如何推理海伦--秦九韶公式？

S=1/2*absinC

=1/2*ab√[1-(cosC)?0?5]

1-(cosC)?0?5=1-[(a?0?5+b?0?5-c?0?5)/(2ab)]?0?5

=[(2ab)?0?5-(a?0?5+b?0?5-c?0?5)^2]/(2ab)?0?5

=(2ab+a?0?5+b?0?5-c?0?5)(2ab-a?0?5-b?0?5+c?0?5)/(2ab)?0?5

=[(a+b)?0?5-c?0?5]{c?0?5-(a-b)?0?5]/(2ab)?0?5

=(a+b+c)(a+b-c)(c+a-b)(c-a+b)/[4(ab)?0?5]

=4*(a+b+c)/2*(a+b-c)/2*(c+a-b)/2*(b+c-a)/2]/(ab)?0?5

p=(a+b+c)/2--->(a+b-c)/2=(a+b+c)/2-c=p-c;(c+a-b)/2)=p-b;(b+c-a)/2=p-a

S=1/2*ab*√[4p(p-a)(p-b)(p-c)/(ab)?0?5]

=√[p(p-a)(p-b)(p-c)]