Question : In $\triangle {PQR}, {PN}$ is the median on ${QR}$. If ${PN}={QN}$, then what is the value of $\angle {QPR}$?
Option 1: $90^\circ$
Option 2: $80^\circ$
Option 3: $60^\circ$
Option 4: $75^\circ$
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Correct Answer: $90^\circ$
Solution :
In $\triangle PQR$, PN is the median.
So, QN = NR
and PN = QN (given)
In $\triangle PQN$
PN = QN
Let $\angle QPN=\angle NQP = \theta$
Similarly, In $\triangle PNR$
PN = NR
Let $\angle NPR = \angle NRP = \alpha$
$\therefore QPR=\theta + \alpha$
In $\triangle PQR$,
$\angle QPR+\angle QRP +\angle RQP=180^\circ$
⇒ $(\theta+\alpha)+\theta + \alpha=180^\circ$
⇒ $2(\theta+\alpha)=180^\circ$
⇒ $(\theta+\alpha)=90^\circ$
So, $\angle QPR = 90^\circ$
Hence, the correct answer is $90^\circ$.
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