Question

Prove that the line segment joining the mid point of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference of these sides ?

Answer 1

Let, $ABCD$ is a trapezium and $CA$ and $DB$ are the diagonals.

$M,N$ be the midpoints of $CA$ and $DB$ respectively.

Construction:

Join $CN$ and extend it to the point $X$ on $AB$.

Now, in $△CND$ and $△XNB$,

$\angle CDN=\angle NBX$ [Since, alternate interior angles of the parallel lines$DC$ and $AB$ with $DB$ as transversal ]

$\angle CND=\angle XNB$ [ vertically opposite angles ]

$DN=BN$ [ $\because N$ is the midpoint of $DB$ ]

$\therefore △CND\cong △XNB$ [ Angle-Side-Angle property ]

By C.P.C.T,

$CD=BX$ and $CN=NX$

$\Rightarrow N$ is also the midpoint of $CX$.

Again, in $△CAX$,

as $M$ and $N$ are the midpoints,

Mid-point theorem: If a line segment formed by mid-points of any two sides then it is parallel to third side and is half of it.

So, by the midpoint theorem,

$MN\parallel AX$ and $MN=\frac{1}{2}AX$

$MN\parallel AB$ [Since, $X$ is a point on $AB$.]---(1)

Now, lets take $MN=\frac{1}{2}AX$

$\Rightarrow MN=\frac{1}{2}(AB-BX)$

$\Rightarrow MN=\frac{1}{2}(AB-CD)$ [Since, $BX=CD$ proved]---(2)

From (1) and (2) it is clear that,

the line segment joining the mid point of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference of these sides

Hence, proved.