### If two sides $AB$ and $BC,$ and the median $AD$ of $\triangle ABC$ are correspondingly equal to the two sides $PQ$ and $QR,$ and the median $PM$ of $\triangle PQR.$ Prove that $\triangle ABC \cong \triangle PQR.$

Step by Step Explanation:
1. We are given that $AB = PQ, \space BC = QR, \text{ and } AD = PM.$

Let us now draw the triangles and mark the equal sides and medians.
2. We need to prove that $\triangle ABC \cong \triangle PQR.$
3. It is given that \begin{aligned} &BC = QR \\ \implies &\dfrac{1}{2} BC = \dfrac{1}{2} QR \\ \implies & BD = QM && \ldots (1) && \text { [As the median from a vertex of a triangle bisects the opposite side.] } \end{aligned} Now, in $\triangle ABD$ and $\triangle PQM,$ we have \begin{aligned} & AD = PM && [\text{Given}] \\ & AB = PQ && [\text{Given}] \\ & BD = QM && [\text{From (1)}] \\ \therefore \space &\triangle ABD \cong \triangle PQM && [\text{By SSS criterion}] \end{aligned}
4. As corresponding parts of congruent triangles are equal, we have $$\angle B = \angle Q \space \space \space \ldots (2)$$
5. In $\triangle ABC$ and $\triangle PQR,$ we have \begin{aligned} & BC = QR && [\text{Given}] \\ & AB = PQ && [\text{Given}] \\ & \angle B = \angle Q && [\text{From (2)}] \\ \therefore \space & \triangle ABC \cong \triangle PQR && [\text{By SAS criterion}] \end{aligned}
Hence Proved.

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