Original answer: Respondents added 20 10-03-07 02:59.
Solution: Connect BC
∫∠BDC = 140
∴∠DBC+∠DCB= 180 -∠BDC=40
∫∠BGC = 1 10。
∴∠GBD+∠GCD=∠BGC-(∠DBC+∠DCB)=30
And CF is the bisector of angle ACD, and BE and cf intersect at point g.
∴∠ABD+∠ACD=30 ×2=60
∴(∠abd+∠acd)+(∠dbc+∠dcb)=60+40 = 100
∴∠a= 180-(∠Abd+∠ACD)+(∠DBC+∠dcb)= 80