MCQ
Choose the correct answer from the given four options : In $\angle\text{BAC}=90^\circ$ and $\text{ AD}\perp\text{BC}.$ Then, Traingles
  • A
    $ \mathrm{BD} \times \mathrm{CD}=\mathrm{BC}^2 $
  • B
    $ \mathrm{AB} \times \mathrm{AC}=\mathrm{BC}^2 $
  • $ \mathrm{BD} \times \mathrm{CD}=\mathrm{AD}^2 $
  • D
    $ \mathrm{AB} \times \mathrm{AC}=\mathrm{AD}^2 $

Answer

Correct option: C.
$ \mathrm{BD} \times \mathrm{CD}=\mathrm{AD}^2 $

$\angle\text{D}=\angle\text{D}=90^\circ$
$\angle\text{DBA}=\angle\text{DAC}\ \ \big[$each equal to $ 90^\circ-<\text{c}\big]$
$\therefore\triangle\text{ADB}\sim\triangle\text{ADC}\ \ \big[$by $\text{AAA}$ simillariy criterion $\big]$
$\therefore\frac{BD}{AD}=\frac{AD}{CD}$
$\Rightarrow\text{BD}\times\text{CD}=\text{AD}^2$

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