Question

Suppose ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$. Prove that $\angle A=\angle B$ and $\angle C=\angle D$.

Answer 1

In given figure ABCD is a trapezium,

$\Rightarrow$ Join AC and BD. Extent AB and draw a line through C parallel to DA meeting AB produced at E.

$\Rightarrow$ AB$\parallel$DC ----- ( 1 ) [Given]

$\Rightarrow$ and AD$\parallel$CE ---- ( 2 ) [Construction]

$\therefore$ ADCE is a parallelogram [Opposite pairs of sides are parallel]

$\Rightarrow$ $\angle$A + $\angle$E = 180${}^{\circ }$ --- ( 3 ) [Consecutive interior angles]

$\Rightarrow$ $\angle$B + $\angle$CBE = 180${}^{\circ }$ ---( 4 ) [Linear pair]

$\Rightarrow$ AD = CE ------ ( 5 ) [Opposite sides of a parallelogram.]

$\Rightarrow$ AD = BC ------ ( 6 ) [Given]

$\Rightarrow$ BC = CE [From ( 5 ) and ( 6 )]

$\Rightarrow$ $\angle$E = $\angle$CBE ---- ( 7 ) [Angles opposite to equal sides]

$\therefore$ $\angle$B + $\angle$E = 180${}^{\circ }$ --- ( 8 ) [From (4) and (7)]

Now from (3) and (8) we have,

$\Rightarrow$ $\angle$A + $\angle$E = $\angle$B + $\angle$E

$\therefore$ $\angle A=\angle B$ [Proved]

$\Rightarrow$ $\angle$A + $\angle$ D = 180${}^{\circ }$

$\Rightarrow$ $\angle$B + $\angle$C = 180${}^{\circ }$

$\Rightarrow$ $\angle$A + $\angle$D = $\angle$B + $\angle$C [$\because$ $\angle$A = $\angle$B]

$\therefore$ $\angle C=\angle D$ [Proved]