Question

In figure 5.15, In $\square$PQRS the points A, B, C and D are the mid points of side PQ, side QR, side RS and side SP respectively.

Answer 1

Construction:
Join QS.

$$\begin{gathered} \mathrm{Consider}∆\mathrm{PQS}. \\ \mathrm{A}\mathrm{and}\mathrm{D}\mathrm{are}\mathrm{the}\mathrm{midpoints}\mathrm{of}\mathrm{P}Q\mathrm{and}\mathrm{PS}\mathrm{respectively}. \\ \mathrm{By}\mathrm{midpoint}\mathrm{theorem}, \\ \mathrm{AD}=\frac{1}{2}\mathrm{QS}...\left(1\right) \\ \mathrm{and}\mathrm{AD}\parallel \mathrm{QS}...\left(2\right) \\ \mathrm{Now},\mathrm{consider}∆\mathrm{QRS}. \\ \mathrm{B}\mathrm{and}\mathrm{C}\mathrm{are}\mathrm{the}\mathrm{midpoints}\mathrm{of}\mathrm{RS}\mathrm{and}\mathrm{RQ}\mathrm{respectively}. \\ \mathrm{By}\mathrm{midpoint}\mathrm{theorem}, \\ \mathrm{BC}=\frac{1}{2}\mathrm{QS}...\left(3\right) \\ \mathrm{and}\mathrm{BC}\parallel \mathrm{QS}...\left(4\right) \\ \mathrm{Using}\mathrm{equation}s\left(1\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get}: \\ \mathrm{AD}=\mathrm{BC} \\ Now, \\ \mathrm{Using}\mathrm{equation}s\left(2\right)\mathrm{and}\left(4\right),\mathrm{we}\mathrm{get}: \\ \mathrm{AD}\parallel \mathrm{BC} \\ \mathrm{Since},\mathrm{the}\mathrm{opposite}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{quadrilateral}\mathrm{are}\mathrm{parallel}\mathrm{and}\mathrm{equal},\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{parallelogram}. \end{gathered}$$