《2025年轻松作业本八年级数学上册苏科版》

15. 如图,在$\triangle ABC和\triangle AEF$中,$AB= AC$,$AE= AF$,$∠CAB= ∠EAF$。$BE交FC于点O$,交$AC于点D$。
(1)求证：$BE= CF$；
(2)当$∠BAC= 70^{\circ}$时,求$∠BOC$的度数。
(1) 证明：$\because \angle CAB = \angle EAF$，$\therefore \angle CAB + \angle CAE = \angle EAF + \angle CAE$。$\therefore \angle BAE = \angle CAF$。在$\triangle BAE$和$\triangle CAF$中，$\begin{cases} AB = AC, \\ \angle BAE = \angle CAF, \\ AE = AF. \end{cases}$ $\therefore \triangle BAE \cong \triangle CAF(SAS)$。
$\therefore BE = CF$；
(2) 解：$\because \triangle BAE \cong \triangle CAF$，$\therefore \angle EBA = \angle FCA$，即$\angle DBA = \angle OCD$。$\because \angle BDA = \angle ODC$，
$\therefore \angle BAD = \angle COD$。$\because \angle BAC = 70^{\circ}$，$\therefore \angle BAD = 70^{\circ}$。
$\therefore \angle COD = 70^{\circ}$，即$\angle BOC = $70°。

答案：
(1) 证明：$\because \angle CAB = \angle EAF$，$\therefore \angle CAB + \angle CAE = \angle EAF + \angle CAE$。$\therefore \angle BAE = \angle CAF$。在$\triangle BAE$和$\triangle CAF$中，$\begin{cases} AB = AC, \\ \angle BAE = \angle CAF, \\ AE = AF. \end{cases}$ $\therefore \triangle BAE \cong \triangle CAF(SAS)$。
$\therefore BE = CF$；
(2) 解：$\because \triangle BAE \cong \triangle CAF$，$\therefore \angle EBA = \angle FCA$，即$\angle DBA = \angle OCD$。$\because \angle BDA = \angle ODC$，
$\therefore \angle BAD = \angle COD$。$\because \angle BAC = 70^{\circ}$，$\therefore \angle BAD = 70^{\circ}$。
$\therefore \angle COD = 70^{\circ}$，即$\angle BOC = 70^{\circ}$。

16. 如图,在$\triangle ABC$中,点$D在边AC$上,$DB= BC$,$E是CD$的中点,$F是AB$的中点。求证：$EF= \frac{1}{2}AB$。
证明：连接$BE$，在$\triangle BCD$中，$DB = BC$，$E$是$CD$的中点，$\therefore BE \perp CD$。在$Rt\triangle ABE$中，$EF$是斜边$AB$上的中线，$\therefore EF = \frac{1}{2}AB$。

答案： 证明：连接$BE$，在$\triangle BCD$中，$DB = BC$，$E$是$CD$的中点，$\therefore BE \perp CD$。在$Rt\triangle ABE$中，$EF$是斜边$AB$上的中线，$\therefore EF = \frac{1}{2}AB$。

17. (2024·南充)如图,在$\triangle ABC$中,点$D为BC$边的中点,过点$B作BE// AC交AD的延长线于点E$。
(1)求证：$\triangle BDE\cong \triangle CDA$；
证明：$\because$点$D$为$BC$的中点，$\therefore BD = CD$。$\because BE // AC$，$\therefore \angle EBD = \angle C$，$\angle E = \angle CAD$，在$\triangle BDE$和$\triangle CDA$中，$\begin{cases} \angle EBD = \angle C, \\ \angle E = \angle CAD, \\ BD = CD, \end{cases}$ $\therefore \triangle BDE \cong \triangle CDA$(AAS)；
(2)若$AD⊥BC$,求证：$BA= BE$。
证明：$\because$点$D$为$BC$的中点，$AD \perp BC$，$\therefore$直线$AD$为线段$BC$的垂直平分线。$\therefore BA = CA$。由(1)可知：$\triangle BDE \cong \triangle CDA$，$\therefore BE = CA$。$\therefore BA = BE$。

答案： 证明：
(1) $\because$点$D$为$BC$的中点，$\therefore BD = CD$。$\because BE // AC$，$\therefore \angle EBD = \angle C$，$\angle E = \angle CAD$，在$\triangle BDE$和$\triangle CDA$中，$\begin{cases} \angle EBD = \angle C, \\ \angle E = \angle CAD, \\ BD = CD, \end{cases}$ $\therefore \triangle BDE \cong \triangle CDA(AAS)$；
(2) $\because$点$D$为$BC$的中点，$AD \perp BC$，$\therefore$直线$AD$为线段$BC$的垂直平分线。
$\therefore BA = CA$。由
(1)可知：$\triangle BDE \cong \triangle CDA$，$\therefore BE = CA$。
$\therefore BA = BE$。

