设PA =a,PB= b,PC =c,
则(s1)^2 +(s2)^2 +(s3)^2 = (1/4)[(a^2)(b^2) +(b^2)(c^2) +(c^2)(a^2)] (2)
AB ^2 = a^2 + b^2,BC^2 = b^2 + c^2,CA^2 = c^2 + a^2 (勾股定理)
由余弦定理:cos角BAC = [AB^2 + AC^2 - BC^2]/[2*AB*AC ]=(a^2)/根号[(a^2 +b^2)(a^2 +c^2)]
sin角BAC ={根号[ (a^2 +b^2)(a^2 +c^2)-a^4)] }/ 根号[(a^2 +b^2)(a^2 +c^2)]
= {根号[ (a^2)(b^2) +(b^2)(c^2) +(c^2)(a^2)]} / 根号[(a^2 +b^2)(a^2 +c^2)]
三角形ABC的面积S = (1/2)AB* AC*sin角BAC
S^2 = (1/4) [(a^2 +b^2)(a^2 +c^2)]*[sin角BAC]^2
=(1/4)[ (a^2)(b^2) +(b^2)(c^2) +(c^2)(a^2)] (2)
比较(1)(2),知:S^2 =(s1)^2 +(s2)^2 +(s3)^2.
命题得到证明.