《2025年一遍过八年级数学上册苏科版》

9 如图,在$\triangle ABC$中,$AB = AC$,$AD⊥BC于点D$,$DE⊥AB于点E$,$BF⊥AC于点F$。若$DE = 2$,则$BF$的长为(B)
A.$3$
B.$4$
C.$5$
D.$6$

答案： B 在△ABC中,
∵AB = AC,AD⊥BC于点D,
∴CD = BD,
∴S△ABC = 2S△ABD,
∴$\frac{1}{2}$AC×BF = 2×$\frac{1}{2}$AB×DE.
∵AC = AB,DE = 2,
∴BF = 2DE = 4.

10 一题多解 在螳螂的示意图中,$AB// DE$,$\triangle ABC$是等腰三角形,$∠ABC = 124^{\circ}$,$∠CDE = 72^{\circ}$,则$∠ACD = $( )

A.$16^{\circ}$
B.$28^{\circ}$
C.$44^{\circ}$
D.$45^{\circ}$

答案：
C 通解 如图1,延长ED,交AC于点F.
∵△ABC是等腰三角形,∠ABC = 124°,
∴∠A = ∠ACB = 28°.
∵AB//DE,
∴∠CFD = ∠A = 28°.
∵∠CDE = 72°,
∴∠ACD = ∠CDE - ∠CFD = 72° - 28° = 44°.
另解1 如图2,过点C作CF//AB.
∵AB//DE,
∴CF//DE,
∴∠DCF + ∠CDE = 180°.
∵∠CDE = 72°,
∴∠DCF = 180° - 72° = 108°.
∵△ABC是等腰三角形,∠ABC = 124°,
∴∠A = 28°.
∵CF//AB,
∴∠ACF + ∠A = 180°,
∴∠ACF = 180° - 28° = 152°,
∴∠ACD = ∠ACF - ∠DCF = 44°.
另解2 如图3,延长CD交AB的延长线BH于点G.
∵DE//AB,
∴∠CGH = ∠CDE = 72°.
∵AB = BC,∠ABC = 124°,
∴∠A = ∠ACB = 28°,
∴∠ACD = ∠CGH - ∠A = 72° - 28° = 44°.

11 [2025 扬州邗江实验中学期中]$\triangle ABC的两边AB$,$AC的垂直平分线分别交直线BC于D$,$E$,连接$AD$,$AE$。若$∠BAC + ∠DAE = 150^{\circ}$,则$∠BAC$的度数为______。

答案：
30°或110° 当∠BAC是锐角时,如图1.
∵△ABC的两边AB,AC的垂直平分线分别交直线BC于点D,E,
∴AD = BD,AE = CE,
∴∠ABD = ∠BAD = ∠BAC + ∠CAD,∠CAE = ∠ACE.
∵∠DAE = 150° - ∠BAC,
∴∠CAE = ∠ACE = 150° - ∠BAC - ∠CAD.
∵∠ABC + ∠BAC + ∠ACB = 180°,
∴(∠BAC + ∠CAD) + ∠BAC + (150° - ∠BAC - ∠CAD) = 180°,解得∠BAC = 30°.当∠BAC是直角时,不符合题意;当∠BAC是钝角时,如图2.
∵△ABC的两边AB,AC的垂直平分线分别交直线BC于点D,E,
∴AD = BD,AE = CE,
∴∠ABD = ∠BAD = ∠BAC - ∠CAD,∠CAE = ∠ACE.
∵∠DAE = 150° - ∠BAC,
∴∠CAE = ∠ACE = ∠CAD - (150° - ∠BAC) = ∠CAD - 150° + ∠BAC.
∵∠ABC + ∠BAC + ∠ACB = 180°,
∴(∠BAC - ∠CAD) + ∠BAC + (∠CAD - 150° + ∠BAC) = 180°,解得∠BAC = 110°.综上,∠BAC的度数为30°或110°.

12 [2024 南京联合体期末]如图,$\triangle AOB和\triangle COD$是等腰直角三角形,$∠AOB = ∠COD = 90^{\circ}$,连接$AD$,$BC$。若$OA = 1$,$OD = 2$,则四边形$ABCD$面积的最大值为______。