18. 如图,直线$a$、$b相交于点A$,$C$、$E分别是直线b$、$a$上的点,且$BC⊥a$,$DE⊥b$,点$M$、$N分别是EC$、$DB$的中点。求证：
(1)$MD= MB$；
(2)$MN⊥BD$。

答案： 证明：
(1) $\because BC \perp a$，$DE \perp b$，$\therefore \angle CDE = \angle CBE = 90^{\circ}$。$\therefore \triangle CBE$、$\triangle CDE$为直角三角形。$\because$点$M$是$EC$中点，$\therefore DM = BM = \frac{1}{2}EC$。$\therefore DM = BM$；
(2) $\because DM = BM$，$\therefore \triangle MDB$为等腰三角形。又$\because N$为$BD$的中点，$\therefore MN$为$BD$边上的中线。$\therefore MN \perp BD$（三线合一）。

19. 如图,$AD为线段BC$的垂直平分线,在线段$AD上取一点E$,使得$∠ACE= 20^{\circ}$,在线段$CE上取一点F$,使得$∠FBC= 10^{\circ}$,连接$BE$、$AF$。若$∠ABC= 50^{\circ}$,求证：$BE⊥AF$。
证明：$\because AD$为线段$BC$的垂直平分线，$\therefore AB = AC$，$EB = EC$，$\angle ADB = 90^{\circ}$。$\therefore \angle ABC = \angle ACB = 50^{\circ}$，$\angle EBC = \angle ECB$。
$\therefore \angle ACE = \angle ABE$。$\because \angle ACE = 20^{\circ}$，$\angle FBC = 10^{\circ}$，
$\therefore \angle ABE = 20^{\circ}$。$\therefore \angle FBE = \angle ABC - \angle ABE - \angle FBC = 50^{\circ} - 20^{\circ} - 10^{\circ} = 20^{\circ}$。$\therefore \angle EBC = \angle FBE + \angle FBC = 20^{\circ} + 10^{\circ} = 30^{\circ}$。$\therefore \angle ABF = \angle ABE + \angle FBE = 40^{\circ}$，$\angle ABE = \angle FBE$。$\because \angle BAE = 90^{\circ} - \angle ABD = 40^{\circ}$，$\angle BFE = \angle FBC + \angle FCB = \angle FBC + \angle ECB = 10^{\circ} + 30^{\circ} = 40^{\circ}$，$\therefore \angle BAE =$
$\angle BFE$。在$\triangle ABE$和$\triangle FBE$中，$\begin{cases} \angle ABE = \angle FBE, \\ \angle BAE = \angle BFE, \\ BE = BE. \end{cases}$
$\therefore \triangle ABE \cong \triangle FBE$(AAS)。$\therefore BA = BF$。$\therefore \triangle BAF$是等腰三角形。$\because \angle ABE = \angle FBE$，$\therefore BE$是$\triangle BAF$的顶角的角平分线。$\therefore BE \perp AF$。

答案： 证明：$\because AD$为线段$BC$的垂直平分线，$\therefore AB = AC$，$EB = EC$，$\angle ADB = 90^{\circ}$。$\therefore \angle ABC = \angle ACB = 50^{\circ}$，$\angle EBC = \angle ECB$。
$\therefore \angle ACE = \angle ABE$。$\because \angle ACE = 20^{\circ}$，$\angle FBC = 10^{\circ}$，
$\therefore \angle ABE = 20^{\circ}$。$\therefore \angle FBE = \angle ABC - \angle ABE - \angle FBC = 50^{\circ} - 20^{\circ} - 10^{\circ} = 20^{\circ}$。$\therefore \angle EBC = \angle FBE + \angle FBC = 20^{\circ} + 10^{\circ} = 30^{\circ}$。$\therefore \angle ABF = \angle ABE + \angle FBE = 40^{\circ}$，$\angle ABE = \angle FBE$。$\because \angle BAE = 90^{\circ} - \angle ABD = 40^{\circ}$，$\angle BFE = \angle FBC + \angle FCB = \angle FBC + \angle ECB = 10^{\circ} + 30^{\circ} = 40^{\circ}$，$\therefore \angle BAE =$
$\angle BFE$。在$\triangle ABE$和$\triangle FBE$中，$\begin{cases} \angle ABE = \angle FBE, \\ \angle BAE = \angle BFE, \\ BE = BE. \end{cases}$
$\therefore \triangle ABE \cong \triangle FBE(AAS)$。$\therefore BA = BF$。$\therefore \triangle BAF$是等腰三角形。$\because \angle ABE = \angle FBE$，$\therefore BE$是$\triangle BAF$的顶角的角平分线。$\therefore BE \perp AF$。