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

Let us consider a Cartesian plane having a parallelogram OABC in which O is the origin. 

We have to prove that middle point of the opposite sides of a quadrilateral and the join of the mid-points of its diagonals meet in a point and bisect each other. 

Let coordinates be A(0, 0). 

So other coordinates will be B(x₁ + x₂, y₁), C(x₂, 0) ... refer figure.

Let P, Q, R and S be the mid-points of the sides AB, BC, CD, DA respectively. 

By midpoint formula,

x = \(\frac{x_1+x_2}2\), y = \(\frac{y_1+y_2}2\)

For midpoint P on AB,

 x = \(\frac{x_1+x_2+x_1}2\), y = \(\frac{y_1+y_1}2\)

∴ x = \(\frac{2x_1+x_2}2\), y = \(\frac{2y_1}2\)

∴ Coordinate of P is (\(\frac{2x_1+x_2}2\), \(\frac{y_1}2\))

For R, we can observe that, R lies on x axis. 

∴ Coordinate of R is (\(\frac{x_2}2\),0)

For midpoint S on OA,

x = \(\frac{x_1+0}2\), y = \(\frac{y_1+0}2\)

∴ Coordinate of S is ( \(\frac{x_1}2\), \(\frac{y_1}2\))

For midpoint of PR,

x = \(\frac{\frac{2x_1+x_2}2+\frac{x_2}2}2\), y = \(\frac{y_1+0}2\)

∴ x = \(\frac{x_1+x_2}2\) , y = \(\frac{y_1}2\)

∴ Midpoint of PR is (\(\frac{x_1+x_2}2\), \(\frac{y_1}2\))

Similarly midpoint of QS is  (\(\frac{x_1+x_2}2\), \(\frac{y_1}2\))

Also, 

similarly midpoint of AC and OA is (\(\frac{x_1+x_2}2\), \(\frac{y_1}2\))

Hence, 

midpoints of PR, QS, AC and OA coincide 

∴We say that middle point of the opposite sides of a quadrilateral and the join of the mid-points of its diagonals meet in a point and bisect each other.