Question

$ABCD$ is a trapezium in which $AB\parallel CD$ and $AD=BC$ (see figure). Show that
(i) $\angle A=\angle B$
(ii) $\angle C=\angle D$
(iii) $△ABC\equiv △BAD$
(iv) diagonal $AC=$ diagonal $BD$

Answer 1

Given $ABCD$ is trapezium where $AD=BC$.

(i) To prove: $\angle A=\angle B$

we can see that $AECD$ is a parallelogram, so sum of adjacent angles $={180}^o$

$\to \angle A+\angle E={180}^o$

$\to \angle A+x={180}^o$

$\to \angle A={180}^o-x=\angle B$

Hence proved.

(ii) To prove: $\angle C=\angle D$

sum of adjacent angles in parallelogram is $\pi$, so

$\to \angle D\angle C+{180}^o-2x={180}^o$

$\to \angle C+\angle D=2x$

Now

$\to \angle B+\angle C={180}^o$

$\to {180}^o-x+\angle C={180}^o=0$ $\angle C=x$, so

$\angle D=x$

And,

$\angle C=\angle D$

Hence proved.

(iii) $\Delta ABC=\Delta BAD$

$\to$ side $AB$ is common.

$\to AD=BC$ (given)

so the angle including both the sides is also same,

$\angle A=\angle B$. So

$\Delta ABC=\Delta BAD$ (By SAS congruent Rule)

Hence proved.

(iv) As $\Delta ABC=\Delta BAD$

The third side of both triangles i.e. diagonals are equal $AC=BD$