Solid geometry! ! !

One is that point P is inside the cube, and the other is that point P is outside the cube.

After P, PO is perpendicular to surfaces ABCD and O. Because P-ABCD is a regular pyramid, the photography of point P on surface ABCD is the center of surface ABCD, that is, point O is the intersection of AB and CD.

From the above, it can be seen that PO is perpendicular to the curved surface ABCD, and since the quadrilateral ABCD is a regular quadrilateral and AO is perpendicular to BD, according to the projective theorem, PA is perpendicular to BD.

In the cube ABCD-A1B1C1D1,BD is parallel to B 1D 1, so PA is perpendicular to B 1D 1.

Haha, give it to me, hehe.