As shown in the figure, P is a point on the diagonal BD of square ABCD

As shown in the figure, P is a point on the diagonal BD of square ABCD

Connect to PC,
⊙ PE ⊥ DC, PF ⊥ BC, ABCD are square,
∴∠PEC=∠PFC=∠ECF=90°,
The quadrilateral PECF is a rectangle,
∴PC=EF,
And ∵ P is any point on BD,
The symmetry of PA and PC with respect to BD,
It can be concluded that PA = PC,
∴EF=AP.

It is known that the cube abcd-a1b1c1d1, O is the intersection of the bottom ABCD diagonals

It is proved that: (1) connect a1c1, let a1c1 ∩ b1d1 = O1, connect AO1, ∩ abcd-a1b1c1d1 is a cube, ∩ a1acc1 is a parallelogram ∩ a1c1 ∩ AC and a1c1 = AC and O1, O are the midpoint of a1c1 and AC, ∩ o1c1 ∩ AO and o1c1 = Ao ∩ aoc1o1 is a parallelogram ∩ C1O ∥ AO1, AO1 ⊂ plane ab1d1

In the square abcd-a1b1c1d1, how many diagonals are there on the surface which is different from the straight line BD and forms a 60 degree angle?

There are four lines: AD1, Ab1, B1C and CD1

How many diagonal lines are there perpendicular to BD1 in square abcd-a1b1c1d1?

First of all, drawing, it is not difficult to find that each surface of the cube has, and only one diagonal line is perpendicular to BD1, that is, six. Then, the interior of the cube can make six planes through each fixed point, and then look at the top, you can find, obviously can not find the diagonal line that meets the conditions, so there are a total of six