2025年中考二轮数学专题训练-几何模型之倍长中线

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2025年中考二轮数学专题训练-几何模型之倍长中线

 

1.已知eqId15c0dbe3c080c4c4636c64803e5c1f76中,

@@@2d79c569-4d5c-446a-845e-a5b081f843b8

(1)如图1,点E为eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的中点,连接eqId68a83fdd2ba72a2dba0b6b10bb3e06b9并延长到点F,使eqId501ee29c1981a063668ce425dc067afd,则eqId274cf35acb4a1748d15c39d15a9bea7b与eqId60ef95894ceebaf236170e8832dcf7e3的数量关系是   .

(2)如图2,若eqId1816fb0bed72025e79aa4b7185055ce5,点E为边eqId60ef95894ceebaf236170e8832dcf7e3上一点,过点C作eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的垂线交eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的延长线于点D,连接eqId03902478df1a55bc99703210bccab910,若eqId4e13b8746bb2d2cd8b06ab849db9c2d6,求证:eqId59f0ad7cca22899a2eaa5f93368cfe5b.

(3)如图3,点D在eqId15c0dbe3c080c4c4636c64803e5c1f76内部,且满足eqId84add90e8bfb8e93f42d0adb76667541,点M在eqId9e52a8f07834cbbbe4224962672fbbb2的延长线上,连接eqIdd50703c46b6153945d718b198f03b4b5交eqIdd40b319212a7e7528b053e1c7097e966的延长线于点N,若点N为eqIdd50703c46b6153945d718b198f03b4b5的中点,求证:eqId6efa18960db26734dbd633a79e115eb6.

2.已知,在正方形eqId5cb3f9a5da641be35117fd35ba07a6aa中,eqId5f2ea13010e2399194be2a681310543e是以点eqIdb6c59076e80b15755ef630fd5f40d1f7为直角顶点的等腰直角三角形,连接eqId5cbe3d509ed0c87eee5034fd2160a88b并取其中点G,连接eqIda174360ec4bde7dc70e806a03166f080、eqId560c48ee30dcf219ccc39c6aab0dad0a.

@@@64a553fd-8a8b-4c16-b6a8-7f682db9b5ad

(1)如图1,若eqId5f2ea13010e2399194be2a681310543e的顶点E在线段eqId83511c460cbd4800bfa13a986c445fea上,则eqIda174360ec4bde7dc70e806a03166f080和eqId560c48ee30dcf219ccc39c6aab0dad0a的关系______;

(2)如图2,若eqId5f2ea13010e2399194be2a681310543e的顶点E在线段eqIdd7d25313067f3e11f68da496344c4956上时,则(1)中的结论是否还成立?请说明理由;

(3)若eqId5f2ea13010e2399194be2a681310543e的顶点E在eqId0764ee446a6280bad108c96c1445f6ca内,如图3位置所示,则(1)中的结论是否还成立?请说明理由.

3.已知eqId90b40c2b0ab8e1cfe5112d428b4b829f和eqId3a0acc93490a6a784eb62201d93dd93d都是等腰直角三角形,eqId0fc75953bf5dcfa4af308c34bf9952d3,连接eqIdb6b0a3fa475b24f57ecd79c681259561,点F为eqId85c4bdfb0db1e31e8459df1d15f9ab55中点.

@@@c4d6465e-02ba-43ec-849e-923bea48c5f3  

(1)如图1,求证:eqId04c243ef4ea81bc314ab9374a81875cf;

(2)将eqId3a0acc93490a6a784eb62201d93dd93d绕C点旋转到如图2所示的位置,连接eqId4e88ee337b7c9082f4fe84fd1752d55a,过C点作eqId23951481e6f657482f2554612fdfa0f4于M点.

①探究eqId68a83fdd2ba72a2dba0b6b10bb3e06b9和eqIdd40b319212a7e7528b053e1c7097e966的关系,并说明理由;

②连接eqId735056c174e8dd7906257a2a50a962a7,求证:F,C,M三点共线.

4.如图,O为四边形ABCD内一点,E为AB的中点,OA=OD,OB=OC,∠AOB+∠COD=eqId95e1dcdc9d50fe147e3924ce30bba519.

(1)若∠BOE=∠BAO,AB=eqId804c767ba8ba0ac1fc157fc345cea965,求OB的长;page number 0

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(2)用等式表示线段OE和CD之间的关系,并证明.

@@@1ac49a41-f0ee-4927-ad71-ff4c9300c2cc

5.某数学兴趣小组在一次活动中进行了探究试验活动,请你来加入.

【探究与发现】

如图1,延长△ABC的边BC到D,使DC=BC,过D作DE∥AB交AC延长线于点E,求证:△ABC≌△EDC.

【理解与应用】

如图2,已知在△ABC中,点E在边BC上且∠CAE=∠B,点E是CD的中点,若AD平分∠BAE.

(1)求证:AC=BD;

(2)若BD=3,AD=5,AE=x,求x的取值范围.

@@@ee686cec-3104-4bc3-9670-9df778454020

6.如图1,在△ABC中,若AB=10,BC=8,求AC边上的中线BD的取值范围.

(1)小聪同学是这样思考的:延长BD至E,使DE=BD,连接CE,可证得△CED≌△ABD.

①请证明△CED≌△ABD;

②中线BD的取值范围是    .

(2)问题拓展:如图2,在△ABC中,点D是AC的中点,分别以AB,BC为直角边向△ABC外作等腰直角三角形ABM和等腰直角三角形BCN,其中,AB=BM,BC=BN,∠ABM=∠NBC=∠90°,连接MN.请写出BD与MN的数量关系,并说明理由.

