Question

The line segments joining the midpoints $M$ and $N$ are parallel sides $AB$ and $DC$ respectively of a trapezium $ABCD$ is perpendicular to both the sides $AB$ and $DC$. Prove that $AD=BC$.

Answer 1

Construct $AN$ and $BN$ at the point $N$

Consider $△ANM$ and $\angle BNM$

We know that $N$ is the midpoint of the line $AB$

So we get

$AM=BM$

From the figure we know that

$\angle AMN=\angle BMN={90}^{\circ }$

$MN$ is common i.e. $MN=MN$

By $SAS$ congruence criterion

$△ANM\cong △BNM$

$AN=BN(c.p.c.t)\dots \left(1\right)$

We know that

$\angle ANM=\angle BNM(c.p.c.t)$

Subtracting $LHS$ and $RHS$ by ${90}^{\circ }$

${90}^{\circ }-\angle ANM={90}^{\circ }-\angle BNM$

So we get

$\angle AND=\angle BNC\dots \left(2\right)$

Now, consider $△AND$ and $△BNC$

$AN=BN$

$\angle AND=\angle BNC$

We know that $N$ is the midpoint of the line $DC$

$DN=CN$

By $SAS$ congruence criterion

$△AND\cong △BNC$

$AD=BC(c.p.c.t)$

Therefore, it is proved that $AD=BC$.