![](https://img.fx361.cc/images/2023/0413/6f1ef022987b0d5408455947ce87d8c4979a884f.webp)
![](https://img.fx361.cc/images/2023/0413/358c97bca37d7d9e32d3edc584611d9a899b01b2.webp)
证明① 在文[1]中已证△MNP为正三角形.
∵BC=a,BM=CN=b,∠C=60°,∴CM=a-b,
![](https://img.fx361.cc/images/2023/0413/71e57cfc132a43798bad0a4184609301d7bf9096.webp)
![](https://img.fx361.cc/images/2023/0413/15a31bd190c09640d9d38b8e6522ad4b69ba73d3.webp)
特别的,当点M,N,P为中点时,即b=0.5a,△MNP的面积为△ABC面积的0.25倍.
![](https://img.fx361.cc/images/2023/0413/5abb9d5323391e1fde297c9bbad777961ecb2a1b.webp)
![](https://img.fx361.cc/images/2023/0413/c1ae3772e21884a05a817f2a9ce96d06bfdcd01a.webp)
图3-1
![](https://img.fx361.cc/images/2023/0413/fbe749de7c6e01ca46a5def4c243ba7e9ef14ae8.webp)
![](https://img.fx361.cc/images/2023/0413/df5b5aa158681ac8d4e8eeb9208452442c610435.webp)
证明① 在文[1]中已证∠BQM=60°,△EFQ为正三角形.
![](https://img.fx361.cc/images/2023/0413/6079271ab779c77350bbf5930eee59acef22d918.webp)
∵在△QBM和△CBN中,∠BQM=∠BCN,∠QBM=∠CBN,∴△QBM∽△CBN,
![](https://img.fx361.cc/images/2023/0413/e626bdc92cc1bb59167619e988aca30e61cb8317.webp)
易证△BMQ≌△CNE(ASA),∴BM=CN,QM=EN,
![](https://img.fx361.cc/images/2023/0413/8d32dd2ce0e45daa823b4687d4d532bc88883136.webp)
(1)
![](https://img.fx361.cc/images/2023/0413/e693822ce8de30c78108e760c9a04ec88234a06f.webp)
∴BN2=a2+b2-2a·bcos60°=a2+b2-ab>0,
![](https://img.fx361.cc/images/2023/0413/2c6ef80ad5dc45ce0a3249592944fd17c85df664.webp)
(2)
![](https://img.fx361.cc/images/2023/0413/5732881f6062479f1dfc6752867aae215fbe0e46.webp)
![](https://img.fx361.cc/images/2023/0413/3b2ef938ad55cf292ac4bc878d4bd4fee9289440.webp)
特别的当b=0时,也满足上述式子,此时△EFQ即为△ABC.
![](https://img.fx361.cc/images/2023/0413/77ed82d4e7dd7900f1718e75ac7c4448b56e7ae5.webp)
图3-2
![](https://img.fx361.cc/images/2023/0413/0badca13c957dc7004842fcd4d210e16d7a6dc67.webp)
∵EQ=BN-NQ-BE.
![](https://img.fx361.cc/images/2023/0413/6e6f79c6c8cbf92ef23a2f84f65b51675b855c41.webp)
![](https://img.fx361.cc/images/2023/0413/c0fb69a931fbf5e51a07154bc27f84c70015c23c.webp)
![](https://img.fx361.cc/images/2023/0413/e6854b718b92957363f5b8bebc99a8c5c9279029.webp)
特别的当b=a时,也满足上述式子,此时△EFQ即为△ABC.
![](https://img.fx361.cc/images/2023/0413/c797344cdd1738d1b198d6843293a7a8a78f735d.webp)
![](https://img.fx361.cc/images/2023/0413/e67ff4a5891f0082e66e738a11360a0023471146.webp)
![](https://img.fx361.cc/images/2023/0413/df5b5aa158681ac8d4e8eeb9208452442c610435.webp)
探究命题4 已知,如图4,点M,N,P分别在正三角形ABC(边长为a)的BC,CA,AB的延长线上,且BM=CN=AP=b(b>a),AM,BN交于点Q,BN,CP交于点E,CP,AM交于点F.
![](https://img.fx361.cc/images/2023/0413/af279801fb4eb87f89680048d424d704f06514e1.webp)
图4
![](https://img.fx361.cc/images/2023/0413/a34f27464775c82989c0aabff120cff55f7ca776.webp)
![](https://img.fx361.cc/images/2023/0413/1697e10578155dc41fd4deda821958a7e0748ae3.webp)
证明① 在文[1]中已证△EFQ为正三角形.由题意可得:BC=a(a>0),BM=AP=b(b>a),∠BCN=60°,∴∠PBC=120°,
∴在△CBP中,CP2=BC2+BP2-2BC·BP·cos∠PBC(余弦定理),
∴CP2=a2+(b-a)2-2a·(b-a)cos120°=a2+b2-ab>0,
易证△BPE≌△CMF(AAS),∴BP=CM=b-a,EP=FM,易证△BCP∽△FCM,
![](https://img.fx361.cc/images/2023/0413/086996838dd7fcc2b00a6b51dac3550cf7324a70.webp)
(3)
(4)
![](https://img.fx361.cc/images/2023/0413/7700438edbde7e44ff6cc7470262e684172ae9ab.webp)
![](https://img.fx361.cc/images/2023/0413/62f94c6d9a58925eaa0bc53a756e7bd3189bb539.webp)
![](https://img.fx361.cc/images/2023/0413/a34f27464775c82989c0aabff120cff55f7ca776.webp)
![](https://img.fx361.cc/images/2023/0413/ff7487b9c85da7d8b8264c5ba0cc5910cc47293d.webp)
![](https://img.fx361.cc/images/2023/0413/1697e10578155dc41fd4deda821958a7e0748ae3.webp)
二、结束语
一题多变,抛砖引玉,希望能开阔学生的视野,找到解题的灵感,使类似的问题迎刃而解.有纰漏之处,敬请读者指正.
[1]杨川.发现之旅:由正三角形“衍生”出正三角形[J].考试与评价,2016(8).
[2]程峰.探究与分点有关的两个三角形面积的比值[J].初中数学教与学,2011(23).