9. （2025·无锡期末）如图，在$△ ABC$中，$∠ B = 20^{\circ}$，点$D$为边$BC$上一点，将$△ ADC$沿直线$AD$折叠后，点$C$落到点$E$处，$∠ ADC = 100^{\circ}$。
（1）求证：$AB// DE$；
（2）若$AE$恰好平分$∠ BAD$，求$∠ E$的度数。
（1）求证：$AB// DE$；
（2）若$AE$恰好平分$∠ BAD$，求$∠ E$的度数。
答案：9. (1)由折叠可知∠ADE = ∠ADC = 100°,
∴∠ADB = 180° - ∠ADC = 180° - 100° = 80°,
∴∠BDE = ∠ADE - ∠ADB = 100° - 80° = 20°,
∴∠BDE = ∠B = 20°,
∴AB//DE.
(2)
∵∠ADC是△ABD的外角,
∴∠ADC = ∠B + ∠BAD,
∴100° = 20° + ∠BAD,
∴∠BAD = 80°.
∵AE平分∠BAD,
∴∠EAD = $\frac{1}{2}$∠BAD = $\frac{1}{2}$×80° = 40°.在△ADE中,∠ADE + ∠EAD + ∠E = 180°,
∴∠E = 180° - ∠ADE - ∠EAD = 180° - 100° - 40° = 40°.
∴∠ADB = 180° - ∠ADC = 180° - 100° = 80°,
∴∠BDE = ∠ADE - ∠ADB = 100° - 80° = 20°,
∴∠BDE = ∠B = 20°,
∴AB//DE.
(2)
∵∠ADC是△ABD的外角,
∴∠ADC = ∠B + ∠BAD,
∴100° = 20° + ∠BAD,
∴∠BAD = 80°.
∵AE平分∠BAD,
∴∠EAD = $\frac{1}{2}$∠BAD = $\frac{1}{2}$×80° = 40°.在△ADE中,∠ADE + ∠EAD + ∠E = 180°,
∴∠E = 180° - ∠ADE - ∠EAD = 180° - 100° - 40° = 40°.
10. (2025·商丘校级月考)如图所示，在$△ ABC$中，点$D$是$BC$边上一点，$∠ 1 = ∠ 2$，$∠ 3 = ∠ 4$，$∠ BAC = 63^{\circ}$，则$∠ DAC =$(
A.$78^{\circ}$
B.$39^{\circ}$
C.$24^{\circ}$
D.$48^{\circ}$
C
)A.$78^{\circ}$
B.$39^{\circ}$
C.$24^{\circ}$
D.$48^{\circ}$
答案：10. C 解析:设∠1 = ∠2 = x,则∠ADB = 180° - (∠1 + ∠2)=180° - 2x,
∴∠3 = ∠4 = 180° - ∠ADB = 2x.
∵∠BAC = 63°,
∴∠2 + ∠4 = 117°,即x + 2x = 117°,
∴x = 39°,
∴∠3 = ∠4 = 78°,
∴∠DAC = 180° - ∠3 - ∠4 = 24°,故选C.
∴∠3 = ∠4 = 180° - ∠ADB = 2x.
∵∠BAC = 63°,
∴∠2 + ∠4 = 117°,即x + 2x = 117°,
∴x = 39°,
∴∠3 = ∠4 = 78°,
∴∠DAC = 180° - ∠3 - ∠4 = 24°,故选C.
11. 如图，两面镜子$AB$，$BC$的夹角为$∠ α$，光线经过镜子后反射，$∠ 1 = ∠ 2$，$∠ 3 = ∠ 4$。若$∠ α = 70^{\circ}$，则$∠ β$的度数是(
A.$30^{\circ}$
B.$35^{\circ}$
C.$40^{\circ}$
D.$45^{\circ}$
C
)A.$30^{\circ}$
B.$35^{\circ}$
C.$40^{\circ}$
D.$45^{\circ}$
答案：
11. C 解析:如图,由题意得,∠5 = 180° - (∠1 + ∠2)=180° - 2∠2,∠6 = 180° - (∠3 + ∠4)=180° - 2∠3.
∵∠α = 70°,
∴∠2 + ∠3 = 180° - ∠α = 110°.
∵∠β = 180° - (∠5 + ∠6),
∴∠β = 180° - (180° - 2∠2 + 180° - 2∠3)=2×(∠2 + ∠3) - 180° = 2×110° - 180° = 220° - 180° = 40°.故选C.
