《2025年53精准练八年级数学下册北师大版山西专版》

1. 如图，在△ABC中，AB = BC，DE⊥BC于点D，交AC于点E.若E是AC的中点，求证：∠CED = $\frac{1}{2}$∠B.

答案：
证明:如图，连接BE.
∵AB = BC,E是AC的中点,
∴BE⊥AC,∠ABE = ∠CBE = $\frac{1}{2}$∠ABC,
∴∠BEC = 90°,
∴∠CED + ∠BED = 90°.
∵ED⊥BC,
∴∠BDE = 90°,
∴∠CBE + ∠BED = 90°,
∴∠CED = ∠CBE,
∴∠CED = $\frac{1}{2}$∠ABC.

2. 如图，在Rt△ABC中，AC = BC，∠C = 90°，O为AB的中点，OE⊥OF，且OE，OF分别交AC，BC于点E，F.求证：OE = OF.

答案：
证明:如图，连接OC.
∵AC = BC,∠ACB = 90°,O为AB的中点,
∴∠B = ∠ACO = ∠BCO = 45°,CO⊥AB,
∴OC = OB,∠COB = 90°,
∴∠COF + ∠FOB = 90°.
∵∠EOF = 90°,
∴∠EOC + ∠COF = 90°,
∴∠FOB = ∠EOC.在△EOC和△FOB中,$\begin{cases} \angle EOC = \angle FOB, \\ OC = OB, \\ \angle OCE = \angle B, \end{cases}$
∴△EOC≌△FOB(ASA),
∴OE = OF.

3. 如图，已知∠AOB = 60°，点P在边OA上，OP = 12，点M、N在边OB上，PM = PN，若OM = 5，求MN的长.

答案：
解:如图,作PC⊥OB于C.
∵PM = PN,PC⊥OB,
∴CM = CN在△OPC中,∠PCO = 90°,∠POC = 60°,OP = 12,
∴∠OPC = 30°,
∴OC = $\frac{1}{2}$OP = 6,
∵OM = 5,
∴CM = OC - OM = 6 - 5 = 1,
∴CN = CM = 1,
∴MN = CM + CN = 1 + 1 = 2.

4. [2024运城期中]如图，$△ABC$中，$AD$为$∠BAC$的平分线，作$BD$垂直$AD$于$D$，$△ACD$的面积为8，则$△ABC$的面积为________.

答案：
16 详解:如图所示,延长BD交AC于E,
∵AD为∠BAC的平分线,AD⊥BD,
∴∠BAD = ∠EAD,∠ADB = ∠ADE = 90°,又
∵AD = AD,
∴△ADB≌△ADE(ASA),
∴BD = DE,
∴$S_{\triangle ADB}=S_{\triangle ADE}$,$S_{\triangle EDC}=S_{\triangle BDC}$,
∵$S_{\triangle ABC}=S_{\triangle ADB}+S_{\triangle ADE}+S_{\triangle EDC}+S_{\triangle BDC}$,
∴$S_{\triangle ADE}+S_{\triangle EDC}=\frac{1}{2}S_{\triangle ABC}$,即$S_{\triangle ABC}=2S_{\triangle ACD}=16$.

5. 如图，在△ABC中，∠BAC = 120°，AD⊥BC于D，且AB + BD = DC，求∠C的度数.

答案：
解: 如图, 在DC上截取DH, 使得DH = DB,连接AH.
∵BD = DH,AD⊥BH,
∴AB = AH.
∵AB + BD = DC,DC = DH + HC,
∴AB = CH = AH,
∴∠B = ∠AHD,∠C = ∠HAC,设∠C = x,则∠AHD = ∠B = 2x,
∵∠B + ∠C + ∠BAC = 180°,
∴3x + 120° = 180°,
∴x = 20°,
∴∠C = 20°.