《2025年学霸题中题八年级数学上册人教版》

1. (2024·东莞期中)如图①,在$\triangle ABC$中,$∠BAC = 90^{\circ},AB = AC,∠ABC = 45^{\circ},MN$是经过点$A$的直线,$BD⊥MN$于点$D,CE⊥MN$于点$E$.
(1)求证:$BD = AE$.
(2)若将$MN$绕点$A$旋转,使$MN$与$BC$相交于点$G$,如图②,其他条件不变,求证:$BD = AE$.
(3)在(2)的情况下,若$CE$的延长线过$AB$的中点$F$(如图③),连接$GF$,求证:$∠1 = ∠2$.

答案：

(1) $\because ∠BAC = 90^{\circ}, \therefore ∠BAD + ∠EAC = 90^{\circ}. \because BD ⊥ MN$ 于点 $D$，$CE ⊥ MN$ 于点 $E, \therefore ∠ADB = ∠AEC = 90^{\circ}, \therefore ∠ECA + ∠EAC = 90^{\circ}$，$\therefore ∠ECA = ∠BAD. \because AB = AC, \therefore △ABD ≌ △CAE(AAS), \therefore BD = AE$.
(2) $\because BD ⊥ MN$ 于点 $D, CE ⊥ MN$ 于点 $E, \therefore ∠BDA = ∠CEA = 90^{\circ}$.
$\because ∠BAD + ∠CAE = 90^{\circ}, ∠ACE + ∠CAE = 90^{\circ}, \therefore ∠ACE = ∠BAD$.
在 $△ABD$ 和 $△CAE$ 中，$\left\{\begin{array}{l} ∠BDA = ∠AEC, \\ ∠BAD = ∠ACE, \\ AB = CA, \end{array}\right. \therefore △ABD ≌ △CAE$ (AAS), $\therefore BD = AE$.
(3) 如图，过点 $B$ 作 $BP // AC$ 交 $MN$ 于点 $P, \because BP // AC, \therefore ∠PBA + ∠BAC = 180^{\circ}. \because ∠BAC = 90^{\circ}, \therefore ∠PBA = ∠BAC = 90^{\circ}$，由
(2) 可得，$∠BAD = ∠ACE$，在 $△BAP$ 和 $△ACF$ 中，$\left\{\begin{array}{l} ∠PBA = ∠FAC, \\ AB = AC, \\ ∠BAP = ∠ACF, \end{array}\right. \therefore △BAP ≌ △ACF(ASA), \therefore ∠1 = ∠BPA, AF = BP. \because BF = AF, \therefore BF = BP$.
$\because △ABC$ 是等腰直角三角形，$\therefore ∠ABC = 45^{\circ}. \because ∠PBA = 90^{\circ}$，
$\therefore ∠PBG = 45^{\circ}, \therefore ∠ABG = ∠PBG$，在 $△BFG$ 和 $△BPG$ 中，
$\left\{\begin{array}{l} BF = BP, \\ ∠FBG = ∠PBG, \\ BG = BG, \end{array}\right. \therefore △BFG ≌ △BPG(SAS), \therefore ∠BPG = ∠2$.
$\because ∠BPG = ∠1, \therefore ∠1 = ∠2$.

2. (2024·平顶山期末)如图,在等腰直角三角形$ABC$中,$AB = AC,∠BAC = 90^{\circ}$,点$D$在$BC$边上,连接$AD,AE⊥AD,AE = AD$,连接$CE,DE$.
(1)求证:$∠ACE = ∠B = 45^{\circ}$.
(2)求$∠BCE$的度数.
(3)点$A$关于直线$CE$的对称点为$A_1$,连接$CA_1,EA_1$.补全图形,判断$∠EA_1C$与$∠BAD$之间的数量关系并说明理由.

答案：

(1) $\because AB = AC, ∠BAC = 90^{\circ}, \therefore ∠B = ∠ACB = 45^{\circ}. \because ∠BAC = ∠DAE = 90^{\circ}, \therefore ∠BAD + ∠DAC = ∠CAE + ∠DAC, \therefore ∠BAD = ∠CAE$.
又 $\because AB = AC, AD = AE, \therefore △ABD ≌ △ACE(SAS), \therefore ∠ACE = ∠B = 45^{\circ}$.
(2) 由
(1) 可知，$∠ACE = 45^{\circ}, ∠ACB = 45^{\circ}, \therefore ∠BCE = ∠ACE + ∠ACB = 45^{\circ} + 45^{\circ} = 90^{\circ}$.
(3) 如图，$∠EA_1C = ∠BAD$，理由如下：

$\because$ 点 $A$ 与 $A_1$ 关于 $CE$ 对称，$\therefore EA = EA_1, CA = CA_1, CE = CE$，
$\therefore △ACE ≌ △A_1CE(SSS), \therefore ∠EA_1C = ∠EAC. \because △ABD ≌ △ACE$，
$\therefore ∠CAE = ∠BAD, \therefore ∠EA_1C = ∠BAD$.

