The two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \(\triangle{PQR}\) (see figure). Show that
(i) \(\triangle{ABM}\) \(\displaystyle \cong \) \(\triangle{PQN}\)
(ii) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{PQR}\)

We have, AM is the median of \(\triangle{ABC}\),
So, BM = MC = (\(\frac{1}{2} \) ) BC ...(i)(Since, M Bisects BC)

Similarly, PN is the median of \(\triangle{PQR}\).
So, QN = NR = \(\frac{1}{2} \) QR ...(ii)(Since, N Bisects QR)

Now, we have,
BC = QR ...(Given)
Multiplying both sides by \(\frac{1}{2} \), we get,
\(\frac{1}{2} \) BC = \(\frac{1}{2} \) QR
\(\therefore \) BM = QN ...(iii)(From eq.(i) and (ii))

In \(\triangle{ABM}\) and \(\triangle{PQN}\), we have,
AB = PQ ...(Given)
AM = PN ...(Given)
and BM = QN ...(From Eq.(iii))
\(\triangle{ABM}\) \(\displaystyle \cong \) \(\triangle{PQN}\) ...(By SSS congruency test)
Hence, part (i) is proved.

In \(\triangle{ABC}\) and \(\triangle{PQR}\), we have,
AB = PQ ...(Given)
\(\angle{B}\) = \(\angle{Q}\) ...(Since, \(\triangle{ABM}\) \(\displaystyle \cong \) \(\triangle{PQN}\))
and BC = QR ...(Given)
\(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{PQR}\) ...(By SAS congruency test)
Hence, proved.