假设AC与BE 的交点为P,∠EBC为∠1 , ∠EAC为∠2, ∠APC = ∠BPC = ∠3
∠E=180°-∠2- ∠3
因为 ∠3=180° - ∠1 - 78°
∠FAC=2∠1 +78° ∠FAC = 2∠2
所以2∠2 = 2∠1 +78°
所以∠2 = ∠1 +39°
所以∠E=180°-(∠1 +39°)- (180° - ∠1 - 78° )
去括号得出∠E=180°-∠1 -39° -180° +∠1 + 78°
合并的出∠E = 39°
∠FAC=180-∠BAC
∠C=180-∠BAC-∠ABC
∠EAB=180-1/2∠FAC
∠ABE=1/2∠ABC
∠E=180-∠EAB-∠ABE=180-(180-1/2∠FAC)-1/2∠ABC
=1/2(∠FAC-∠ABC)
=1/2(180-∠BAC-∠ABC)=1/2∠C=78/2=39