由內分比定理
BD:CD = AB:AC = c:b
BD = ac/(b + c),CD = ab/(b + c)
同理
AF = bc/(a + b),BF = ac/(a + b)
AE = bc/(a + c),CE = ab/(a + c)
△AEF/△ABC = (AE * AF)/(AB * AC) = bc/[(a + b)(a + c)]
△BDF/△ABC = (BD * BF)/(AB * AC) = ac/[(a + b)(b + c)]
△CDE/△ABC = (CD * CE)/(AB * AC) = ab/[(a + c)(b + c)]
△DEF/△ABC = 1 - bc/[(a + b)(a + c)] - ac/[(a + b)(b + c)] - ab/[(a + c)(b + c)] = 2abc/[(a + b)(b + c)(a + c)]
這樣做的話,後面的題目也不用做了