答案：
$\frac{9}{2}$ 解题思路:连接AC,BD交于点E,BD交OA于点F,先证△AOC≌△BOD,得出AC = BD,∠OBF = ∠FAE,再由三角形内角和定理推出AC⊥BD,S四边形ABCD = $\frac{1}{2}$AC²,然后由当AC = OA + OC = 3时,AC取最大值,此时S四边形ABCD的值最大,即可得出结果.
如图,连接AC,BD交于点E,BD交OA于点F(等腰直角手拉手).
∵△AOB和△COD是等腰直角三角形,∠AOB = ∠COD = 90°,OA = OB,OC = OD,
∴∠COD + ∠AOD = ∠AOB + ∠AOD,即∠AOC = ∠BOD.在△AOC和△BOD中,OA = OB,∠AOC = ∠BOD,OC = OD,
∴△AOC≌△BOD(SAS),
∴AC = BD,∠FAE = ∠OBF.又
∵∠BFO = ∠AFE,
∴∠BOF = ∠AEF = 90°,
∴AC⊥BD,
∴S四边形ABCD = S△ABC + S△ACD = $\frac{1}{2}$BE·AC + $\frac{1}{2}$DE·AC = $\frac{1}{2}$AC·(BE + DE) = $\frac{1}{2}$AC·BD = $\frac{1}{2}$AC².
∵OA + OC ≥ AC,
∴当AC = OA + OC = 1 + 2 = 3时,AC取最大值,此时S四边形ABCD的值最大,
∴四边形ABCD面积的最大值为$\frac{1}{2}$×3² = $\frac{9}{2}$.

13 [2024 南通海安海陵中学期中]在$\triangle ABC$中,$AB = AC$,$∠BAC = 45^{\circ}$。
(1)如图,点$D在AB$边上,点$E在AC$边上,$BD = CE$,$BE与CD交于点F$。求证：$BF = CF$。
(2)若$D是AB$边上的一个动点,$E是AC$边上的一个动点,且$BD = CE$,$BE与CD交于点F$。当$\triangle BFD$是等腰三角形时,求$∠FBD$的度数。

答案：
(1)证明:
∵AB = AC,
∴∠ABC = ∠ACB.在△BCD和△CBE中,BC = CB,∠DBC = ∠ECB,BD = CE,
∴△BCD≌△CBE(SAS),
∴∠FCB = ∠FBC,
∴BF = CF.
(2)解:
∵AB = AC,∠BAC = 45°,
∴∠ABC = ∠ACB = $\frac{1}{2}$(180° - ∠BAC) = 67.5°.由
(1)知∠FBC = ∠FCB,
∴∠DBF = ∠ECF.设∠FBD = ∠ECF = x,则∠FBC = ∠FCB = 67.5° - x,∠BDF = ∠ECF + ∠BAC = x + 45°,∠DFB = 2∠FBC = 2(67.5° - x) = 135° - 2x.
∵△BFD是等腰三角形,
∴分三种情况讨论:①当BD = BF时,∠BDF = ∠DFB,
∴x + 45° = 135° - 2x,解得x = 30°,即∠FBD = 30°;②当BD = DF时,∠FBD = ∠DFB,
∴x = 135° - 2x,解得x = 45°,即∠FBD = 45°;③当BF = DF时,∠FBD = ∠FDB,
∴x = x + 45°,不合题意,舍去.综上所述,∠FBD的度数为30°或45°.

14 推理能力 教材复习题变式 问题：如图,在$\triangle ABD$中,$BA = BD$。在$BD的延长线上取点E$,$C$,作$\triangle AEC$,使$EA = EC$。若$∠BAE = 90^{\circ}$,$∠B = 45^{\circ}$,求$∠DAC$的度数。
答案：$∠DAC = 45^{\circ}$。
思考：
(1)如果把以上“问题”中的条件“$∠B = 45^{\circ}$”去掉,其余条件不变,那么$∠DAC$的度数会改变吗？说明理由。
(2)如果把以上“问题”中的条件“$∠B = 45^{\circ}$”去掉,再将“$∠BAE = 90^{\circ}$”改为“$∠BAE = n^{\circ}$”,其余条件不变,求$∠DAC$的度数。

答案： 解:
(1)∠DAC的度数不会改变.理由如下:
∵BA = BD,
∴∠BAD = ∠BDA,
∴∠BAD = $\frac{1}{2}$(180° - ∠B).
∵EA = EC,
∴∠EAC = ∠ECA = $\frac{1}{2}$(180° - ∠AEC) = $\frac{1}{2}$∠BEA = $\frac{1}{2}$(180° - ∠B - ∠BAE),
∴∠DAC = ∠BAE - ∠BAD + ∠EAC = ∠BAE - $\frac{1}{2}$(180° - ∠B) + $\frac{1}{2}$(180° - ∠B - ∠BAE) = $\frac{1}{2}$∠BAE = $\frac{1}{2}$×90° = 45°.
(2)设∠B = m°,则∠BAD = $\frac{1}{2}$(180° - m°) = 90° - $\frac{1}{2}$m°,∠AEB = 180° - n° - m°,
∴∠DAE = ∠BAE - ∠BAD = n° - 90° + $\frac{1}{2}$m°.
∵EA = EC,
∴∠CAE = $\frac{1}{2}$∠AEB = 90° - $\frac{1}{2}$n° - $\frac{1}{2}$m°,
∴∠DAC = ∠DAE + ∠CAE = n° - 90° + $\frac{1}{2}$m° + 90° - $\frac{1}{2}$n° - $\frac{1}{2}$m° = $\frac{1}{2}$n°.