平面几何中有一个结论: 在Rt△ABC中,两直角边AC=b,BC= a,斜边AB上的高为h,则1/h2=1/a2+1/b2. 该结论的证明很简单．类比它,在立体几何中有何发现? 我们猜想,在立体几何中,也有类似的一个公式: 在三棱锥V-ABC中,若三条侧棱VA、 VB、VC两两垂直,且长度分别为a、b、C,顶点V到底面ABC的距离VH=h,则1/h2=1/a2+1/b2+1/c2. 注意这只是由类比得到的一个猜想,是否成立还须证明． There is a conclusion in plane geometry: In Rt △ ABC, two right-angled edges AC=b, BC=a, and the heights on the hypotenuse AB are h, then 1/h2=1/a2+1/b2. Very simple. Analogous to it, what was found in the solid geometry? We guessed that in the three-dimensional geometry, there is a similar formula: In the triangular pyramid V-ABC, if the three side edges VA, VB, VC are perpendicular to each other, and the length is respectively a, b, C, the distance from the vertex V to the bottom surface ABC is VH=h, then 1/h2=1/a2+1/b2+1/c2. Note that this is only a conjecture obtained by analogy.