加练21 几何探究之探究性问题-【一战成名新中考】2025辽宁中考数学中考必考知识点专题特训

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几何探究之探究性问题

1.【问题初探】

(1)数学活动课上,李老师和同学们共同探究平行线的作用.李老师给出如下问题:

eqId8a11029ca6b4b9e7f777af0280cf163c,点eqIddad2a36927223bd70f426ba06aea4b45为eqId9d78abbad68bbbf12af10cd40ef4c353下方一点,连接eqIda1443bfe022f648f813fb1e15b2d78b6,得到eqIdac3c1375c64dceef45846308a418cf7f,试探究eqIdac3c1375c64dceef45846308a418cf7f与eqIdf42a5f457b561608b86205dc8058dca4的数量关系.

(1)小红的做法是:如图2,过点eqIddad2a36927223bd70f426ba06aea4b45作eqId06a0147f6a16b64d6940505633c1f16a.

(2)小明的做法是:如图3,设eqIdbd33764ff4efddfe11a98a609753715c交eqId9d78abbad68bbbf12af10cd40ef4c353于点eqId2a30f3a8b673cc28bd90c50cf1a35281,过点eqId2a30f3a8b673cc28bd90c50cf1a35281作eqId9d8e1eaccee83bf4158da0f0a4141318.

请你选择一名同学的做法,写出证明过程.

@@@bd2ca5eb-0e36-4a98-96d3-c5e113e71206

【归纳总结】

(2)李老师和同学们发现,在解决题目的过程中,都运用了作平行线的方法,平行线起到了构造等角的作用.为了帮助学生更好的体会平行线的作用,李老师提出了下面问题,请你解答.

如图4,直线eqId8a11029ca6b4b9e7f777af0280cf163c,点eqId2a30f3a8b673cc28bd90c50cf1a35281在eqIdb5c4cd264c97c1f261229925cc5a6761之间,点eqIddad2a36927223bd70f426ba06aea4b45在eqId9d78abbad68bbbf12af10cd40ef4c353下方,连eqId2ec8858389f4c3156a946ba8bf0d8a7b.延长eqId6c8ffe24cf9f327aeb241225ab15ab1a至eqIdf7da7c9981afb256f74c4162846db6d6和eqId55bdea5402c61b5b273656bd74b1ac01的角平分线相交于点eqId2a30f3a8b673cc28bd90c50cf1a35281.探究eqId435f30095577dc714634354f5ad27715与eqIdc45a8a837c11c07073da3ff751d70278的数量关系;

【学以致用】

(3)如图5,eqId5f56df4070fd684f4daa9343d41efe15和eqIde764d6c177965e35b04b6ee72cd1576e的角平分线相交于点eqId2a30f3a8b673cc28bd90c50cf1a35281.作eqId438f34bc8b04e8c494b91306ac6fe352平分eqId3e3311cf0812fc0b3a0cf5b183c69d7e交eqId9ce6c0e9de83f2e64ae33609fc08459d的延长线于点eqIdacc290b44635265137fdf13146b6a6d9,若eqId1b25fc1e76483da7def1369275486d47,求eqIdae6ee579a2565dae0809c25493243135的度数.

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2.【问题初探】如图1,在eqId15c0dbe3c080c4c4636c64803e5c1f76中,eqIdfc34db5860990e51ba31edc8cdd077c2,eqIdf1682d306c38087d9e6f7efb9cec596a,eqId8455657dde27aabe6adb7b188e031c11是eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的中点,求eqId0dc5c9827dfd0be5a9c85962d6ccbfb1边上的中线eqId03902478df1a55bc99703210bccab910的取值范围;

@@@749af49b-bfb8-4251-87db-74ecb38bd909

(1)我们可以用如下的方法解决问题:延长eqId03902478df1a55bc99703210bccab910到点E,使eqId0ad98ad714864041a632ca949308e417,连接eqId85c4bdfb0db1e31e8459df1d15f9ab55,可以判定eqId55e022d6bfafe4ae064677ade0b30f5b_________,所以eqIdd4f76e37d6e6b182896ccaaa187cfe07_________,在eqIde742966e3711cfa53dce04022acf4bcc中,可求出eqId68a83fdd2ba72a2dba0b6b10bb3e06b9的取值范围是_________,则中线eqId03902478df1a55bc99703210bccab910的取值范围是_________;

【类比分析】

(2)如图2,eqId40e7ba6e267dde1a65a98f9f36b585ce的直角顶点eqId8455657dde27aabe6adb7b188e031c11与eqIddd967903ed5a6f640a5b801ec8be0070直角边eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的中点重合,连接eqId4eedae8d316c76e3d0b451256de03fb9,若eqId8c86a77ba954cceb081bf4b37a153c07,eqId39a284cedefae0e06c3b6b4b4f7babe3,eqId16e0c5cb53fd85b7a23f0580df6bb49a,求eqId68a83fdd2ba72a2dba0b6b10bb3e06b9的长.

【方法迁移】

(3)如图3,将一副三角板按如图所示的方式摆放,其中eqIdf73fbc2f33d3402cd96a776a376d9835的直角顶点eqId8455657dde27aabe6adb7b188e031c11与eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的中点重合,边eqIde6e490f703eb6c9bb1278c78ebc2d661和边eqIdf52a58fbaf4fea03567e88a9f0f6e37e交于点eqId895dc3dc3a6606ff487a4c4863e18509,边eqIdd004d2d115b477ade6af7ddb93db0df8和边eqId60ef95894ceebaf236170e8832dcf7e3交于点eqId73465a1f9aa03481295bf6bd3c6903ac,连接eqIde42887d9bf31c1dd99f13c39e63c9ab9.请判断线段eqId06e322e0c87479bba874db9ae9ba36b5,线段eqIda6655e2fa64a32cd12fe0279afd65d73和线段eqIde42887d9bf31c1dd99f13c39e63c9ab9的等量关系,并说明理由.

