《2025年高效精练八年级数学上册苏科版》

12. (2024 秋·枣阳市期末)如图,$AD=BC,AC=BD$,求证:$\triangle EAB$是等腰三角形.

证明:在$\triangle ADB$和$\triangle BCA$中,$AD = BC$,$AC = BD$,____$AB = BA$____,$\therefore \triangle ADB \cong \triangle BCA$(SSS),$\therefore$____$\angle DBA = \angle CAB$____,$\therefore$____$AE = BE$____,$\therefore \triangle EAB$是等腰三角形.

答案： 证明:在$\triangle ADB$和$\triangle BCA$中,$AD = BC$,$AC = BD$,$AB = BA$,$\therefore \triangle ADB \cong \triangle BCA$(SSS),$\therefore \angle DBA = \angle CAB$,$\therefore AE = BE$,$\therefore \triangle EAB$是等腰三角形.

13. 如图,在$\triangle ABC$中,$AB=AC$,点 D,E,F 分别在 AB,BC,AC 边上,且$BE=CF,BD=CE.$
(1) 求证:$\triangle DEF$是等腰三角形;
(2) 当$∠A=40^{\circ }$时,求$∠DEF$的度数.

答案：
证明:$\because AB = AC$,$\therefore \angle ABC = \angle ACB$,在$\triangle DBE$和$\triangle ECF$中,$\begin{cases} BE = CF, \\ \angle ABC = \angle ACB, \\ BD = CE, \end{cases}$$\therefore \triangle DBE \cong \triangle ECF$(SAS),$\therefore DE = EF$,$\therefore \triangle DEF$是等腰三角形;
(2) 如图,$\because \triangle DBE \cong \triangle ECF$,$\therefore \angle 1 = \angle 3$,$\angle 2 = \angle 4$.$\because \angle A + \angle B + \angle C = 180^{\circ}$,$\therefore \angle B = \frac{1}{2}(180^{\circ} - 40^{\circ}) = 70^{\circ}$.$\therefore \angle 1 + \angle 2 = 110^{\circ}$,$\therefore \angle 3 + \angle 2 = 110^{\circ}$,$\therefore \angle DEF = 70^{\circ}$.

14. (2023 秋·枣阳市期末)如图,在$\triangle ABC$中,AD 平分$∠BAC$,E 是 BC 上一点,$BE=CD$,$EF// AD$交 AB 于 F 点,交 CA 的延长线于 P,$CH// AB$交 AD 的延长线于点 H.
(1) 求证:$\triangle APF$是等腰三角形;
(2) 猜想 AB 与 PC 的大小有什么关系? 证明你的猜想.

答案：

(1) 如图,证明:$\because EF // AD$,$\therefore \angle 1 = \angle 4$,$\angle 2 = \angle P$.$\because AD$平分$\angle BAC$,$\therefore \angle 1 = \angle 2$,$\therefore \angle 4 = \angle P$,$\therefore AF = AP$,即$\triangle APF$是等腰三角形;
(2)$AB = PC$.理由如下:证明:$\because CH // AB$,$\therefore \angle 5 = \angle B$,$\angle H = \angle 1$.$\because EF // AD$,$\therefore \angle 1 = \angle 3$,$\therefore \angle H = \angle 3$,在$\triangle BEF$和$\triangle CDH$中,$\begin{cases} \angle 5 = \angle B, \\ \angle H = \angle 3, \\ BE = CD, \end{cases}$$\therefore \triangle BEF \cong \triangle CDH$(AAS),$\therefore BF = CH$.$\because AD$平分$\angle BAC$,$\therefore \angle 1 = \angle 2$,$\therefore \angle 2 = \angle H$,$\therefore AC = CH$,$\therefore AC = BF$.$\because AB = AF + BF$,$PC = AP + AC$,$\therefore AB = PC$.

