Question

In figure,prove AB $\parallel$ EF.

Answer 1

$ExtendEFtoPasshowninthefigureabove.Now\angle BCD=\angle BCE+\angle ECD={30}^0+{35}^0={65}^0⟶\left(1\right)Given,\angle ABC={65}^0⟶\left(2\right)from\left(1\right)and\left(2\right)\angle ABC=\angle BCD={65}^0\left(alternateangle\right)\therefore AB\parallel CD⟶\left(3\right)Now\angle PEC={180}^0-{145}^0={35}^0\angle PEC=\angle ECD={35}^0\left(alternateangle\right)\therefore PEcSincePEistheentendedlinesegmentofEF\therefore EF\parallel CD⟶\left(4\right)from\left(3\right)and\left(4\right)AB\parallel EFHenceproved.$