Question

In the given figure, RS = QT and QS = RT. Then which of the following options is correct?

• A. $PQ=PR$
• B. $PR=TR$
• C. $PQ=QS$
• D. none of these

Answer 1

The correct option is A

$PQ=PR$

In $\Delta RTQ$ and $\Delta QSR$
QT = RS [given]
RT = SQ [given]
RQ = RQ [common]
Thus, $\Delta RTQ$ $\cong$ $\Delta QSR$ (by S.S.S. axiom)
$\begin{array}{cc}\angle 1=\angle 2& \left[cpct\right]\\ \angle 3=\angle 4& \left[verticallyoppositeangles\right]\\ \angle 5=\angle 6& \end{array}$
Now, in $\Delta PSQ$ and $\Delta PTR$
QS = RT[given]
$[\because \angle 1+\angle 7=\angle 2+\angle 8(supplementary\angle s\left)\right]$
$\angle 7=\angle 8$ ($\because \angle 1=\angle 2$)
$\angle 6=\angle 5$ (proved above)
$\therefore \Delta PSQ\cong \Delta PTR$ [by A.A.S. congruence rule]
$\Rightarrow$ PQ = PR [cpct]