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# 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.