Given ABCD is a qudrilateral such that AB||DC and AD = BC.
Construction Extend AB to E and draw a line CE parallel to AD.
Proof Since, AD||CE and transversal AE cuts them at A and E, respectively.
`therefore" "angleA+angleE=180^(@)" "`[since, sum of cointerior angles is `180^(@)`]
`rArr" "angleA=180^(@)-angleE" "...(i)`
Since, `" "AB||CD and AD||CE`
So, quadrilateral AECD is a parallelogram.
`rArr" "AD=CErArrBC=CE" "[becauseAD=BC,given]`
Now, in `Delta`BCE
`" "CE=BC" "`[proved above]
`rArr" "angleCBE=angleCEB` ltBrgt `" "` [opposite angles of equal side are equal]
`rArr" "180^(@)-angleB=angleE" "[becauseangleB+angleCBE=180^(@)]`
`rArr" "180^(@)-angleE=angleB" "...(ii)`
From Eqs. (i) and (ii), `" "angleA=angleB" "` Henced proved.