证明当p＝1／2(a＋b＋c）时,三角形面积为S△＝√p(p－a)(p－b)(p－c)
S=1/2*absinC
=1/2*ab√[1-(cosC)²]
1-(cosC)²=1-[(a²+b²-c²)/(2ab)]²
=[(2ab)²-(a²+b²-c²)^2]/(2ab)²
=(2ab+a²+b²-c²)(2ab-a²-b²+c²)/(2ab)²
=[(a+b)²-c²]{c²-(a-b)²]/(2ab)²
=(a+b+c)(a+b-c)(c+a-b)(c-a+b)/[4(ab)²]
=4*(a+b+c)/2*(a+b-c)/2*(c+a-b)/2*(b+c-a)/2]/(ab)²
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)²]
=√[p(p-a)(p-b)(p-c)]