11. C 解析:如图,由题意得,∠5 = 180° - (∠1 + ∠2)=180° - 2∠2,∠6 = 180° - (∠3 + ∠4)=180° - 2∠3.
∵∠α = 70°,
∴∠2 + ∠3 = 180° - ∠α = 110°.
∵∠β = 180° - (∠5 + ∠6),
∴∠β = 180° - (180° - 2∠2 + 180° - 2∠3)=2×(∠2 + ∠3) - 180° = 2×110° - 180° = 220° - 180° = 40°.故选C.
12. 如图，在$△ ABC$中，$∠ A = 78^{\circ}$，$∠ ACD$是$△ ABC$的一个外角，$∠ EBC = \frac{1}{3}∠ ABC$，$∠ ECD = \frac{1}{3}∠ ACD$，则$∠ E$为
26
$^{\circ}$。答案：12. 26 解析:如图,
∵∠1 + ∠E = ∠2,
∴∠E = ∠2 - ∠1.
∵∠EBC = $\frac{1}{3}$∠ABC,∠ECD = $\frac{1}{3}$∠ACD,
∴∠ABC = 3∠1,∠ACD = 3∠2.
∵∠A + ∠ABC = ∠ACD,
∴∠A + 3∠1 = 3∠2,
∴∠A = 3∠2 - 3∠1 = 3(∠2 - ∠1)=3∠E = 78°,
∴∠E = 26°.
∵∠1 + ∠E = ∠2,
∴∠E = ∠2 - ∠1.
∵∠EBC = $\frac{1}{3}$∠ABC,∠ECD = $\frac{1}{3}$∠ACD,
∴∠ABC = 3∠1,∠ACD = 3∠2.
∵∠A + ∠ABC = ∠ACD,
∴∠A + 3∠1 = 3∠2,
∴∠A = 3∠2 - 3∠1 = 3(∠2 - ∠1)=3∠E = 78°,
∴∠E = 26°.
13. (2025·龙岩期末)在$△ ABC$中，$∠ A = 60^{\circ}$，$∠ B = 20^{\circ}$，点$D$在$AB$边上，连接$CD$，若$△ ACD$为直角三角形，则$∠ BCD$的度数为
70°或10°
。答案：
13. 70°或10° 解析:分两种情况:如图①,当∠ADC = 90°时,
∵∠B = 20°,∠B + ∠BCD = 90°,
∴∠BCD = 90° - ∠B = 90° - 20° = 70°;
如图②,当∠ACD = 90°时,
∵∠A = 60°,∠B = 20°,∠A + ∠B + ∠ACB = 180°,
∴∠ACB = 180° - ∠B - ∠A = 180° - 20° - 60° = 100°,
∴∠BCD = ∠ACB - ∠ACD = 100° - 90° = 10°.综上所述,∠BCD的度数为70°或10°.
13. 70°或10° 解析:分两种情况:如图①,当∠ADC = 90°时,
∵∠B = 20°,∠B + ∠BCD = 90°,
∴∠BCD = 90° - ∠B = 90° - 20° = 70°;
如图②,当∠ACD = 90°时,
∵∠A = 60°,∠B = 20°,∠A + ∠B + ∠ACB = 180°,
∴∠ACB = 180° - ∠B - ∠A = 180° - 20° - 60° = 100°,
∴∠BCD = ∠ACB - ∠ACD = 100° - 90° = 10°.综上所述,∠BCD的度数为70°或10°.
14. 新题型 双空题 如图，在$△ ABC$中，$∠ BAC = 90^{\circ}$，点$D$是$BC$上一点，将$△ ABD$沿$AD$翻折后得到$△ AED$，边$AE$交$BC$于点$F$。若$∠ C - ∠ B = 50^{\circ}$，$∠ BAD = x^{\circ}(0 ＜ x≤ 45)$。
(1)则$∠ B$的度数为
(2)若$△ DEF$中有两个角相等，则$x =$
(1)则$∠ B$的度数为
20
$^{\circ}$；(2)若$△ DEF$中有两个角相等，则$x =$
30
。答案：14. (1)20 解析:
∵∠C - ∠B = 50°,∠C + ∠B = 90°,
∴∠C = 70°,∠B = 20°.
(2)30 解析:
∵∠BAD = x°,
∴∠ADF = ∠B + ∠BAD = (20 + x)°.
∴∠ADB = ∠ADE = 180° - ∠ADF = (160 - x)°,
∴∠FDE = ∠ADE - ∠ADF = (140 - 2x)°.
∵∠B = ∠E = 20°,
∴∠DFE = 180° - ∠E - ∠FDE = (2x + 20)°.当∠FDE = ∠DFE时,140 - 2x = 2x + 20,解得x = 30.当∠DFE = ∠E = 20°时,2x + 20 = 20,解得x = 0.