3. (2024·南京月考)如图,在$Rt\triangle ABC$中,$∠BCA = 90^{\circ},AC = BC$,过点$A$作$MN⊥AB$,点$D$在$AB$上(不与点$A,B$重合),作$∠DCE = 45^{\circ}$,$∠DCE$的边$CE$交直线$MN$于点$E$,连接$DE$.
(1)如图①,当点$E$在射线$AM$上时,作$CF⊥CE,CF$交$AB$于点$F$,求证:$CE = CF$;
(2)如图②,当点$E$在射线$AN$上时,写出线段$AE,DE,BD$之间的数量关系,并说明理由;
(3)在(2)的基础上,若$CE$与$AD$交于点$P$,当$P$为$CE$的中点,且四边形$AEDC$的面积比$\triangle BCD$的面积大$16$时,直接写出$\triangle ABC$的面积.

答案：

(1) $\because ∠BCA = 90^{\circ}, AC = BC, \therefore ∠CAB = ∠CBF = 45^{\circ}. \because MN ⊥ AB$，
$\therefore ∠MAB = 90^{\circ}, \therefore ∠CAE = ∠MAB - ∠CAB = 90^{\circ} - 45^{\circ} = 45^{\circ}$，
$\therefore ∠CAE = ∠CBF. \because CF ⊥ CE, \therefore ∠ECF = 90^{\circ}, \therefore ∠ECA + ∠ACF = 90^{\circ}. \because ∠ACF + ∠FCB = ∠BCA = 90^{\circ}, \therefore ∠ECA = ∠FCB$.
在 $△ACE$ 和 $△BCF$ 中，$\left\{\begin{array}{l} ∠CAE = ∠CBF, \\ AC = BC, \\ ∠ECA = ∠FCB, \end{array}\right. \therefore △ACE ≌ △BCF$ (ASA), $\therefore CE = CF$.
(2) 线段 $AE, DE, BD$ 之间的数量关系为 $AE + BD = DE$，理由如下：
如图①，过点 $C$ 作 $CF ⊥ CE$ 交 $AB$ 延长线于点 $F$，同
(1) 得 $△ACE ≌ △BCF, \therefore AE = BF, CE = CF. \because ∠DCE = 45^{\circ}, ∠FCE = 90^{\circ}, \therefore ∠DCF = ∠FCE - ∠DCE = 90^{\circ} - 45^{\circ} = 45^{\circ}, \therefore ∠DCF = ∠DCE$. 又 $\because CD = CD$，
$\therefore △CDE ≌ △CDF(SAS), \therefore DE = DF. \because BF + BD = DF, AE = BF, \therefore AE + BD = DE$.
　　
(3) $S_{△ABC} = 16$. 解析：如图②，过点 $C$ 作 $CH ⊥ AB$ 于点 $H$，则 $∠CHP = 90^{\circ}$，由
(2) 可知，$S_{△ACE} = S_{△BCF}, S_{△CDE} = S_{△CDF}$，
$\therefore S_{四边形AEDC} = S_{△ACE} + S_{△CDE} = S_{△BCF} + S_{△DCF} = S_{△ACE} + S_{△BCF} + S_{△BCD} = 2S_{△ACE} + S_{△BCD}, \therefore S_{四边形AEDC} - S_{△BCD} = 2S_{△ACE}$.
$\because S_{四边形AEDC} - S_{△BCD} = 16, \therefore 2S_{△ACE} = 16, \therefore S_{△ACE} = 8. \because P$ 为 $CE$ 的中点，$\therefore S_{△APE} = S_{△APC} = \frac{1}{2}S_{△ACE} = 4. \because MN ⊥ AB, \therefore ∠EAP = 90^{\circ} = ∠CHP$. 又 $\because ∠CPH = ∠EPA, \therefore △HPC ≌ △APE(AAS)$，
$\therefore S_{△HPC} = S_{△APE} = 4, \therefore S_{△ACH} = S_{△APC} + S_{△HPC} = 4 + 4 = 8. \because AC = BC$，
$CH ⊥ AB, \therefore AH = BH, \therefore S_{△ABC} = 2S_{△ACH} = 2 × 8 = 16$.