Question

$ABCD$ is a trapezium in which $AB\parallel CD$ and $AD=BC$ show that
I) $\angle A=\angle B$
II) $\angle C=\angle D$
III) triangle $ABC$ congruent to triangle $BAD$
IV) diagonal $AC$ =diagonal $BD$
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Answer 1

Let us extend $AB$. Then, draw a line through $C$, which is parallel to $AD$, intersecting $AE$ at point $E$. It is clear that $AECD$ is a parallelogram.

(i) $AD=CE$ (Opposite sides of parallelogram $AECD$)
However, $AD=BC$ (Given)
Therefore, $BC=CE$
$\angle CEB=\angle CBE$ (Angle opposite to equal sides are also equal)
Consider parallel lines $AD$ and $CE$.
$AE$ is the transversal line for them.
So, $\angle A+\angle CEB={180}^{\circ }$ (Angles on the same side of transversal)
$\angle A+\angle CBE={180}^{\circ }$ (Using the relation $\angle CEB=\angle CBE$) ... (1)
However, $\angle B+\angle CBE={180}^{\circ }$ (Linear pair angles) ... (2)
From equations (1) and (2), we obtain $\angle A+\angle CEB=\angle B+\angle CBE$
$\therefore \angle A=\angle B$


(ii) Given $AB\parallel CD$
$\angle A+\angle D={180}^{\circ }$ (Angles on the same side of the transversal)
Also, $\angle C+\angle B={180}^{\circ }$ (Angles on the same side of the transversal)
$\therefore \angle A+\angle D=\angle C+\angle B$
However, $\angle A=\angle B$ [Using the result obtained in (i)]
$\therefore \angle C=\angle D$

(iii) In $\Delta ABC$ and $\Delta BAD$,

$AB=BA$ (Common side)
$BC=AD$ (Given)
$\angle B=\angle A$ (Proved before)
$\therefore \Delta ABC\cong \Delta BAD$ (SAS congruence rule)

(iv) We had observed that, $\Delta ABC\cong \Delta BAD$
$\therefore AC=BD$ (By corresponding parts of congruent triangles)