∵0 ＜ x ≤ 45,
∴不符合题意,舍去.当∠FDE = ∠E = 20°时,140 - 2x = 20,解得x = 60.
∵0 ＜ x ≤ 45,
∴不符合题意,舍去.综上可知,x = 30.
∵∠C - ∠B = 50°,∠C + ∠B = 90°,
∴∠C = 70°,∠B = 20°.
(2)30 解析:
∵∠BAD = x°,
∴∠ADF = ∠B + ∠BAD = (20 + x)°.
∴∠ADB = ∠ADE = 180° - ∠ADF = (160 - x)°,
∴∠FDE = ∠ADE - ∠ADF = (140 - 2x)°.
∵∠B = ∠E = 20°,
∴∠DFE = 180° - ∠E - ∠FDE = (2x + 20)°.当∠FDE = ∠DFE时,140 - 2x = 2x + 20,解得x = 30.当∠DFE = ∠E = 20°时,2x + 20 = 20,解得x = 0.
∵0 ＜ x ≤ 45,
∴不符合题意,舍去.当∠FDE = ∠E = 20°时,140 - 2x = 20,解得x = 60.
∵0 ＜ x ≤ 45,
∴不符合题意,舍去.综上可知,x = 30.
15. 如图①，$△ ABC$各角的平分线$AD$，$BE$，$CF$相交于点$O$。
（1）试说明$∠ BOC = 90^{\circ} + \frac{1}{2}∠ BAC$；
（2）如图②，过点$O$作$OG⊥ BC$于点$G$，试判断$∠ BOD$与$∠ COG$的大小关系（大于，小于或等于），并说明理由。
（1）试说明$∠ BOC = 90^{\circ} + \frac{1}{2}∠ BAC$；
（2）如图②，过点$O$作$OG⊥ BC$于点$G$，试判断$∠ BOD$与$∠ COG$的大小关系（大于，小于或等于），并说明理由。
答案：15. (1)
∵BO,CO分别平分∠ABC,∠ACB,
∴∠OBC = $\frac{1}{2}$∠ABC,∠OCB = $\frac{1}{2}$∠ACB.
∴∠BOC = 180° - ∠OBC - ∠OCB = 180° - $\frac{1}{2}$(∠ABC + ∠ACB)=180° - $\frac{1}{2}$(180° - ∠BAC)=180° - 90° + $\frac{1}{2}$∠BAC = 90° + $\frac{1}{2}$∠BAC.
(2)∠BOD = ∠COG.理由如下:
∵△ABC各角的平分线AD,BE,CF相交于点O,
∴∠ABO = $\frac{1}{2}$∠ABC,∠BAO = $\frac{1}{2}$∠BAC,∠OCG = $\frac{1}{2}$∠ACB.
∵∠BOD + ∠AOB = 180°,且∠ABO + ∠BAO + ∠AOB = 180°,
∴∠BOD = ∠ABO + ∠BAO = $\frac{1}{2}$(∠ABC + ∠BAC)= $\frac{1}{2}$(180° - ∠ACB)=90° - ∠OCG.
∵OG⊥BC于点G,
∴∠OGC = 90°,
∴∠COG = 90° - ∠OCG,
∴∠BOD = ∠COG.
∵BO,CO分别平分∠ABC,∠ACB,
∴∠OBC = $\frac{1}{2}$∠ABC,∠OCB = $\frac{1}{2}$∠ACB.
∴∠BOC = 180° - ∠OBC - ∠OCB = 180° - $\frac{1}{2}$(∠ABC + ∠ACB)=180° - $\frac{1}{2}$(180° - ∠BAC)=180° - 90° + $\frac{1}{2}$∠BAC = 90° + $\frac{1}{2}$∠BAC.
(2)∠BOD = ∠COG.理由如下:
∵△ABC各角的平分线AD,BE,CF相交于点O,
∴∠ABO = $\frac{1}{2}$∠ABC,∠BAO = $\frac{1}{2}$∠BAC,∠OCG = $\frac{1}{2}$∠ACB.
∵∠BOD + ∠AOB = 180°,且∠ABO + ∠BAO + ∠AOB = 180°,
∴∠BOD = ∠ABO + ∠BAO = $\frac{1}{2}$(∠ABC + ∠BAC)= $\frac{1}{2}$(180° - ∠ACB)=90° - ∠OCG.
∵OG⊥BC于点G,
∴∠OGC = 90°,
∴∠COG = 90° - ∠OCG,
∴∠BOD = ∠COG.