MCQ
$D$ is a point on side $\text{BC}$ of a $\triangle ABC$ such that $\angle ADC=\angle BAC$, then $CA^2=$
  • A
    $B C \times A D$
  • B
    $B C^2$
  • C
    $A B^2$
  • $B C \times C D$

Answer

Correct option: D.
$B C \times C D$
In $\triangle A B C$ and $\triangle D A C$,
$\angle B A C=\angle A D C ($ Given$)$
$\angle B C A=\angle A C D($Common$)$
$\therefore \triangle A B C \sim \triangle D A C$
$($By $AA$ similarity criterion$)$
$\Rightarrow \frac{B C}{C A}=\frac{C A}{C D}$
$\Rightarrow C A^2=B C \times C D$
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