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在立体几何中 ,有一个常见的模型 :图 1 图 2如图 1,已知直线a、b、l与平面α满足a α ,b α ,a∩b =P ,P∈l ,l与a、b成相等的角θ ,在l上任取异于点P的Q点 ,过Q作QK⊥α于K ,那么K点到直线a、b的距离相等 ,即K点落在∠APB(或其补角 )的平分线所在的直线上 ,?
In stereo geometry, there is a common model: Figure 1 Figure 2 As shown in Figure 1, the known straight lines a, b, l and plane α satisfy a α, b α, a∩b = P, P∈l, l and a , b is equal to the angle θ. Any point on l that is different from point P is Q. If Q is QK⊥α is in K, then the distance from point K to line a, b is equal, that is, point K falls on APB (or What is the complement of the bisector of the line?