初二几何题，在线求解，有附图

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2 | 2011-11-7 11:47:24 | 显示全部楼层 |阅读模式

因为正三角形fc=ec/2af=ac-ec/2db=eb/2ad=ab-db=ab-eb/2d到af的距离=√3/2 ad =√3/2 (ab-eb/2)Sadf=√3/4 (ab-eb/2)*(ac-ec/2)
=√3/4 (ab-eb/2)*[ab-(bc-be)/2]
=√3/4 (ab-eb/2)*[ab-(ab-eb)/2]
=√3/4 (ab-eb/2)*(ab/2+eb/2)
=√3/16 (2ab-eb)*(ab+eb)
=√3/16 (2ab2+ab*eb-eb2)Sbcfd=Sabc-Sadf=√3/4...

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