Given, ABC is an isosceles triangle in which AB = AC
AEDC is a parallelogram
∠CDF=70∘ and ∠BFE=100∘
Since AEDC is a parallelogram
∠ACD+70∘=180∘
⇒∠ACD=110∘→(1)
⇒∠ACD+∠GCB=180∘ (∵ linear pair )
⇒∠GCB=180∘−110∘=70∘→(2)
∠GFD+∠BFE=180∘ (linear pair)
⇒∠GFD=180∘−100∘=80∘→(3)
In ∠BFD,∠FBD=180∘−(80∘+70∘)=30∘
SinceAB=AC,∠ABC=∠ACB
∴∠ABC=70∘
⇒∠ABF=∠ABC−∠FBD=70∘−30∘=40∘