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

Answer

Correct option: C.
$B D \times C D=A D^2$

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

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