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  • In Fig., \angle BAC=90^{o} and AD\perp BC. Then, (a) BD.CD=BC^{2} (b) AB.AC=BC^{2} (c) BD.CD=AD^{2} (d) AB.AC=AD^{2}

In Fig., $\angle B A C=90^{\circ}$ and $A D \perp B C$. Then,

(a) $B D \cdot C D=B C^2$
(b) $A B \cdot A C=B C^2$
(c) $B D \cdot C D=A D^2$
(d) $A B \cdot A C=A D^2$

Answers (1)

Given:- $\angle B A C=90^{\circ}$
First of all, we have to find $\angle D B A$ and $\angle D A C$ in triangles $A B D$ and $A D C$ respectively
In $\triangle A B D$ and $\angle A D C$
$\angle A D B=90^{\circ}$
$\angle A D C=90^{\circ}$
In $\triangle A B C$
$\angle B A C=90^{\circ}$ (given)

Now let us find angle CAD from triangle ADC
Here $\angle A D C+\angle D C A+\angle C A D=180^{\circ}$
(sum of interior angles of triangle $=180^{\circ}$ )

$
\begin{aligned}
& 90^{\circ}+\angle C+\angle C A D=180^{\circ} \\
& \angle C A D=180^{\circ}-90^{\circ}-\angle C \\
& \angle C A D=90^{\circ}-\angle C
\end{aligned}
$

Now let us find angle DBA using triangle ABC

$
\begin{aligned}
& \text { Here } \angle A B C+\angle B C A+\angle C A B=180^{\circ} \\
& \text { \{sum of interior angles of triangle } \left.=180^{\circ}\right\} \\
& \angle A B C+\angle C+90^{\circ}=180^{\circ} \\
& \angle A B C=90^{\circ}-\angle C \\
& \angle A B C=\angle D B A=90^{\circ}-\angle C \\
& \text { In } \triangle A B D \text { and } \triangle A D C \text { we get } \\
& \angle A D B=\angle A D C\left\{90^{\circ}\right. \text { each\} } \\
& A D=A D\{\text { common side }\} \\
& \left.\angle C A D=\angle D B A \text { \{from equation (1) and (2) it is clear that both is } 90^{\circ}-\angle C\right\} \\
& \therefore \triangle A B D \sim \triangle A D C \text { (by ASA similarity) } \\
& B D \frac{B D}{A D}=\frac{A D}{C D} \\
& B D \cdot C D=A D \cdot A D\{\text { by cross multiplication }\} \\
& B D \cdot C D=A D^2
\end{aligned}
$
Hence option (C) is correct.

Posted by

infoexpert23

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