Question

prove that the line segments joining the midpoints of opposite sides of a quadrilateral bisects 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.

Answer 2

Step-by-step explanation:

P IS MIDPOINT OF AB

Q IS MIDPOINT OF BC

BY MIDPOINT THEOREM WE GET

==> PQ || AC AND

PQ =

$\frac{1}{2}$

IN TRIANGLE ADC

S IS MIDPOINT OF AD

R IS MIDPOINT OF DC

BY MIDPOINT THEOREM WE HAVE

= SR || AC AND

SR =

$\frac{1}{2}$

AC ---(2)

FROM EQUATION (1) AND (2) WE HAVE

= PQ RS IS A PARALLELOGRAM

WE KNOW IN PARALLELOGRAM DIAGONALS

BISECT EACH OTHER

SO ,

IT MEANS SQ AND PR BISECTS EACH OTHER

HENCE,

THE LINE SEGMENT JOINING THE MIDPOINT OF THE OPPOSITE SIDE OF A QUADRILATERAL BISECTS EACH OTHER.

HOPE THIS HELPS YOU ALL

THANKS❤

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