AA-criterion of similarity, we obtain
$
\begin{array}{l}
\quad \triangle B D A \sim \triangle B A C \qquad\ldots(i)\\
\text { Similarly, } \triangle A D C \sim \triangle B A C\qquad\ldots(ii)
\end{array}
$
From (i) and (ii), we obtain
$
\triangle B D A \sim \triangle A D C \Rightarrow \frac{B D}{A D}=\frac{D A}{D C}=\frac{B A}{A C} \Rightarrow B D \times D C=A D^2
$
ALIEE In a right triangle perpendicular drawn from the vertex of right angle to the hypotenuse, divides the triangle into two similar triangles.
$
\therefore \quad \triangle A D B \sim \triangle C D A \Rightarrow \frac{A D}{C D}=\frac{D B}{D A} \Rightarrow A D^2=B D \times C D .
$

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