《2025年三维数字课堂八年级数学上册人教版》

12. 如图15-12所示,在△ABC中,AD平分∠BAC,过线段CD上一点E作EG//AD,交AC于点F,交BA的延长线于点G.
(1)求证:△AFG是等腰三角形.
(2)若CE= EF,∠BAC= 80°,求∠B的度数.

答案：
(1)证明:
∵AD平分∠BAC,
∴∠BAD = ∠CAD,
∵EG//AD,
∴∠BAD = ∠G,∠CAD = ∠AFG,
∴∠G = ∠AFG,
∴AF = AG,
∴△AFG是等腰三角形.
(2)解:
∵CE = EF,
∴∠CFE = ∠C.
∵∠AFG = ∠CFE,∠AFG = ∠CAD,
∴∠C = ∠CAD.
∵∠BAC = 80°,AD平分∠BAC,
∴∠C = ∠CAD = 40°,
∴∠B = 180° - ∠BAC - ∠C = 60°.

13. 如图15-13,在△ABC中,AB的垂直平分线EF交BC于点E,交AB于点F,D为线段CE的中点,且BE= AC.
(1)求证AD⊥BC；
(2)若∠C= 70°,求∠BAC的度数.

答案：
(1)证明:连接AE.
∵EF是AB的垂直平分线，
∴BE = AE.
∵BE = AC,
∴AE = AC.
∵D为线段CE的中点，
∴AD⊥BC.
(2)解:
∵AE = BE,
∴∠B = ∠BAE.
∵∠AEC是△ABE的外角，
∴∠AEC = ∠B + ∠BAE = 2∠B.
∵AE = AC,
∴∠AEC = ∠C = 2∠B.
∵∠C = 70°,
∴∠B = 35°.
∵∠B + ∠C + ∠BAC = 180°,
∴∠BAC = 180° - 35° - 70° = 75°.

14. 如图15-14,在△ABC中,已知∠ABC= 3∠C,AD平分∠BAC,BE⊥AD于点E.求证BE= $\frac{1}{2}$(AC-AB).

答案： 证明:延长BE交AC于点F.
∵BF⊥AD,
∴∠AEB = ∠AEF.
∵AD平分∠BAC,
∴∠BAE = ∠FAE.在△ABE和△AFE中,
$\left\{\begin{array}{l} ∠AEB = ∠AEF,\\ AE = AE,\\ ∠BAE = ∠FAE,\end{array}\right.$
∴△ABE≌△AFE(ASA).
∴∠ABF = ∠AFB,AB = AF,BE = EF.
∵∠C + ∠CBF = ∠AFB = ∠ABF,∠ABF + ∠CBF = ∠ABC = 3∠C,
∴∠C + 2∠CBF = 3∠C.
∴∠CBF = ∠C.
∴BF = CF.
∴BE = $\frac{1}{2}$BF = $\frac{1}{2}$CF.
∵CF = AC - AF = AC - AB,
∴BE = $\frac{1}{2}$(AC - AB).

15. 如图15-15,已知E是∠AOB的平分线上一点,EC⊥OB,ED⊥OA,垂足分别为C,D,连接CD,交OE于点F.
(1)求证:OE是CD的垂直平分线.
(2)若∠AOB= 60°,请探究OE,EF之间有什么数量关系,并证明你的结论.

答案：
(1)证明:
∵OE是∠AOB的平分线，EC ⊥OB,ED⊥OA,
∴DE = CE.
在Rt△ODE和Rt△OCE中,$\left\{\begin{array}{l} OE = OE,\\ DE = CE,\end{array}\right.$
∴Rt△ODE≌Rt△OCE(HL).
∴OD = OC.
∴△DOC是等腰三角形.
∵OE是∠AOB的平分线，
∴OF⊥CD,CF = DF.
∴OE是CD的垂直平分线.
(2)解:OE = 4EF.
证明:
∵OE是∠AOB的平分线，∠AOB = 60°,
∴∠AOE = ∠BOE = 30°.
∵EC⊥OB,ED⊥OA,OE⊥CD,
∴OE = 2DE,∠ODF = ∠OED = 60°.
∴∠EDF = 30°.
∴DE = 2EF.
∴OE = 4EF.