[ Area of a Quadrilateral ]

Given a convex quadrilateral $A B C D$ with area $s$ and a point $M$ inside it. Points $P, Q, R, S$ are symmetric to point $M$ with respect to the midpoints of the sides of quadrilateral $A B C D$. Find the area of quadrilateral $P Q R S$.
Answer: 2s. The area of quadrilateral $P Q R S$ is 4 times the area of the quadrilateral formed by the midpoints of the sides of quadrilateral $A B C D$. Let $P_{1}, Q_{1}, R_{1}, S_{1}$ be the midpoints of sides $A B, B C, C D, D A$. Then $S_{\mathrm{P}_{1} \mathrm{BQ}_{1}}+S_{\mathrm{R}_{1} \mathrm{DS}_{1}}=\frac{1}{4} S_{\mathrm{ABC}}+\frac{1}{4} S_{\mathrm{ADC}}=\frac{1}{4} s$ and $S_{\mathrm{Q}_{1} \mathrm{CR}_{1}}+S_{\mathrm{P}_{1} \mathrm{AS}_{1}}=\frac{1}{4} s$. Therefore, $S_{\mathrm{P}_{1} \mathrm{Q}_{1} \mathrm{R}_{1} \mathrm{~S}_{1}}=s-S_{\mathrm{P}_{1} \mathrm{BQ}_{1}}-S_{\mathrm{R}_{1} \mathrm{DS}_{1}}-S_{\mathrm{Q}_{1} \mathrm{CR}_{1}}$ $-S_{\mathrm{P}_{1} \mathrm{AS}_{1}}=\frac{1}{2} s$.