由题意知PC、PA、PB分别垂直于PAB、PBC、PAC三个平面.连接CH,且延长交AB于D,连接PD.那么有题意知PH⊥CH,且PC⊥PD,CD和PD均⊥AB.那么有PH^2/PC^2=sin^2(角PCD)=PD^2/CD^2=HD*CD/CD^2=HD/CD=[HD*AB/2]/[CD*AB/2]=S△HAB/S△CAB.同理可知PH^2/PA^2=S△HBC/S△CAB和PH^2/PB^2=S△HAC/S△CAB.所以PH^2/PA^2+PH^2/PB^2+PH^2/PC^2=（S△HBC+S△HAC+S△HAB)/S△CAB=1.得证.
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