Question

$A(1,1),B(5,4),C(3,8),D(-1,2)$ are the vertices of $\square ABCD$. If $P,Q,R,S$ are the mid-points of $\overline{AB},\overline{BC},\overline{CD},\overline{DA}$ respectiely, show that $PQRS$ is parallelogram.

Answer 1

$ABCD$ is a parallelogram so co-ordinates of $P$

$\Rightarrow (\frac{1+5}{2},\frac{1+9}{2})$

$P(3,\frac{5}{2})$

$Q=(\frac{5+3}{2},\frac{4+8}{2})\Rightarrow (4,6)$

$R=(\frac{3+(-1)}{2},\frac{4+8}{2})\Rightarrow (1,5)$

$S=(\frac{-1+1}{2},\frac{2+1}{2})\Rightarrow (0,\frac{3}{2})$

$PQ=\sqrt{(4-3{)}^2+(6-{\frac{5}{2}}^2)}$

$=\sqrt{1^2+\left({\frac{7}{2}}^2\right)}\Rightarrow \sqrt{\frac{53}{4}}$

$QR=\sqrt{(1-4{)}^2+(5-6{)}^2}$

$=\sqrt{(-3{)}^2+(-1{)}^2}=\sqrt{10}$

$RS=\sqrt{(0-1{)}^2+{(\frac{3}{2}-5)}^2}$

$=\sqrt{(-1{)}^2+{\left(\frac{-7}{2}\right)}^2}=\sqrt{\frac{53}{4}}$

$SP=\sqrt{(3-0{)}^2+{(\frac{5}{2},\frac{-3}{2})}^2}$

$=\sqrt{(3{)}^2+(1{)}^2}=\sqrt{10}$

So $PQ=RS$ and $QR=SP$ as in parallelogram opposite sides are equal.

$\therefore PQRS$ is parallelogram

Hence showed.