《2025年名校题库八年级数学上册北师大版》

7.(金牛区期末)在$\triangle ABC$中,$∠BAC= 135^{\circ },AB= AC$,D为BC边上一点.
(1)如图1,若$AD= AM,∠DAM= 135^{\circ }$.
①求证:$BD= CM;$
②若$∠CMD= 90^{\circ }$,求$\frac {DM}{DC}$的值.
(2)如图2,E为线段CD上一点,且$CE= 1,BC= 4,∠DAE= 67.5^{\circ }$,求DE的长.

答案：

(1)证明:
∵∠DAB = ∠DBA,
∴DA = DB.在△ACD和△BCD中,$\begin{cases} AC = BC, \\ DA = DB, \\ CD = CD, \end{cases}$
∴△ACD ≌ △BCD(SSS).
(2)①证明:
∵∠DAB = ∠DBA = 30°,
∴∠ADB = 120°.
∵△ACD ≌ △BCD,
∴∠ADC = ∠BDC = 120°,∠ACD = ∠BCD = 45°.
∵∠ACB = 90°,AC = BC,
∴∠CAB = ∠CBA = 45°,
∴∠CAE = ∠CBD = 15°,
∴∠CEA = 75°,
∴∠CDE = ∠BDE = 60°.在△CDE和△FDE中,$\begin{cases} CD = FD, \\ \angle CDE = \angle FDE, \\ DE = DE, \end{cases}$
∴△CDE ≌ △FDE(SAS),
∴∠AEF = ∠AEC = 75°.
∵∠DAB = 30°,
∴∠AGE = 180° - ∠AEG - ∠DAB = 180° - 75° - 30° = 75°,
∴∠AEG = ∠AGE,
∴AE = AG. ②2$\sqrt{3} + 2$ $\sqrt{3} + 1$ [解析]如图,连接AF,在AC上取点H,连接EH,使HE = HA,
∴∠HEA = ∠HAE.由①知∠CAE = 15°,
∴∠CHE = ∠HEA + ∠HAE = 30°.
∵CE = 2,
∴EH = 4,
∴CH = $\sqrt{EH^{2} - CE^{2}}$ = 2$\sqrt{3}$,HA = EH = 4,
∴CB = CA = CH + HA = 2$\sqrt{3} + 4$,
∴BE = BC - CE = 2$\sqrt{3} + 2$.
∵△CDE ≌ △FDE,
∴CE = FE.在△ACE和△AFE中,$\begin{cases} CE = FE, \\ \angle AEC = \angle AEF, \\ AE = AE, \end{cases}$
∴△ACE ≌ △AFE(SAS),
∴∠AFE = ∠ACE = 90°.
∵AE = AG,
∴EF = FG,$S_{\triangle AGF} = S_{\triangle AEF} = S_{\triangle ACE}$,
∴$S_{\triangle BGE} = S_{\triangle ABC} - 3S_{\triangle ACE} = \frac{1}{2}×(2\sqrt{3} + 4)^{2} - 3×\frac{1}{2}×(2\sqrt{3} + 4)×2 = 2\sqrt{3} + 2$.
∵EF = FG,
∴$S_{\triangle BGF} = \frac{1}{2}S_{\triangle BGE} = \sqrt{3} + 1$.

8.(树德实验)如图,在$Rt\triangle ABC$中,$∠ACB= 90^{\circ },AC= BC,∠DAB= ∠DBA= 30^{\circ }$,延长AD交BC于点E.
(1)如图1,求证:$\triangle ACD\cong \triangle BCD$.
(2)如图2,F为线段DB上一点,且$CD= FD$,连接EF并延长交AB于点G.
①求证:$AE= AG;$
②若$CE= 2$,填空:$BE= $____,$S_{\triangle BGF}= $____.(不写过程)

答案：

(1)证明:
∵∠BAC = 135° = ∠DAM,
∴∠BAD = ∠CAM.又
∵AB = AC,AD = AM,
∴△ABD ≌ △ACM(SAS),
∴BD = CM. ②解:
∵∠BAC = 135°,AB = AC,
∴∠B = ∠ACB = 22.5°.
∵△ABD ≌ △ACM,
∴∠ACM = ∠B = 22.5°,
∴∠DCM = 45°.又
∵∠CMD = 90°,
∴△DCM是等腰直角三角形,
∴CD = $\sqrt{2}DM$,
∴$\frac{DM}{CD}$ = $\frac{\sqrt{2}}{2}$.
(2)解:
∵CE = 1,BC = 4,
∴BE = 3.
∵∠DAE = 67.5°,∠BAC = 135°,
∴∠BAD + ∠CAE = 67.5° = ∠DAE.如图,将△AEC绕点A顺时针旋转135°,得到△AHB,连接DH,过点D作DN⊥BH于点N,
∴△AEC ≌ △AHB,
∴BH = EC = 1,∠C = ∠ABH = 22.5°,AE = AH,∠EAC = ∠BAH,
∴∠HBD = 45°.
∵DN⊥BH,
∴△BDN是等腰直角三角形,
∴BN = DN,BD = $\sqrt{2}BN$.
∵∠BAD + ∠CAE = 67.5° = ∠BAD + ∠BAH = ∠DAH,
∴∠DAE = ∠DAH.又
∵AD = AD,AE = AH,
∴△ADE ≌ △ADH(SAS),
∴HD = DE.设BD = x,则BN = DN = $\frac{\sqrt{2}}{2}x$,DH = DE = 3 - x,
∴HN = $\frac{\sqrt{2}}{2}x - 1$.
∵在Rt△DNH中,DH² = HN² + DN²,
∴(3 - x)² = ($\frac{\sqrt{2}}{2}x - 1$)² + ($\frac{\sqrt{2}}{2}x$)²,解得$x = \frac{24 + 4\sqrt{2}}{17}$,
∴DE = $\frac{27 - 4\sqrt{2}}{17}$.