In Fig 8.109, if AB ||CD and CD||EF, find ∠ACE.
Since,
EF ‖ CD
Therefore,
∠EFC + ∠LEC = 180o(Co. interior angles)
∠ECD = 180o – 130o
= 50o
Also, BA ‖ CD
∠BAC = ∠ACD = 70o(Alternate angles)
But, ∠ACE + ∠ECD = 70o
∠ACE = 70o -50o
=20o