若三角形面积为S,求三角形三条中线所围成三角形的面积

试试.
先推导一下三角形的中线公式.
设△ABC的三个角A,B,C所对的边分别是a,b,c,它们的中点依次为D,E,F,则AD的长可以这样求：在△ABC中,cosB=(a²+c²-b²)/(2ac),在△ABD中,cosB=(a²/4+c²-AD²)/(ac),所以(a²+c²-b²)/(2ac)=(a²/4+c²-AD²)/(ac),
a²+c²-b²=a²/2+2c²-2AD²,2AD²=b²+c²-a²/2=(2b²+2c²-a²)/2,
AD²=(2b²+2c²-a²)/4,
AD=√(2b²+2c²-a²)/2,同理
BE=√(2a²+2c²-b²)/2,
CF=√(2a²+2b²-c²)/2.
设(a+b+c)/2=p,则△ABC的面积S=√[p(p-a)(p-b)(p-c)]
=√[(a+b+c)(b+c-a)(a+c-b)(a+b-c)]/4
=√[(b²+2bc+c²-a²)(a²-b²-c²+2bc)]/4
=√{[2bc+(b²+c²-a²)][2bc-(b²+c²-a²)]}/4
=√[4b²c²-(b²+c²-a²)²]/4
=√[4b²c²-(b²)²-(c²)²-(a²)²-2b²c²+2a²b²+2a²c²]/4
=√[2b²c²+2a²b²+2a²c²-(b²)²-(c²)²-(a²)²]/4.
设(AD+BE+CF)/2=p′,则△ABC三条中线围成的三角形的面积
S′=√[p′(p′-AD)(p′-BE)(p′-CF)]
=√[(AD+BE+CF)(BE+CF-AD)(AD+CF-BE)(AD+BE-CF)]/4
=√{[(AD+BE+CF)(BE+CF-AD)][(AD+CF-BE)(AD+BE-CF)]}/4
=√{[BE²+2BE×CF+CF²-AD²][AD²-BE²-CF²+2BE×CF]}/4
=√{[2BE×CF+(BE²+CF²-AD²)][2BE×CF-(BE²+CF²-AD²)]}/4
=√{4BE²×CF²-(BE²+CF²-AD²)²}/4
=√{4BE²×CF²-[(BE²)²+(CF²)²+(AD²)²+2BE²×CF²-2BE²×AD²-2CF²×AD²]}/4
=√{4BE²×CF²-(BE²)²-(CF²)²-(AD²)²-2BE²×CF²+2BE²×AD²+2CF²×AD²]}/4
=√{2BE²×CF²+2BE²×AD²+2CF²×AD²-(BE²)²-(CF²)²-(AD²)²}/4
=√{BE²(2CF²-BE²)+AD²(2BE²-AD²)+CF²(2AD²-CF²)}/4
=√{(2a²+2c²-b²)/4×(4a²+4b²-2c²-2a²-2c²+b²)/4+(2b²+2c²-a²)/4×(4a²+4c²-2b²-2b²-2c²+a²)4+(2a²+2b²-c²)/4×(4b²+4c²-2a²-2a²-2b²+c²)/4}/4
=√{(2a²+2c²-b²)/4×(2a²+5b²-4c²)/4+(2b²+2c²-a²)/4×(5a²+2c²-4b²)/4+(2a²+2b²-c²)/4×(2b²+5c²-4a²)/4}/4
=√{(2a²+2c²-b²)(2a²+5b²-4c²)+(2b²+2c²-a²)(5a²+2c²-4b²)+(2a²+2b²-c²)(2b²+5c²-4a²)}/16
=√{4(a²)²+8a²b²-4a²c²-5(b²)²+14b²c²-8(c²)²+4(c²)²+8c²a²-4c²b²-5(a²)²+14a²b²-8(b²)²+4(b²)²+8b²c²-4b²a²-5(c²)²+14a²c²-8(a²)²}/16
=√{18a²b²+18b²c²+18c²a²-9(a²)²-9(b²)²-9(c²)²}/16
=3/4×√{2a²b²+2b²c²+2c²a²-(a²)²-(b²)²-(c²)²}/4.
前面已证S=√{2a²b²+2b²c²+2c²a²-(a²)²-(b²)²-(c²)²}/4,所以
S′=3/4×S.
终于做出来了,不知你是否有耐心看完.你的问题使我在情急之下推导出了三角形中线公式.已知三角形三边求三角形面积的公式叫海伦公式,不知道你们学过没有.推导海伦公式,可从S=absinC/2入手.如需帮助请告诉我.