如图1,△ABC中,∠ABC的角平分线与∠ACB的外角∠ACD的平分线交于E.

(1)设∠ABE=∠CBE=X(度);∠ACE=∠DCE=Y(度).则:
∠ECD=∠CBE+∠E,即:Y=X+∠E.则:2Y=2X+2∠E, 2∠E=2Y-2X;
又∠A+∠ABC=∠ACD,即:∠A+2X=2Y, ∠A=2Y-2X.
∴∠A=2∠E.
(2)∠P=180-(∠PCF+∠PFC)=180-(1/2)*(∠ACF+∠AFC)=180-(1/2)∠A=180-∠E.
∴∠P=180-∠E.(即∠P与∠E互补)
(3)∠A'MB+∠AMA'=∠A'MB+2∠AMN=180;
同理:∠A'NC+2∠ANM=180;
则:∠A'MB+∠A'NC+2(∠AMN+∠ANM)=360;
又∠AMN+∠ANM=180-∠A,则2(∠AMN+∠ANM)=360-2∠A.
∴∠A'MB+∠A'NC+(360-2∠A)=∠360,化简得:∠A'MB+∠A'NC=2∠A-----------------------------(1)
设GE'交CE于P,则:
∠E'GB=∠GPE+∠E=∠E'HC+∠E'+∠E=∠E'HC+2∠E;故:∠E'GB-∠E'HC=2∠E.---------------(2)
故:∠A'MB+∠A'NC=2∠E'GB-2∠E'HC.