is a right triangle right - angled at A and AD . Then, BD DC = - Vidyadip

MCQ

$\triangle\text{ABC}$ is a right triangle right $-$ angled at $A$ and $\text{AD}\perp\text{BC}.$ Then, $\frac{\text{BD}}{\text{DC}}=$

✓
$\Big(\frac{\text{AB}}{\text{AC}}\Big)^2$
B
$\frac{\text{AB}}{\text{AC}}$
C
$\Big(\frac{\text{AB}}{\text{AD}}\Big)^2$
D
$\frac{\text{AB}}{\text{AD}}$

✓

Answer

Correct option: A.

$\Big(\frac{\text{AB}}{\text{AC}}\Big)^2$

In right angled $\triangle\text{ABC},\ \angle\text{A}=90^\circ$
$\text{AD}\perp\text{BC}$
$\therefore\triangle\text{ABD}\sim\triangle\text{ABC}$
$\therefore\frac{\text{AB}}{\text{BC}}=\frac{\text{BD}}{\text{AB}}$
$\Rightarrow\text{AB}^2=\text{BD}\times\text{BC}\ \ ...(\text{i})$
Similarly $\triangle\text{ACD}\sim\triangle\text{ABC}$
$\therefore\frac{\text{AC}}{\text{BC}}=\frac{\text{DC}}{\text{AC}}$
$\Rightarrow\text{AC}^2=\text{DC}\times\text{BC}\ \ ...(\text{ii})$
Dividing $(ii)$ by $(i)$
$\frac{\text{BD}\times\text{BC}}{\text{DC}\times{\text{BC}}}=\frac{\text{AB}^2}{\text{AC}^2}$
$\Rightarrow\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}^2}{\text{AC}^2}$
Hence $\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}^2}{\text{AC}^2}$

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