As shown in the figure, BE is the bisector of ∠ABD, CF is the bisector of ∠ACD, and BE and CF intersect at G point, ∠ BDC = 140, ∠ BGC = 1 10. Find the degree of ∠ a

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