Question

P是三角形ABC所在平面外一点,PA PB PC两两互相垂直,三角形PAB,三角形PBC,三角形PAC的面积分别是s1

Answer 1

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.