唔明呀!!唔明呀!!

希羅公式是 Heron's Formula 吧！

Area of triangle ABC = (ab x sin C) / 2
cos A = (b ^2 + c ^2 - a ^2) / 2bc
sin A = √(1 - cos ^2 A)
= √ { 1 - [ ( b ^2 + c ^2 - a ^2) / 2bc ] ^2 }
= √{ [ 1 - (b ^2 + c ^2 - a ^2) / 2bc ] [ 1 + (b ^2 + c ^2 - a ^2) / 2bc] }
= √{ [ (2bc - b ^2 - c ^2 + a ^2) / 2bc ] [ (2bc + b ^2 +c ^2 - a^2) / 2bc] }
= 1 / 2bc x √{ [ a ^2 - (b ^2 - 2bc + c ^2) ] [ (b ^2 + 2bc + c ^2) - a ^2] }
= 1 / 2bc x √{ [ a ^2 - (b - c) ^2] [ (b + c) ^2 - a ^2]}
= 1 / 2bc x √[ (a + b - c) (a - b + c) ( -a + b + c) (a + b + c) ]
= 1 / 2bc x √[ (a + b + c) (b + c - a) ( a + c - b) ( a + b - c)]
= 1 / 2bc x √[2s x 2(s - a) x 2(s - b) x 2 (s - c) ] [s = (a + b + c) / 2]
= 1 / 2bc x ^4√[ s (s - a) (s - b) (s - c) ]
= 2 / bc x √[ s (s - a) (s - b) (s - c) ]
所以
(bc x sin A) / 2 =√[ s (s - a) (s - b) (s - c) ]

2007-01-12 00:04:24 補充：
多謝你選我的答案

希羅公式是 Heron's Formula 吧！

Area of triangle ABC = (ab x sin C) / 2
cos A = (b ^2 + c ^2 - a ^2) / 2bc
sin A = √(1 - cos ^2 A)
= √ { 1 - [ ( b ^2 + c ^2 - a ^2) / 2bc ] ^2 }
= √{ [ 1 - (b ^2 + c ^2 - a ^2) / 2bc ] [ 1 + (b ^2 + c ^2 - a ^2) / 2bc] }
= √{ [ (2bc - b ^2 - c ^2 + a ^2) / 2bc ] [ (2bc + b ^2 +c ^2 - a^2) / 2bc] }
= 1 / 2bc x √{ [ a ^2 - (b ^2 - 2bc + c ^2) ] [ (b ^2 + 2bc + c ^2) - a ^2] }
= 1 / 2bc x √{ [ a ^2 - (b - c) ^2] [ (b + c) ^2 - a ^2]}
= 1 / 2bc x √[ (a + b - c) (a - b + c) ( -a + b + c) (a + b + c) ]
= 1 / 2bc x √[ (a + b + c) (b + c - a) ( a + c - b) ( a + b - c)]
= 1 / 2bc x √[2s x 2(s - a) x 2(s - b) x 2 (s - c) ] [s = (a + b + c) / 2]
= 1 / 2bc x ^4√[ s (s - a) (s - b) (s - c) ]
= 2 / bc x √[ s (s - a) (s - b) (s - c) ]
所以
(bc x sin A) / 2 =√[ s (s - a) (s - b) (s - c) ]