22.(10分)(2025·菏泽期末)如图,在$△ ABC$中,$∠ A=\frac{1}{2}∠ ABC=\frac{1}{6}∠ ACB$,BD是$∠ ABC$的平分线,交AB边上的高CE于点F.
求:(1)$∠ ACE$的度数;
(2)$∠ BFC$的度数.
求:(1)$∠ ACE$的度数;
(2)$∠ BFC$的度数.
答案：22.解:(1)$∵∠ A=\frac{1}{2}∠ ABC=\frac{1}{6}∠ ACB$,
$\therefore ∠ ABC=2∠ A,∠ ACB=6∠ A$.
$∵∠ A+∠ ABC+∠ ACB=180°$,
$\therefore ∠ A+2∠ A+6∠ A=180°$,解得$∠ A=20°$.
$∵CE⊥AB,∴∠ AEC=∠ BEC=90°$,
$\therefore ∠ ACE=180°-∠ A-∠ AEC=180°-20°-90°=70°$.
(2)$∵BD$平分$∠ ABC,∠ ABC=2∠ A=2×20°=40°$,
$\therefore ∠ ABD=\frac{1}{2}∠ ABC=\frac{1}{2}×40°=20°$.
$∵∠ BFC$是$△ BEF$的外角,
$\therefore ∠ BFC=∠ ABD+∠ BEC=20°+90°=110°$.
$\therefore ∠ ABC=2∠ A,∠ ACB=6∠ A$.
$∵∠ A+∠ ABC+∠ ACB=180°$,
$\therefore ∠ A+2∠ A+6∠ A=180°$,解得$∠ A=20°$.
$∵CE⊥AB,∴∠ AEC=∠ BEC=90°$,
$\therefore ∠ ACE=180°-∠ A-∠ AEC=180°-20°-90°=70°$.
(2)$∵BD$平分$∠ ABC,∠ ABC=2∠ A=2×20°=40°$,
$\therefore ∠ ABD=\frac{1}{2}∠ ABC=\frac{1}{2}×40°=20°$.
$∵∠ BFC$是$△ BEF$的外角,
$\therefore ∠ BFC=∠ ABD+∠ BEC=20°+90°=110°$.
23.(10分)如图,在△ABC中,AD是高,AE,BF分别是∠BAC,∠ABC的平分线,它们相交于点O,∠BAC=50°,∠C=∠BAC+20°,求∠DAC和∠BOA的度数.
答案：23.解:$∵∠ BAC=50°,∠ C=∠ BAC+20°,∴∠ C=70°$.
$∵AD⊥BC,∴∠ ADC=90°$,
$\therefore ∠ DAC=180°-∠ C-∠ ADC=180°-70°-90°=20°$.
$∵∠ BAC=50°,∠ C=70°$,
$\therefore ∠ ABC=180°-∠ BAC-∠ C=180°-50°-70°=60°$.
$∵AE,BF$分别是$∠ BAC,∠ ABC$的平分线,
$\therefore ∠ BAE=\frac{1}{2}∠ BAC=25°,∠ ABF=\frac{1}{2}∠ ABC=30°$,
$\therefore ∠ BOA=180°-∠ BAE-∠ ABF=180°-25°-30°=125°$.
$∵AD⊥BC,∴∠ ADC=90°$,
$\therefore ∠ DAC=180°-∠ C-∠ ADC=180°-70°-90°=20°$.
$∵∠ BAC=50°,∠ C=70°$,
$\therefore ∠ ABC=180°-∠ BAC-∠ C=180°-50°-70°=60°$.
$∵AE,BF$分别是$∠ BAC,∠ ABC$的平分线,
$\therefore ∠ BAE=\frac{1}{2}∠ BAC=25°,∠ ABF=\frac{1}{2}∠ ABC=30°$,
$\therefore ∠ BOA=180°-∠ BAE-∠ ABF=180°-25°-30°=125°$.
