EF, GH two parallel lines cut by transversal BC

Prove ^EBC = ^HCB

Draw AD through midpoint of BC and perpendicular to EF.

^BAD = 1 rt. ang.        (construct)

^CDA = 1 rt ang.        (straight line perp. to one parallel is perp. to other parallel)

^COD  = ^BOA           (opp. ang.s of intersecting lines)

^DCO = ^1rt ang - ^COD   (angles of tri. add to 2 rt. ang.s)

               = ^OBA

^DCO = ^HCB and ^OBA = ^EBC, so 

^EBC = ^HCB