Given, G is the mid-point of the side PQ of \(\triangle \)PQR,
\(\therefore\) PG = GQ or PQ = 2PG .....(1)
Now, since GH || QR.
Using Basic Proportionality theorem, we get,
\(\frac{PG}{GQ} = \frac{PH}{HR}\)
\(\frac{PG}{PQ} = \frac{PH}{PR}\)
⇒ \(\frac{PG}{2PG} = \frac{PH}{PR}\) [Using (1)]
⇒ \(\frac12= \frac{PH}{PR}\)
⇒ PR = 2PH
⇒ PH + HR = 2PH
⇒ HR = PH
Hence, H is the mid-point of PR.