《2025年轻松作业本八年级数学上册人教版》

13. 如图,在△ABC中,AB= AC,点D、E在BC上,连接AD、AE.如果只添加一个条件使∠DAB= ∠EAC,那么添加的条件不能为 (C)

A. BD= CE
B. AD= AE
C. DA= DE
D. BE= CD

答案： C

14. 如图,在△ABC中,AB= AC,D、E是△ABC内两点,AD平分∠BAC,∠EBC= ∠E= 60°.若BE= 6,DE= 2,则BC的长是 (B)

A. 6
B. 8
C. 9
D. 10

答案： B

15. 如图$,S_{△ABC}= 8 cm^2,AD$平分∠BAC,且AD⊥BD于点D,则$S_{△ADC}= $4$cm^2.$

答案： 4

16. (2024·内江中考)如图,在△ABC中,∠DCE= 40°,AE= AC,BC= BD,则∠ACB的度数为______100°.

答案： $100^{\circ}$

17. 如图,等边△ABC的边长为4,AD是BC边上的中线,F是AD边上的动点,E是AC边上一点,若AE= 2,当EF+CF取得最小值时,∠ECF= 30°.

答案： $30^{\circ}$

18. (2025·泗洪一模)如图,AB= AC,DB= DC,点E在AD上.求证:EB= EC.

答案： 解：$\because AB = AC$，$DB = DC$，$\therefore AD$ 是线段 $BC$ 的垂直平分线，$\because$ 点 $E$ 在 $AD$ 上，$\therefore EB = EC$.

19. 如图,在△ABC中,∠ACB= 90°,CM⊥AB于点M,AT平分∠BAC交CM于点D,交BC于点T,过点D作DE//AB交BC于点E.求证:CT= BE.
证明：过点 $T$ 作 $TF \perp AB$ 于点 $F$，$\because AT$ 平分 $\angle BAC$，$\angle ACB = 90^{\circ}$，$\therefore$ CT = TF，$\because \angle ACB = 90^{\circ}$，$CM \perp AB$，$\therefore \angle ADM + \angle DAM = 90^{\circ}$，$\angle ATC + \angle CAT = 90^{\circ}$，$\because AT$ 平分 $\angle BAC$，$\therefore \angle DAM = \angle CAT$，$\therefore \angle ADM = \angle ATC$，$\because \angle ADM = \angle CDT$，$\therefore \angle CDT = \angle CTD$，$\therefore$ CD = CT，又 $\because CT = TF$（已证），$\therefore$ CD = TF，$\because CM \perp AB$，$DE // AB$，$\therefore \angle CDE = 90^{\circ}$，$\angle B = \angle DEC$，在 $\triangle CDE$ 和 $\triangle TFB$ 中，$\begin{cases} \angle DEC = \angle B, \\ \angle CDE = \angle TFB = 90^{\circ}, \\ CD = TF, \end{cases}$ $\therefore$ △CDE ≌ △TFB(AAS)，$\therefore$ CE = TB，$\therefore CE - TE = TB - TE$，即 $CT = BE$.

答案： 证明：过点 $T$ 作 $TF \perp AB$ 于点 $F$，$\because AT$ 平分 $\angle BAC$，$\angle ACB = 90^{\circ}$，$\therefore CT = TF$，$\because \angle ACB = 90^{\circ}$，$CM \perp AB$，$\therefore \angle ADM + \angle DAM = 90^{\circ}$，$\angle ATC + \angle CAT = 90^{\circ}$，$\because AT$ 平分 $\angle BAC$，$\therefore \angle DAM = \angle CAT$，$\therefore \angle ADM = \angle ATC$，$\because \angle ADM = \angle CDT$，$\therefore \angle CDT = \angle CTD$，$\therefore CD = CT$，又 $\because CT = TF$（已证），$\therefore CD = TF$，$\because CM \perp AB$，$DE // AB$，$\therefore \angle CDE = 90^{\circ}$，$\angle B = \angle DEC$，在 $\triangle CDE$ 和 $\triangle TFB$ 中，$\begin{cases} \angle DEC = \angle B, \\ \angle CDE = \angle TFB = 90^{\circ}, \\ CD = TF, \end{cases}$ $\therefore \triangle CDE \cong \triangle TFB(AAS)$，$\therefore CE = TB$，$\therefore CE - TE = TB - TE$，即 $CT = BE$.