Question

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

Answer 1

Proved below.

Step-by-step explanation:

Given:

Let ABCD be the quadrilateral such that diagonal AC is along x axis. Suppose the coordinates A,B,C and D be $(0,0), (x_1,x_2),(x_1,0),(x_3,y_3)$ 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 $(\frac{0+x_1}{2} )\frac{0+0}{2}=\frac{x_1}{2}$

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

Coordinates of mid point of GH are  $(\frac{\frac{x_1}{2}+\frac{x_2+x_3}{2} }{2} ,\frac{\frac{y_2}{2}+\frac{y_3}{2}}{2} )=(\frac{x_1+x_2+x_3}{4},\frac{y_2+y_3}{4})$

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

Hence, proved.