24.(12分)(2025·如皋期末)在△ABC中,AD为∠BAC的平分线,点E在AD上(不与点A,D重合),∠BED=∠BDE,延长BE交AC于点F.
(1)如图①,若∠ABC=90°,求∠BFC的度数;
(2)当∠ABC≠90°时,求证:∠BFC+∠ABC=180°;
(3)如图②,∠ACB的平分线交BF于点G,请用一个等式表示∠CGF,∠ABC,∠ACB之间的数量关系,并说明理由.
(1)如图①,若∠ABC=90°,求∠BFC的度数;
(2)当∠ABC≠90°时,求证:∠BFC+∠ABC=180°;
(3)如图②,∠ACB的平分线交BF于点G,请用一个等式表示∠CGF,∠ABC,∠ACB之间的数量关系,并说明理由.
答案：24.(1)解:$∵AD$为$∠ BAC$的平分线,
$\therefore ∠ BAD=∠ FAD$.
$∵∠ BED=∠ BDE,∠ BED=∠ AEF$,
$\therefore ∠ AEF=∠ BDE$,
$\therefore ∠ BFC=∠ AEF+∠ FAD=∠ BDE+∠ BAD$.
$∵∠ BDE+∠ BAD+∠ ABC=180°,∠ ABC=90°$,
$\therefore ∠ BDE+∠ BAD=180°-∠ ABC=90°$,
$\therefore ∠ BFC=90°$.
(2)证明:$∵AD$为$∠ BAC$的平分线,
$\therefore ∠ BAD=∠ FAD$.
$∵∠ BED=∠ BDE,∠ BED=∠ AEF$,
$\therefore ∠ AEF=∠ BDE$,
$\therefore ∠ BFC=∠ AEF+∠ FAD=∠ BDE+∠ BAD$.
$∵∠ BDE+∠ BAD+∠ ABC=180°$,
$\therefore ∠ BFC+∠ ABC=180°$.
(3)解:$∠ ABC=∠ CGF+\frac{1}{2}∠ ACB$.理由如下:
由(2),得$∠ ABC=180°-∠ BFC$,
$∵∠ CGF+∠ GCF=180°-∠ BFC$,
$\therefore ∠ ABC=∠ CGF+∠ GCF$.
$∵CG$平分$∠ ACB$,
$\therefore ∠ GCF=\frac{1}{2}∠ ACB,∴∠ ABC=∠ CGF+\frac{1}{2}∠ ACB$.
$\therefore ∠ BAD=∠ FAD$.
$∵∠ BED=∠ BDE,∠ BED=∠ AEF$,
$\therefore ∠ AEF=∠ BDE$,
$\therefore ∠ BFC=∠ AEF+∠ FAD=∠ BDE+∠ BAD$.
$∵∠ BDE+∠ BAD+∠ ABC=180°,∠ ABC=90°$,
$\therefore ∠ BDE+∠ BAD=180°-∠ ABC=90°$,
$\therefore ∠ BFC=90°$.
(2)证明:$∵AD$为$∠ BAC$的平分线,
$\therefore ∠ BAD=∠ FAD$.
$∵∠ BED=∠ BDE,∠ BED=∠ AEF$,
$\therefore ∠ AEF=∠ BDE$,
$\therefore ∠ BFC=∠ AEF+∠ FAD=∠ BDE+∠ BAD$.
$∵∠ BDE+∠ BAD+∠ ABC=180°$,
$\therefore ∠ BFC+∠ ABC=180°$.
(3)解:$∠ ABC=∠ CGF+\frac{1}{2}∠ ACB$.理由如下:
由(2),得$∠ ABC=180°-∠ BFC$,
$∵∠ CGF+∠ GCF=180°-∠ BFC$,
$\therefore ∠ ABC=∠ CGF+∠ GCF$.
$∵CG$平分$∠ ACB$,
$\therefore ∠ GCF=\frac{1}{2}∠ ACB,∴∠ ABC=∠ CGF+\frac{1}{2}∠ ACB$.