Question

$ABCD$ is quadrilateral in which $AB\parallel CD$. If $AD=BC$, show that $\angle A=\angle B$ and $\angle C=\angle D$

Answer 1

Here, AB$\parallel$CD [Given]

$\Rightarrow$ and AD$\parallel$EC [By construction]

$\therefore$ AECD is a parallelogram.

$\Rightarrow$ AD = EC [Opposite sides of parallelogram are equal]

$\Rightarrow$ But AD = EC [Given]

$\therefore$ EC = BC

$\therefore$ $\angle$CBE = $\angle$CEB ---- ( 1 )

$\Rightarrow$ $\angle$B + $\angle$CBE = 180${}^{\circ }$ [Linear pair] ---- ( 2 )

$\Rightarrow$ AD$\parallel$EC [By construction]

$\Rightarrow$ and transeversal AE intersects them

$\therefore$ $\angle$A + $\angle$CEB = 180${}^{\circ }$ ---- ( 3 )

[Sum of adjacent angles of parallelogram is supplementary ]

$\Rightarrow$ $\angle$B + $\angle$CEB = 180${}^{\circ }$ [From ( 2 ) and ( 3 )]

$\Rightarrow$ But $\angle$CBE = $\angle$CEB [From ( 1 )]

$\therefore$ $\angle A=\angle B$ [Proved] -- ( 4 )

$\Rightarrow$ $\because$ AB$\parallel$CD

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

Supplementary angles of parallelogram is 180${}^{\circ }$]

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

$\therefore$ $\angle$A + $\angle$D = $\angle$B + $\angle$C

$\Rightarrow$ But $\angle$A = $\angle$B [From ( 4 )]

$\therefore$ $\angle C=\angle D$