Since the adjacent sides of trapezium are supplementry,
ADC + ABC= 180°.
3ABC = 180°
Also ABC =60°, BAD=60° And ADC=120° and CBA = 120°.
Draw a perpendicular DM bisecting AB, then triangle ADM is formed with DAM =60°. cos60 = 1/2= AM/AD
Hence AM/a=1/2 and AM=a/2.
Similarly the other side perpendicular CN is drawn and cos60 BN = a/2.
The remaining part is same as `b`.
AB = a/2 + a/2 + b = (a+b).