Question

$ABCD$ is a trapezium in which $AB||CD$ and $AD=BC.$ Show that$\angle C=\angle D$

Answer 1

Given:- $ABC$ isi a trapezium where $AB\parallel CD$ and $AD=BC$

To prove:- $\angle C=\angle D$

Construction:- Extend $AB$ and draw a line through $C$ parallel to $AD$ intersecting $AB$ produced at $E$.

Proof:-

$AD\parallel CE\left(\text{Fro construction}\right)$

$AE\parallel CD(\text{As}AB\parallel CD,\&AB\text{produced at}E)$

In $AECD$, both pairs of opposite sides are parallel.

$\therefore AECD$ is a parallelogram.

$\therefore AD=CE.....\left(1\right)(\because \text{Opposite sides of a parallelogram are equal})$

$AD=BC.....\left(2\right)\left(\text{Given}\right)$

From equation $\left(1\right)\&\left(2\right)$, we have

$\therefore BC=CE$

$\Rightarrow \angle CEB=\angle CBE.....\left(3\right)(\because \text{Angle opposite to equal sides are equal})$

Now, for $AD\parallel CE$ and $AE$ is transversal,

$\angle A+\angle CEB=180°$

$\Rightarrow \angle A=180°-\angle CEB.....\left(4\right)$

Also $AE$ is a line,

$\angle B+\angle CBE=180°\left(\text{Linear pair}\right)$

$\Rightarrow \angle B+\angle CEB=180°\left(\text{From}\left(3\right)\right)$

$\Rightarrow \angle B=180°-\angle CEB.....\left(5\right)$

Now, from equation $\left(4\right)\&\left(5\right)$, we get

$\angle A=\angle B.....\left(6\right)$

For $AB\parallel CD$ and $AD$ is the transversal,

$\angle A+\angle D=180°\left(\text{Interior angle on same side of transversal is supplementary}\right)$

$\Rightarrow \angle D=180°-\angle A.....\left(7\right)$

Now, for $AB\parallel CD$ and $BC$ is transversal,

$\angle B+\angle C=180°\left(\text{Interior angle on same side of transversal are supplementary}\right)$

$\Rightarrow \angle C+\angle A=180°\left(\text{From}\left(6\right)\right)$

$\Rightarrow \angle C=180°-\angle A.....\left(8\right)$

From equation $\left(7\right)\&\left(8\right)$, we get

$\angle C=\angle D$

Hence proved.