《2025年轻松作业本八年级数学上册人教版》

17. 如图,△ABC中,AD是∠BAC的平分线,E、F分别为AB、AC上的点,连接DE、DF,∠EDF+∠BAC= 180°.求证：DE= DF.

答案：
证明：在$AB$上截取$AG = AF$，连接$DG$，如图所示。$\because AD$是$\angle BAC$的平分线，$\therefore \angle 1 = \angle 2$。在$\triangle ADG$与$\triangle ADF$中，$\begin{cases}AG = AF\\\angle 1 = \angle 2\\AD = AD\end{cases}$，$\therefore \triangle AGD \cong \triangle AFD$（SAS）。$\therefore \angle AGD = \angle AFD$，$DG = DF$。又$\because \angle AED + \angle EDF + \angle DFA + \angle FAE = 360^{\circ}$，$\angle EDF + \angle BAC = 180^{\circ}$，$\therefore \angle AED + \angle AFD = 180^{\circ}$。又$\angle 4 + \angle AGD = 180^{\circ}$，$\therefore \angle 4 = \angle 3$。$\therefore DE = DG$。$\therefore DE = DF$。

18. 如图,在四边形ABCD中,AD//BC,E为CD的中点,连接AE、BE,BE⊥AE,延长AE,交BC的延长线于点F.求证：
(1)FC= AD；
证明：(1)∵E是CD的中点，∴DE = CE。∵AD // BC，∴∠ADE = ∠FCE，∠DAE = ∠CFE。∴△ADE ≌ △FCE。∴FC = AD；
(2)AB= BC+AD.
证明：(2)∵△ADE ≌ △FCE，∴AE = FE。∵BE ⊥ AE，∴∠AEB = ∠FEB = 90°。又∵BE = BE，∴△ABE ≌ △FBE。∴AB = BF。又∵FB = BC + CF = BC + AD，∴AB = BC + AD。

答案： 证明：（1）$\because E$是$CD$的中点，$\therefore DE = CE$。$\because AD // BC$，$\therefore \angle ADE = \angle FCE$，$\angle DAE = \angle CFE$。$\therefore \triangle ADE \cong \triangle FCE$。$\therefore FC = AD$；
（2）$\because \triangle ADE \cong \triangle FCE$，$\therefore AE = FE$。$\because BE \perp AE$，$\therefore \angle AEB = \angle FEB = 90^{\circ}$。又$\because BE = BE$，$\therefore \triangle ABE \cong \triangle FBE$。$\therefore AB = BF$。又$\because FB = BC + CF = BC + AD$，$\therefore AB = BC + AD$。

19. 如图①,在△ABC中,∠A= 120°,∠C= 20°,BD平分∠ABC,交AC于点D.
(1)求证：BD= CD；
(2)如图②,若∠BAC的角平分线AE交BC于点E,求证：AB+BE= AC；
(3)如图③,若∠BAC的外角平分线AE交CB的延长线于点E,则(2)中的结论是否成立？若成立,给出证明；若不成立,写出正确的结论.

答案：
（1）证明：$\because \angle A = 120^{\circ}$，$\angle C = 20^{\circ}$，$\therefore \angle ABC = 180^{\circ} - 120^{\circ} - 20^{\circ} = 40^{\circ}$。$\because BD$平分$\angle ABC$，$\therefore \angle ABD = \angle DBC = \frac{1}{2} \angle ABC = 20^{\circ}$。$\therefore \angle DBC = \angle C = 20^{\circ}$。$\therefore BD = CD$；
（2）证明：如图①，过点$E$作$EF // BD$交$AC$于点$F$，$\therefore \angle FEC = \angle DBC = 20^{\circ}$。$\therefore \angle FEC = \angle C = 20^{\circ}$。$\therefore \angle AFE = 40^{\circ}$，$FE = FC$。$\therefore \angle AFE = \angle ABC$。$\because AE$是$\angle BAC$的平分线，$\therefore \angle BAE = \angle FAE$。在$\triangle ABE$和$\triangle AFE$中，$\begin{cases}\angle BAE = \angle FAE\\\angle ABE = \angle AFE\\AE = AE\end{cases}$，$\therefore \triangle ABE \cong \triangle AFE$（AAS）。$\therefore BE = EF$。$\therefore BE = EF = FC$，$AB = AF$。$\therefore AB + BE = AF + FC = AC$；
（3）（2）中的结论不成立，正确的结论是$BE - AB = AC$。理由如下：如图③，过点$A$作$AF // BD$交$BE$于点$F$，$\therefore \angle AFC = \angle DBC = 20^{\circ}$。$\therefore \angle AFC = \angle C = 20^{\circ}$。$\therefore AF = AC$。$\because AE$是$\angle BAC$的外角平分线，$\therefore \angle EAB = \frac{1}{2}(180^{\circ} - \angle BAC) = 30^{\circ}$。$\because \angle ABC = 40^{\circ}$，$\therefore \angle E = \angle ABC - \angle EAB = 10^{\circ}$。$\because BD$平分$\angle ABC$，$\therefore \angle ABD = \frac{1}{2} \angle ABC = 20^{\circ}$。$\because AF // BD$，$\therefore \angle FAB = \angle ABD = 20^{\circ}$。$\therefore \angle FAE = \angle EAB - \angle FAB = 10^{\circ}$。$\therefore \angle E = \angle FAE = 10^{\circ}$。$\therefore FE = AF$。$\therefore FE = AF = AC$，$\therefore BE - AB = BE - BF = EF = AC$。