Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see the given figure). Show that:

(i) $\triangle \mathrm{ABM} \cong \triangle \mathrm{PQN}$

(ii) $\triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}$

Solution:

(i) In $\triangle \mathrm{ABC}, \mathrm{AM}$ is the median to $\mathrm{BC}$.

$\therefore B M=\frac{1}{2} B C$

In $\triangle \mathrm{PQR}, \mathrm{PN}$ is the median to $\mathrm{QR}$.

$\therefore \mathrm{QN}=\frac{1}{2} \mathrm{QR}$

However, $B C=Q R$

$\therefore \frac{1}{2} B C=\frac{1}{2} Q R$

\Rightarrow B M=Q N \ldots(1)

In $\triangle \mathrm{ABM}$ and $\triangle \mathrm{PQN}$,

$A B=P Q$ (Given)

$\mathrm{BM}=\mathrm{QN}$ [From equation (1)]

AM = PN (Given)

$\therefore \triangle \mathrm{ABM} \cong \triangle \mathrm{PQN}(\mathrm{SSS}$ congruence rule)

$\angle \mathrm{ABM}=\angle \mathrm{PQN}($ By CPCT $)$

$\angle \mathrm{ABC}=\angle \mathrm{PQR} \ldots(2)$

(ii) In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$,

$\mathrm{AB}=\mathrm{PQ}$ (Given)

$\angle \mathrm{ABC}=\angle \mathrm{PQR}$ [From equation (2)]

$\mathrm{BC}=\mathrm{QR}$ (Given)

$\Rightarrow \triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}$ (By $S A S$ congruence rule)
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