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.^@
Answer:
- 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. - We need to prove that ^@ \triangle ABC \cong \triangle PQR.^@
- 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}@^
- As corresponding parts of congruent triangles are equal, we have @^\angle B = \angle Q \space \space \space \ldots (2)@^
- 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.