3.【问题初探】

(1)在数学活动课上,王老师给出如下问题:如图eqIdbdaa19de263700a15fcf213d64a8cd57,正方形eqId411b38a18046fea8e9fab1f9f9b80a5f中,点eqId2a30f3a8b673cc28bd90c50cf1a35281是线段eqId0dc5c9827dfd0be5a9c85962d6ccbfb1上一点(不与点eqId7f9e8449aad35c5d840a3395ea86df6d、eqIdc5db41a1f31d6baee7c69990811edb9f重合),连接eqId1642eec556eb252de9c1ab7bb5ca90b3.将eqId1642eec556eb252de9c1ab7bb5ca90b3绕点eqId2a30f3a8b673cc28bd90c50cf1a35281顺时针旋转eqIdf7a0387fc1258f31e44a10068c0ccfd0得到eqId49b50357a6545cae8348e3059312f520,连接eqId4cae70b8a9d2d2e96dea62c00ced04b9,求eqIda6fc8bfdbf2a7e4ab5ef262b20611785的度数.

①小聪同学给出如下解题思路:过点eqIda0ed1ec316bc54c37c4286c208f55667作eqIda7699dd30e08702bfc6499eb9c89e54a交eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的延长线于点eqId895dc3dc3a6606ff487a4c4863e18509,得到eqIdc63e36329f5e0979f5ee776ac5d06327与eqId85c4bdfb0db1e31e8459df1d15f9ab55的数量关系,进而求出eqIda6fc8bfdbf2a7e4ab5ef262b20611785的度数.

@@@61b8a311-a0fe-4128-adeb-44154710eca4page number 1

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②小慧同学给出另一种解题思路:在eqIdf52a58fbaf4fea03567e88a9f0f6e37e上截取eqIde69d2b798744645af88a4fa411344a83,使得eqId9c7f8da1f66714eba7f2867a19d02479,连接eqId8b46c607b3deac746c0ef3389ad8f65c,得到eqId48ea8be9535f942121f6f6d873d03272与eqIdbeb3748300134501953d6b1c3a905026之间的数量关系,进而求出eqIda6fc8bfdbf2a7e4ab5ef262b20611785的度数.

请你选择一名同学的解题思路,写出证明过程.

【类比分析】

(2)王老师发现之前两名同学都根据图形的特点运用了构造全等的方法,体现了转化的数学思想;为了帮助学生更好地发现数学直觉,感悟转化思想,培养几何直观,王老师将图eqIdbdaa19de263700a15fcf213d64a8cd57进行变换并提出了下面问题,请你解答.

如图eqId61128ab996360a038e6e64d82fcba004,菱形eqId411b38a18046fea8e9fab1f9f9b80a5f中,eqId597ce705d3a2fe04d29de9e81ec6250d,点eqId2a30f3a8b673cc28bd90c50cf1a35281是线段eqId0dc5c9827dfd0be5a9c85962d6ccbfb1上一点(不与点eqId7f9e8449aad35c5d840a3395ea86df6d、eqIdc5db41a1f31d6baee7c69990811edb9f重合),连接eqId1642eec556eb252de9c1ab7bb5ca90b3,将eqId1642eec556eb252de9c1ab7bb5ca90b3绕点eqId2a30f3a8b673cc28bd90c50cf1a35281顺时针旋转eqIde170f206fdbbd834aad7580c727e2cc6得到eqId49b50357a6545cae8348e3059312f520,连接eqId735056c174e8dd7906257a2a50a962a7,探究eqIda6fc8bfdbf2a7e4ab5ef262b20611785与eqIde170f206fdbbd834aad7580c727e2cc6的数量关系.

@@@b7452f9f-e463-4d93-8b3f-1e6c2e0633aa

【学以致用】

(3)如图eqId5ca7d1107389675d32b56ec097464c14,在eqId15c0dbe3c080c4c4636c64803e5c1f76中,eqIded10df4140819d5451773a45de66201b,eqIdef0402dd5ae3db10281f9f1e11738bcb,eqIda8a7b5adfcac0f46a4cd19da4ebb4a2b,eqId8455657dde27aabe6adb7b188e031c11为eqId60ef95894ceebaf236170e8832dcf7e3上一点,满足eqId1748064cf09e888d16f961d7dbea29ac.eqId2a30f3a8b673cc28bd90c50cf1a35281为eqId0dc5c9827dfd0be5a9c85962d6ccbfb1延长线上一点,eqIdb62dcef170114cd71b99f6646d364ef0,eqId68a83fdd2ba72a2dba0b6b10bb3e06b9与eqIdd40b319212a7e7528b053e1c7097e966延长线交于点eqIda0ed1ec316bc54c37c4286c208f55667,求eqIdd004d2d115b477ade6af7ddb93db0df8的长.

@@@cb84d5ed-5a72-4e87-904e-ad8d0bd3ee64

4.【问题初探】

在数学活动课上,王老师提示出了一个问题:如下图1,在平行四边形eqId411b38a18046fea8e9fab1f9f9b80a5f中,点E在eqIdf52a58fbaf4fea03567e88a9f0f6e37e上,点P在eqId9d78abbad68bbbf12af10cd40ef4c353上,eqId755a11be3b49d05b9215f2f8d1f51c35交eqIdfc5adb5eb60ae4435a12d93854066298于点F,eqIda605f3d0cfadd02238c57f67e56ec1ae,eqIdd3625f344d7c82544a8d12e686fc1324.page number 2

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