Question

Prove that the lines joining the middle points of opposite sides of a quadrilateral and the line joining the middle points of its diagonals meet in a point and bisect one another.

Answer 1

Let $ABCD$ be the quadrilateral such that diagonal $AC$ is along $x$ axis. Suppose the coordinates $A,B,C$ and $D$ be $(0,0),(x_2,y_2)(x_1,0)$ and $(x_3,y_3)$ respectively.

$E$ and $F$ are the mid points of sides $AD$ and $BC$ respectively and $G$ and $H$ are the mid point of daigonals $AC$ and $BD$ and the point of intersection of $EF$ and $GH$ is $I$

Coordinates of $E$ are $(\frac{0+x_3}{2},\frac{0+y_3}{2})=(\frac{x_3}{2},\frac{y_3}{2})$

Coordinates of $F$ are $(\frac{x_1+x_2}{2},\frac{0+y_2}{2})=(\frac{x_1+x_2}{2},\frac{y_2}{2})$

Coordinates of mid point of $EF$ are

$(\frac{\frac{x_3}{2}+\frac{x_1+x_2}{2}}{2},\frac{\frac{y_3}{2}+\frac{y_2}{2}}{2})(\frac{x_1+x_2+x_3}{4},\frac{y_2+y_3}{4})$

$G$ and $H$ are the mid points of diagonal $AC$ and $BD$ respectively then

Coordinates of $G$ are $(\frac{0+x_1}{2},\frac{0+0}{2})=(\frac{x_1}{2},0)$

Coordinates of $H$ are $(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})$

Coordinates of mid point of $GH$ are

$(\frac{\frac{x_1}{2}+\frac{x_2+x_3}{2}}{2},\frac{\frac{y_2}{2}+\frac{y_2+y_3}{2}}{2})(\frac{x_1+x_2+x_3}{4},\frac{y_2+y_3}{4})$

As you can see mid points of both $EF$ and $GH$ are same. So, $EF$ and $GH$ meet and bisect each other.

Hence, proved.