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(i) \(\triangle{ABM}\) \(\displaystyle \cong \) \(\triangle{PQN}\)

(ii) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{PQR}\)

Answer :

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.

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