Question

show that the line segment joining the midpoint of the opposite side of a quadrilateral bisect each other​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that ABCD be a quadrilateral such that P, Q, R, S are the midpoints of the sides of a quadrilateral AB, BC, CD, DA respectively.

Construction: Join PQ, QR, RS, SP, PR, QS and AC

Now, In triangle ABC

$\sf P \: is \: midpoint \: of \: AB$

$\sf Q \: is \: midpoint \: of \: BC$

So, By Midpoint Theorem, we get

$\implies\sf \: PQ \: \parallel \: AC \: and \: PQ \: = \: \dfrac{1}{2}AC - - - (1)$

Now, In triangle ADC

$\sf S \: is \: midpoint \: of \: AD$

$\sf R \: is \: midpoint \: of \: DC$

So, By Midpoint Theorem, we have

$\implies\sf \: SR \: \parallel \: AC \: and \: SR \: = \: \dfrac{1}{2}AC - - - (2)$

From equation (1) and (2), we concluded that

$\implies\sf \: PQRS \: is \: a \: parallelogram.$

[∵ If in a quadrilateral, one pair of opposite sides are equal and parallel, then quadrilateral is a parallelogram ]

Thus, the line segments joining the midpoints of the sides of a quadrilateral is a parallelogram.

Now, we know, in parallelogram, diagonals bisect each other.

So, it means SQ and PR bisects each other.

Hence, the line segment joining the midpoints of the opposite sides of a quadrilateral bisects each other.