Question

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

Answer 1

Given that ABCD is a trapezium,

$AB\parallel CD$ and $AD=BC$

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

(ii) $\angle C=\angle D$

(iii) $△ABC\cong △BAD$

(iv) $AC=BD$

Construction: Draw $CE\parallel AD$ and extend AB to intersect CE at E.

Poof:

$\left(i\right)$ As AECD is a parallelogram.

By construction, CE is parallel to AD and AE is parallel to CD,

$\therefore AD=EC$

But, $AD=BC$ (Given)

$\therefore BC=EC$

$\Rightarrow \angle 3=\angle 4$ (Angle opposite to equal side are equal)

Now, $\angle 1+\angle 4={180}^0$ (consecutive interior angle of parallelogram )

And $\angle 2+\angle 3={180}^0$ (linear pair )

$\Rightarrow \angle 1+\angle 4=\angle 2+\angle 3$ $\Rightarrow \angle 1=\angle 2\because \angle 3=\angle 4$ (so get cancelled with each other )

$\Rightarrow \angle A=\angle B$

$\left(ii\right)$ $AB\parallel CD$

$\therefore \angle A+\angle D=\angle B+\angle C={180}^o$ [Angles on the same side of transversal]

But, $\angle A=\angle B$

Hence, $\angle C=\angle D$

$\left(iii\right)$ In $△ABC$ and $△BAD$, we have

$BC=AD$ (Given)

$AB=BA$ (Common)

$\angle A=\angle B$ (Proved)

Hence, by SAS congruence criterion,

$△ABC\cong △BAD$

$\left(iv\right)$ We have proved that, $△ABC\cong △BAD$

$\therefore AC=BD$ (CPCT)

Hence, all the four required results have been proved.