15. (2024 秋·安庆期末)已知在$\triangle ABC$中,$AB=AC$,点 D 是边 AB 上一点,$∠BCD=∠A.$
(1) 如图 1,试说明$CD=CB$的理由;
$\because AB = AC$,$\therefore \angle ABC = \angle ACB$.$\because \angle BDC$是$\triangle ADC$的一个外角,$\therefore \angle BDC = \angle A + \angle ACD$.$\because \angle ACB = \angle BCD + \angle ACD$,$\angle BCD = \angle A$,$\therefore \angle BDC = \angle ACB$,$\therefore \angle ABC = \angle BDC$,$\therefore CD = CB$
(2) 如图 2,过点 B 作$BE⊥AC$,垂足为点 E,BE 与 CD 相交于点 F.
①试说明$∠BCD=2∠CBE$的理由;
$\because BE \perp AC$,$\therefore \angle BEC = 90^{\circ}$,$\therefore \angle CBE + \angle ACB = 90^{\circ}$,设$\angle CBE = \alpha$,则$\angle ACB = 90^{\circ} - \alpha$,$\therefore \angle ACB = \angle ABC = \angle BDC = 90^{\circ} - \alpha$,$\therefore \angle BCD = 180^{\circ} - \angle BDC - \angle ABC = 180^{\circ} - (90^{\circ} - \alpha) - (90^{\circ} - \alpha) = 2\alpha$,$\therefore \angle BCD = 2 \angle CBE$
②如果$\triangle BDF$是等腰三角形,求$∠A$的度数.
$\because \angle BFD$是$\triangle CBF$的一个外角,$\therefore \angle BFD = \angle CBE + \angle BCD = \alpha + 2\alpha = 3\alpha$,分三种情况:①当$BD = BF$时,$\therefore \angle BDC = \angle BFD = 3\alpha$.$\because \angle ACB = \angle ABC = \angle BDC = 90^{\circ} - \alpha$,$\therefore 90^{\circ} - \alpha = 3\alpha$,$\therefore \alpha = 22.5^{\circ}$,$\therefore \angle A = \angle BCD = 2\alpha = 45^{\circ}$.②当$DB = DF$时,$\therefore \angle DBE = \angle BFD = 3\alpha$.$\because \angle DBE = \angle ABC - \angle CBE = 90^{\circ} - \alpha - \alpha = 90^{\circ} - 2\alpha$,$\therefore 90^{\circ} - 2\alpha = 3\alpha$,$\therefore \alpha = 18^{\circ}$,$\therefore \angle A = \angle BCD = 2\alpha = 36^{\circ}$;③当$FB = FD$时,$\therefore \angle DBE = \angle BDF$,$\because \angle BDF = \angle ABC ＞ \angle DBF$,$\therefore$不存在$FB = FD$,综上所述:如果$\triangle BDF$是等腰三角形,$\angle A$的度数为$45^{\circ}$或$36^{\circ}$

答案：
(1)$\because AB = AC$,$\therefore \angle ABC = \angle ACB$.$\because \angle BDC$是$\triangle ADC$的一个外角,$\therefore \angle BDC = \angle A + \angle ACD$.$\because \angle ACB = \angle BCD + \angle ACD$,$\angle BCD = \angle A$,$\therefore \angle BDC = \angle ACB$,$\therefore \angle ABC = \angle BDC$,$\therefore CD = CB$;
(2) ①$\because BE \perp AC$,$\therefore \angle BEC = 90^{\circ}$,$\therefore \angle CBE + \angle ACB = 90^{\circ}$,设$\angle CBE = \alpha$,则$\angle ACB = 90^{\circ} - \alpha$,$\therefore \angle ACB = \angle ABC = \angle BDC = 90^{\circ} - \alpha$,$\therefore \angle BCD = 180^{\circ} - \angle BDC - \angle ABC = 180^{\circ} - (90^{\circ} - \alpha) - (90^{\circ} - \alpha) = 2\alpha$,$\therefore \angle BCD = 2 \angle CBE$; ②$\because \angle BFD$是$\triangle CBF$的一个外角,$\therefore \angle BFD = \angle CBE + \angle BCD = \alpha + 2\alpha = 3\alpha$,分三种情况:①当$BD = BF$时,$\therefore \angle BDC = \angle BFD = 3\alpha$.$\because \angle ACB = \angle ABC = \angle BDC = 90^{\circ} - \alpha$,$\therefore 90^{\circ} - \alpha = 3\alpha$,$\therefore \alpha = 22.5^{\circ}$,$\therefore \angle A = \angle BCD = 2\alpha = 45^{\circ}$.②当$DB = DF$时,$\therefore \angle DBE = \angle BFD = 3\alpha$.$\because \angle DBE = \angle ABC - \angle CBE = 90^{\circ} - \alpha - \alpha = 90^{\circ} - 2\alpha$,$\therefore 90^{\circ} - 2\alpha = 3\alpha$,$\therefore \alpha = 18^{\circ}$,$\therefore \angle A = \angle BCD = 2\alpha = 36^{\circ}$;③当$FB = FD$时,$\therefore \angle DBE = \angle BDF$,$\because \angle BDF = \angle ABC ＞ \angle DBF$,$\therefore$不存在$FB = FD$,综上所述:如果$\triangle BDF$是等腰三角形,$\angle A$的度数为$45^{\circ}$或$36^{\circ}$.