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