1. (1)如图①,在$△ ABC$中,$∠ A=30°$,$∠ ACB=80°$,$CD$为$AB$边上的高,$BE$平分$∠ ABC$,分别交$CD$,$AC$于点$F$,$E$,求$∠ CFB$的度数.
(2)在(1)的条件下,若$∠ ACB=90°$,如图②,求证:$∠ CFE=∠ CEF$.

1. (1)$\because ∠ A=30°,∠ ACB=80°,\therefore ∠ ABC=180°-∠ A-∠ ACB=70°.\because BE$平分$∠ ABC,\therefore ∠ ABE=∠ CBE=35°,\therefore ∠ CFB=∠ FDB+∠ ABE=125°.$
(2) 如图,$\because ∠ ACB=90°,\therefore ∠ 1+∠ 3= 90°.\because CD ⊥ AB,\therefore ∠ 2+∠ 4=90°.$又$\because BE$平分$∠ ABC,\therefore ∠ 1 = ∠ 2,\therefore ∠ 3 = ∠ 4.$$\because ∠ 4 = ∠ 5,\therefore ∠ 3 = ∠ 5$, 即$∠ CFE = ∠ CEF.$

2. (1)如图①,在$△ ABC$中,$∠ B＜∠ C$,$AE ⊥ BC$,$AD$是$△ ABC$的角平分线,试写出$∠ DAE$,$∠ B$,$∠ C$之间的数量关系,并证明你的结论;
(2)如图②,若点$F$在$AD$的延长线或反向延长线上,作$FE ⊥ BC$,其他条件不变,试写出$∠ F$与$∠ B$,$∠ C$之间的数量关系,并证明你的结论;
(3)如图③,若把(1)中的“$AE ⊥ BC$”改为“$CE ⊥ AD$”,其他条件不变,试写出$∠ ECD$与$∠ B$,$∠ ACB$之间的数量关系,并证明你的结论.

2. (1)$∠ DAE=\frac{1}{2}(∠ C-∠ B)$,证明如下:$\because ∠ ADE$是$△ ABD$的外角,$\therefore ∠ ADE=∠ B+∠ BAD.\because AE ⊥ BC,\therefore ∠ DAE=90°-∠ ADE.$又$\because AD$平分$∠ BAC,\therefore ∠ BAD=\frac{1}{2}∠ BAC,\therefore ∠ DAE=90°-∠ B-\frac{1}{2}∠ BAC.$又$\because ∠ BAC=180°-∠ B-∠ C,\therefore ∠ DAE=90°-∠ B-\frac{1}{2}(180°-∠ B-∠ C),\therefore ∠ DAE=\frac{1}{2}(∠ C-∠ B).$
(2)$∠ F=\frac{1}{2}(∠ C-∠ B)$,证明如下:如图,作$AG ⊥ BC$,垂足为$G$,$\because EF ⊥ BC,\therefore AG // EF$,$\therefore ∠ F=∠ DAG.$由(1)可得$∠ DAG=\frac{1}{2}(∠ C-∠ B),\therefore ∠ F=\frac{1}{2}(∠ C-∠ B).$同理,$∠ F' =\frac{1}{2}(∠ C-∠ B).$
(3)$∠ ECD=\frac{1}{2}(∠ ACB-∠ B).$证明如下:$\because AD$是$△ ABC$的角平分线,$\therefore ∠ BAD=∠ CAD=\frac{1}{2}∠ BAC,\therefore ∠ EDC=∠ B+∠ BAD=∠ B+\frac{1}{2}∠ BAC.\because CE ⊥ AD,\therefore ∠ DEC=∠ AEC=90°,\therefore ∠ EDC+∠ ECD=∠ ACE+∠ DAC,\therefore ∠ B+\frac{1}{2}∠ BAC+∠ ECD=∠ ACE+\frac{1}{2}∠ BAC,\therefore ∠ B+∠ ECD=∠ ACE=∠ ACB-∠ ECD,\therefore 2∠ ECD=∠ ACB-∠ B,\therefore ∠ ECD=\frac{1}{2}(∠ ACB-∠ B).$