Question

Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.

Answer 1

Suppose $\square$PQRS is a square and A, B, C and D are the midpoints of PQ, QR, RS and SP, respectively.
Construction:
Join PR and QS.

$$\begin{gathered} \mathrm{Considering}∆\mathrm{SPQ} \\ \mathrm{A}\mathrm{and}\mathrm{D}\mathrm{are}\mathrm{the}\mathrm{midpoints}\mathrm{of}\mathrm{PQ}\mathrm{and}\mathrm{PS}\mathrm{respectively}. \\ \mathrm{By}\mathrm{the}\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{RQ}\mathrm{and}\mathrm{RS}\mathrm{respectively}. \\ \mathrm{By}\mathrm{the}\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{equations}\left(1\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get}: \\ \mathrm{AD}=\mathrm{BC} \\ \mathrm{Using}\mathrm{equations}\left(2\right)\mathrm{and}\left(4\right),\mathrm{we}\mathrm{get}: \\ \mathrm{AD}\parallel \mathrm{BC} \\ \mathrm{Since}\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}. \\ \mathrm{As}\square \mathrm{PQRS}\mathrm{is}\mathrm{a}\mathrm{rhombus}, \\ \mathrm{SP}=\mathrm{RQ} \\ \frac{1}{2}\mathrm{SP}=\frac{1}{2}\mathrm{RQ}\left(\because \mathrm{D}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{mid}\mathrm{points}\mathrm{of}\mathrm{SP}\mathrm{and}\mathrm{RQ}\mathrm{respectively}\right) \\ \therefore \mathrm{DP}=\mathrm{BQ} \\ \mathrm{In}△\mathrm{DPA}\&△\mathrm{BQA} \\ \mathrm{DP}=\mathrm{BQ}\left(\mathrm{Proved}\mathrm{above}\right) \\ \angle \mathrm{DPA}=\angle \mathrm{BQA}\left(90°\mathrm{each}\right) \\ \mathrm{PA}=\mathrm{QA}\left(\mathrm{A}\mathrm{is}\mathrm{mid}\mathrm{point}\mathrm{of}\mathrm{PQ}\right) \\ \therefore △\mathrm{DPA}\cong △\mathrm{BQA}\left(\mathrm{by}\mathrm{SAS}\mathrm{congruency}\mathrm{criteria}\right) \\ \mathrm{AD}=\mathrm{AB}\left(\mathrm{by}\mathrm{c}.\mathrm{s}.\mathrm{c}.\mathrm{t}\right) \\ \mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{parallelogram}\mathrm{with}\mathrm{AD}=\mathrm{AB} \\ \mathrm{Hence},\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{rhombus}. \end{gathered}$$

Also, since A and C are the midpoints of parallel and equal sides,
AC = QR = PS.
Again, since B and D are the midpoints of parallel and equal sides,
BD = PQ = SR.
So, if the diagonals of a rhombus are equal, then it is a square.
Hence, ABCD is a square.