Question

In $\Delta ABC$, with points P and Q on sides AB and AC, respectively such that $\frac{AP}{PB}=\frac{AQ}{QC}$. If PQ is extended to T such that PT = BC and PB = TC, then x is equal to

• A. ${60}^{\circ }$
• B. ${120}^{\circ }$
• C. ${75}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is A

${60}^{\circ }$

In $\Delta ABC$,
$\frac{AP}{PB}=\frac{AQ}{QC}$
and PT = BC
Since, $\frac{AP}{PB}=\frac{AQ}{QC}\Rightarrow PQ||BC\Rightarrow PT||BC$
Also, PT = BC
$\therefore$ PTCB is a parallelogram,
$\Rightarrow \angle 1=\angle 2$ ...(i)
$\angle 3=\angle 2$ ...(ii) [Interior opposite angles]
In $\Delta APQ$ and $\Delta CTQ$,
$\angle 2=\angle 3$ [from Eq. (ii)]
$\angle 4=\angle 5$ [vertically opposite angles]
$\therefore \Delta APQ\sim \Delta CTQ$ [by AA similarity]
$\Rightarrow \angle x=\angle PAQ\Rightarrow x={60}^{\circ }$