Question

A line from one vertex $A$ of an equilateral $△ABC$ meets the opposite side $BC$ in $P$ and the circumcircle of $△ABC$ in $Q$. If $BQ=4cm$ and $CQ=3cm$, then $PQ$ is equal to

• A. $7cm$
• B. $\frac{4}{3}cm$
• C. $\frac{12}{7}cm$
• D. $2cm$

Answer 1

Answer: (C)

$\frac{12}{7}cm$

$\angle BQA=\angle BCA={60}^{\circ }$
$\angle CQA=\angle CBA={60}^{\circ }$
$\therefore QP$ is angle bisector of $BC$
$\therefore BP:PC=4:3$ (By angle bisector theorem)
Now $△BPQ\sim △APC$
$\frac{PQ}{BQ}=\frac{PC}{AC}$

$\Rightarrow \frac{PQ}{4}=\frac{\frac{3}{7}BC}{BC}$

$\Rightarrow PQ=\frac{12}{7}$.