Question

$ABCD$ is a trapzium in which $AB||CD$ and $AD=BC$. Show that $△ABC\cong △BAD$

Answer 1

Ref. image (I)

Given : ABCD is a trapezium where $AB||CD$ and $AD=BC$

To prove : $△ABC\cong △BAD$

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

Proof :

$AD||CE$ (From construction) & $AE||DC$ (As $AB||DC,\&AB$ is extended)

In $AECD$, both pair of opposite sides are parallel, $AECD$ is a parallelogram

$\therefore AD=CE$ (Opposite sides of parallelogram are equal) ......... Ref. image (III)

But $AD=BC$ (Given)

$\Rightarrow BC=CE$

So, $\angle CEB=\angle CBE$ (In $\Delta BCE$, Angles opposite to equal sides are equal) ...(1)

For $AD||CE$ & $AE$ is the transversal

$\angle A+\angle CEB={180}^{\circ }$.........(Interior angle on same side of transversal is supplementary)

$\angle A={180}^{\circ }-\angle CEB$ ...(2)

Also $AE$ is a line,

So, $\angle B+\angle CBE={180}^{\circ }$ (Linear pair)

$\angle B+\angle CEB={180}^{\circ }$ (From (1))

$\angle B={180}^{\circ }-\angle CEB$ ...(3)

From (2) & (3)

$\angle A=\angle B$ ...............(4)

Now, Ref. image II

In $\Delta ABC$ and $\Delta BAD$

$AB=BA$ (common)

$\angle B=\angle A$ (Proved in part (i))

$BC=AD$ (Given)

$\therefore \Delta ABC\cong \Delta BAD$ (SAS congruence rule)