In the triangle ABC, if the bisector of the inner angle B and C intersects at point E and the bisector of the outer angle intersects at point F, and angle a is equal to 70 degrees, then angle BEC equals? Angle BFC equals?

In the triangle ABC, if the bisector of the inner angle B and C intersects at point E and the bisector of the outer angle intersects at point F, and angle a is equal to 70 degrees, then angle BEC equals? Angle BFC equals?

The results show that: ABC + ACB = 180 degrees - a = 110 degrees; EB and EC are bisectors of angles, then: EBC + ECB = (1 / 2) ∠ ABC + (1 / 2) ∠ ACB = 55 degrees, so: ∠ BEC = 125 degrees. Be and BF are bisectors of inner and outer angles of ABC respectively, so be ⊥ BF can be easily obtained; similarly, EC ⊥ cf. then: FBC + ∠ FCB = (90 ° - ∠ EBC) + (90 ° -}