7.如图,eqId03902478df1a55bc99703210bccab910为eqId15c0dbe3c080c4c4636c64803e5c1f76中eqId0dc5c9827dfd0be5a9c85962d6ccbfb1边上的中线eqIdb0523e3815daddf82022ff865ab18b33.

(1)求证:eqId0d5aef1f9efd3f53b019d04eda78a094;

(2)若eqId7cee49e97546d466ba3ee630e08cdc3c,eqId6f5a152f235942d11c74ef518fca7bba,求eqId03902478df1a55bc99703210bccab910的取值范围.page number 1

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@@@d847ea2d-6ff4-44da-91c6-ea1598706208

8.如图1,在eqId15c0dbe3c080c4c4636c64803e5c1f76中,eqIddb54223bb3fc2fe2497213a4d1f94827是eqIdf52a58fbaf4fea03567e88a9f0f6e37e边的中线,eqId3f63117f593673210c5dbeafee0e0f0f交eqIdf52a58fbaf4fea03567e88a9f0f6e37e延长线于点eqId54a5d7d3b6b63fe5c24c3907b7a8eaa3,eqIdf0aa2665df6d3d72576430f8f1821f43.

 @@@e80b9629183f4c489f6b4d08f282395b

(1)求证eqId60ba1995acdf9b21ef0b8ed91831c362;

(2)如图2,eqId5845ffc7d1af9053b027cf9e726f5367平分eqId8b25f247281afb586b3c686687eae858交eqIddb54223bb3fc2fe2497213a4d1f94827于点eqIddad2a36927223bd70f426ba06aea4b45,交eqId0dc5c9827dfd0be5a9c85962d6ccbfb1于点eqId1dde8112e8eb968fd042418dd632759e,若eqId739dc50373470c8d2d99ee056b94f446,eqId533c67c446f3f34172e24b34fa43c35b,求eqIda50c887867908030d361451a2e85bbe4的值.

9.将一大、一小两个等腰直角三角形拼在一起,eqId2a17c1749dbbbe40b6a8579dd298d047,连接eqId73b3c032441543354c154ee67d744abb.

(1)如图1,若eqIdd322efb382f7cc7a9bb242bcd706d649三点在同一条直线上,则eqId60ef95894ceebaf236170e8832dcf7e3与eqIdd40b319212a7e7528b053e1c7097e966的关系是 ;

@@@23d3da57-7fcb-4dae-8df0-f58e416c4ce9  

(2)如图2,若eqIdd322efb382f7cc7a9bb242bcd706d649三点不在同一条直线上,eqId60ef95894ceebaf236170e8832dcf7e3与eqIdd40b319212a7e7528b053e1c7097e966相交于点eqId2a30f3a8b673cc28bd90c50cf1a35281,连接eqIda299d2b999568e80be8005565ba209a4,猜想eqId028439dc0b4c56ca75e4a734e6bcdef3之间的数量关系,并给予证明;

(3)如图3,在(2)的条件下作eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的中点eqIda0ed1ec316bc54c37c4286c208f55667,连接eqId6cad4595d5352b2884568a59d8d766a4,直接写出eqId03902478df1a55bc99703210bccab910与eqId6cad4595d5352b2884568a59d8d766a4之间的关系.

10.如图1,eqId15c0dbe3c080c4c4636c64803e5c1f76,eqIdc771a4feb150ad9cff8d70431c97eb17都是等腰直角三角形,eqId3522e314fd5b2e39e7280107c6e98bef,eqIdce629f210d757b89a726d2ae851f9ed6,eqIdaada537209e8ab7626e42b17f68907b1,且eqIdc6a46e678bf9d2df5ad4c782b3dc22f5,点eqId8455657dde27aabe6adb7b188e031c11在eqId60ef95894ceebaf236170e8832dcf7e3上,连接eqIdd40b319212a7e7528b053e1c7097e966,eqIdb7cd0d212c71ed7de5f892cc7c9ff72b.

@@@e3f29518-0389-4bdd-b21e-7fd83fd84065

(1)如果eqId78e208452fde94f22b05f291dd28e4e8;

①求eqId2122e3f1e76a635e58e4d54aa594c552的值;

②若eqId0a6936d370d6a238a608ca56f87198de,eqId2c94bb12cee76221e13f9ef955b0aab1是关于eqId81dea63b8ce3e51adf66cf7b9982a248的方程eqId6150565c76b3b2f826ba26ab79d48ded的两根,求eqId294f5ba74cdf695fc9a8a8e52f421328;

(2)如图2,将eqIda25c28359f8d8da9eaf4672a6cf8ae4f绕点eqId5963abe8f421bd99a2aaa94831a951e9逆时针旋转eqId76755791dabbc05f5d842dfbb795a80d.

①在eqId0dc5c9827dfd0be5a9c85962d6ccbfb1上方,与eqId5963abe8f421bd99a2aaa94831a951e9、eqId8455657dde27aabe6adb7b188e031c11、eqId2a30f3a8b673cc28bd90c50cf1a35281同一平面内找一点eqIda0ed1ec316bc54c37c4286c208f55667,使四边形eqId2fa7bbd7831e9ff4f8cffc8889d34f05的面积eqIdcf231f8f86fb922df4ca0c87f044cec3四边形eqIdbbeb51401b008992d4383ae192f75544与四边形eqIda28d6477c85c5a4ac410a884e92fbe53的面积eqIdcf231f8f86fb922df4ca0c87f044cec3四边形eqId01f97f2db9e1d2bca1d9a0b49b203d41相等,并简要说明寻找点eqIda0ed1ec316bc54c37c4286c208f55667的作法;page number 2

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