神马作文网这题的算法、思路.题,如图：
来源：学生作业帮 编辑：神马作文网作业帮 分类：综合作业 时间：2024/11/10 16:08:17
这题的算法、思路.题,如图：
设AE、CD相交于G点、AE、BF相交于H点,CD、BF相交于M点
AF=FC => S(△ABF) = S(CBF),S(AMF) = S(CMF) => S(ABM)=S(BCM)
BD=2AD =>S(BCD)=2S(ACD),S(BMD)=2S(ADM) => S(BCM)=2S(ACM)
S(ABM)+S(BCM)+S(ACM)=S(ABC) = 1
=> S(ABM)=S(BCM)=2/5,S(ACM)=1/5
S(BMD)+S(AMD)=S(ABM),S(BMD)=2S(ADM)
=>S(BMD)=2/3
BD=2AD =>S(BCD)=2S(ACD),S(BGD)=2S(ADG) => S(BCG)=2S(ACG)
同理CE=3BE =>S(ACG)=3S(ABG)
S(BCG)+S(ACG)+S(ABG)=S(ABC)=1
=>S(ABG)=1/10,S(ACG)=3/10,S(BCG)=3/5
BD=2AD => S(BGD)=2S(ADG) => S(ADG)=1/30
同上,S(ABH)=S(BCH),S(ACH)=3S(ABH)
=> S(ABH)=S(BCH)=1/5,S(ACH)=3/5
S(GHM)=S(BMD)+S(ADG)-S(ABH)
=2/3+1/30-3/5
=1/10
算了半天,
AF=FC => S(△ABF) = S(CBF),S(AMF) = S(CMF) => S(ABM)=S(BCM)
BD=2AD =>S(BCD)=2S(ACD),S(BMD)=2S(ADM) => S(BCM)=2S(ACM)
S(ABM)+S(BCM)+S(ACM)=S(ABC) = 1
=> S(ABM)=S(BCM)=2/5,S(ACM)=1/5
S(BMD)+S(AMD)=S(ABM),S(BMD)=2S(ADM)
=>S(BMD)=2/3
BD=2AD =>S(BCD)=2S(ACD),S(BGD)=2S(ADG) => S(BCG)=2S(ACG)
同理CE=3BE =>S(ACG)=3S(ABG)
S(BCG)+S(ACG)+S(ABG)=S(ABC)=1
=>S(ABG)=1/10,S(ACG)=3/10,S(BCG)=3/5
BD=2AD => S(BGD)=2S(ADG) => S(ADG)=1/30
同上,S(ABH)=S(BCH),S(ACH)=3S(ABH)
=> S(ABH)=S(BCH)=1/5,S(ACH)=3/5
S(GHM)=S(BMD)+S(ADG)-S(ABH)
=2/3+1/30-3/5
=1/10
算了半天,