《2025年高效精练八年级数学上册苏科版》

7. 如图,在$\triangle ABC$中,$∠ABC$的平分线交 AC 于点 D,过点 D 作$DE// BC$交 AB 于点 E.
(1) 求证:$BE=DE;$
(2) 若$∠A=80^{\circ },∠C=40^{\circ }$,求$∠BDE$的度数.

(1) 证明:在$\triangle ABC$中,$\angle ABC$的平分线交$AC$于点$D$,$\therefore \angle ABD = \angle CBD$.$\because DE // BC$,$\therefore \angle EDB = \angle CBD$,$\therefore \angle EBD = \angle EDB$,$\therefore BE = DE$.
(2)$\because \angle A = 80^{\circ}$,$\angle C = 40^{\circ}$,$\therefore \angle ABC = 60^{\circ}$.$\because \angle ABC$的平分线交$AC$于点$D$,$\therefore \angle ABD = \angle CBD = \frac{1}{2} \angle ABC = 30^{\circ}$,由(1)知$\angle EDB = \angle EBD =$30°.

答案：
(1) 证明:在$\triangle ABC$中,$\angle ABC$的平分线交$AC$于点$D$,$\therefore \angle ABD = \angle CBD$.$\because DE // BC$,$\therefore \angle EDB = \angle CBD$,$\therefore \angle EBD = \angle EDB$,$\therefore BE = DE$.
(2)$\because \angle A = 80^{\circ}$,$\angle C = 40^{\circ}$,$\therefore \angle ABC = 60^{\circ}$.$\because \angle ABC$的平分线交$AC$于点$D$,$\therefore \angle ABD = \angle CBD = \frac{1}{2} \angle ABC = 30^{\circ}$,由
(1)知$\angle EDB = \angle EBD = 30^{\circ}$.

8. (2024 秋·盐城期末)如图,已知$\triangle ABC$中,$∠ACB=90^{\circ },CD⊥AB$于 D,BF 平分$∠ABC$交CD 于 E,交 AC 于 F. 求证:$CE=CF.$
证明:$\because \angle ACB = 90^{\circ}$,$CD \perp AB$,$\therefore \angle CBF + \angle CFB = \angle DBE + \angle DEB = 90^{\circ}$.$\because BF$平分$\angle ABC$,$\therefore \angle CBF = \angle DBE$,$\therefore \angle CFB = \angle DEB$,又$\because \angle FEC = \angle DEB$,$\therefore \angle CFB = \angle FEC$,$\therefore CE = CF$.

答案： 证明:$\because \angle ACB = 90^{\circ}$,$CD \perp AB$,$\therefore \angle CBF + \angle CFB = \angle DBE + \angle DEB = 90^{\circ}$.$\because BF$平分$\angle ABC$,$\therefore \angle CBF = \angle DBE$,$\therefore \angle CFB = \angle DEB$,又$\because \angle FEC = \angle DEB$,$\therefore \angle CFB = \angle FEC$,$\therefore CE = CF$.

9. 如图,在$\triangle ABC$中,$AB=AC,AD⊥BC$于点 D.
(1) 若$∠C=42^{\circ }$,求$∠BAD$的度数;48°
(2) 若点 E 在边 AB 上,$EF// AC$交 AD 的延长线于点 F. 求证:$AE=FE.$
证明:$\because AB = AC$,$AD \perp BC$于点$D$,$\therefore \angle BAD = \angle CAD$.$\because EF // AC$,$\therefore \angle F = \angle CAD$,$\therefore \angle BAD = \angle F$,$\therefore AE = FE$.

答案：
(1)$\because AB = AC$,$AD \perp BC$于点$D$,$\therefore \angle BAD = \angle CAD$,$\angle ADC = 90^{\circ}$,又$\angle C = 42^{\circ}$,$\therefore \angle BAD = \angle CAD = 90^{\circ} - 42^{\circ} = 48^{\circ}$;
(2)$\because AB = AC$,$AD \perp BC$于点$D$,$\therefore \angle BAD = \angle CAD$.$\because EF // AC$,$\therefore \angle F = \angle CAD$,$\therefore \angle BAD = \angle F$,$\therefore AE = FE$.

10. (2025 春·二七区阶段考)如图,在$\triangle ABC$中,$∠ABC,∠ACB$的平分线相交于 F,过 F 作$DE// BC$,交 AB 于 D,交 AC 于 E,那么下列结论正确的有 (B)
①$\triangle BDF,\triangle CEF$都是等腰三角形;②$DE=DB+CE$;③$\triangle ADE$的周长等于$AB+BC$;④$BF=CF.$

A. 1个
B. 2个
C. 3个
D. 4个

答案： B

11. (2024 秋·通许县期末)如图,在格点中找一点 C,使得$\triangle ABC$是等腰三角形,且 AB 为其中的一条腰,这样的点 C 一共有 (C)

A. 3个
B. 4个
C. 5个
D. 6个

答案： C