梯形abcd ad平行bc s三角形aod为S1 S三角形boc为S2 证s梯形abcd为更号S1加更号S2的平方
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梯形abcd ad平行bc s三角形aod为S1 S三角形boc为S2 证s梯形abcd为更号S1加更号S2的平方
![梯形abcd ad平行bc s三角形aod为S1 S三角形boc为S2 证s梯形abcd为更号S1加更号S2的平方](/uploads/image/z/15221992-40-2.jpg?t=%E6%A2%AF%E5%BD%A2abcd+ad%E5%B9%B3%E8%A1%8Cbc+s%E4%B8%89%E8%A7%92%E5%BD%A2aod%E4%B8%BAS1+S%E4%B8%89%E8%A7%92%E5%BD%A2boc%E4%B8%BAS2+%E8%AF%81s%E6%A2%AF%E5%BD%A2abcd%E4%B8%BA%E6%9B%B4%E5%8F%B7S1%E5%8A%A0%E6%9B%B4%E5%8F%B7S2%E7%9A%84%E5%B9%B3%E6%96%B9)
设AD=a,BC=b,S1的高为h1,S2的高为H2,
∴S梯形=1/2(a+b)(H1+H2)
=1/2·a(H1+H2)+1/2·b(H1+H2)(1),
又(√S1+√S2)²=S1²+2√S1S2+S2²
=1/2·aH1+2√1/2·aH1·1/2·bH2+1/2·bH2,
=1/2·aH1+1/2·bH2+bH1(∵a:b=H1:H2,∴aH2=bH1)
=1/2·aH1+1/2·bH1+1/2·bH2+1/2·bH1,
=1/2·H1(a+b)+1/2·b(H1+H2)(2)
∵(1)中1/2·a(H1+H2)=1/2·aH1+1/2·aH2=1/2·aH1+1/2·bH1,
=1/2·H1(a+b)和(2)相等,
∴(1)=(2).,
∴S梯形=1/2(a+b)(H1+H2)
=1/2·a(H1+H2)+1/2·b(H1+H2)(1),
又(√S1+√S2)²=S1²+2√S1S2+S2²
=1/2·aH1+2√1/2·aH1·1/2·bH2+1/2·bH2,
=1/2·aH1+1/2·bH2+bH1(∵a:b=H1:H2,∴aH2=bH1)
=1/2·aH1+1/2·bH1+1/2·bH2+1/2·bH1,
=1/2·H1(a+b)+1/2·b(H1+H2)(2)
∵(1)中1/2·a(H1+H2)=1/2·aH1+1/2·aH2=1/2·aH1+1/2·bH1,
=1/2·H1(a+b)和(2)相等,
∴(1)=(2).,
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