∵∠1=∠2=∠BAC∴∠BEA=∠AFC,∠EBA=∠1-∠BAE=∠BAC-∠BAE=FAC.
又∵AB=AC,∴△BAE≌△ACF,S△ABE=S△AC政神文危己讲凯候乐F.
∵CD=2BD,S△ABC=9,∴S△ADC=2/3*S△ABC=2/3*9=6
∵S△ABE+S△CDF=S△ACF+S△CDF=S△ADC=6
如图,在来自等腰三角形ABC中,AB=AC,AB>BC.点D在边BC上.CD=2BD
jacky
2024-03-22 04:04
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