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# Two sides $ \mathrm{AB} $ and $ \mathrm{BC} $ and median $ \mathrm{AM} $ of one triangle $ \mathrm{ABC} $ are respectively equal to sides $ \mathrm{PQ} $ and $ \mathrm{QR} $ and median $ \mathrm{PN} $ of $ \triangle \mathrm{PQR} $ (see Fig. 7.40). Show that:

**(i)** $ \triangle \mathrm{ABM} \equiv \triangle \mathrm{PQN} $

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

"

**Given:**

Two sides $AB$ and $BC$ and median $AM$ of one triangle $ABC$ are respectively equal to sides $PQ$ and $QR$ and median $PN$ of $\triangle PQR$.

**To do: **

We have to show that:

(i) $\triangle ABM \cong \triangle PQN$

(ii) $\triangle ABC \cong \triangle PQR$.

**Solution:**

(i) Given,

$AM$ is the median of $\triangle ABC$ and $PN$ is the median of $\triangle PQR$

This implies,

$\frac{1}{2}BC=BM$ and $\frac{1}{2}QR=QN$

and also, $BC=QN$

This implies,

$\frac{1}{2}BC=\frac{1}{2}QR$

Therefore,

$BM=QN$

We know that,

According to the Rule of Side-Angle-Side Congruence:

Triangles are said to be congruent if any pair of corresponding sides and their included angles are equal in both triangles.

In $\triangle ABM$ and $PQN$,

We have, $AM=PN$ and $AB=PQ$

We also proved $BM=QN$

Therefore,

$\triangle ABM \cong \triangle PQN$.

(ii) We know that,

According to the Rule of Side-Angle-Side Congruence:

Triangles are said to be congruent if any pair of corresponding sides and their included angles are equal in both triangles.

In $\triangle ABC$ and $PQR$,

We have, $AB=PQ$ and $BC=QR$

By CPCT we know that,

The corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding angles and sides must be equal.

Therefore,

$\triangle ABC=\triangle PQR$.

Hence, $\triangle ABC \cong \triangle PQR$.

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