however

1 You should be talking about a triangular pyramid.

Check the problem and see if there is any problem.

How can DP intersect with MS on the surface SAB?

Let me tell you a way.

SC and DP can define a plane.

M is the center of gravity of ABC, and m is on DC.

M is also on the plane of DPCS.

Since MS intersects SC, SC is parallel to PD.

Doesn't MS also intersect with PD?

I have a name ~'

Help you get your certificate in order

I thought you just asked 1.

SC⊥ surface SAB

SC‖PD

Then spread on the surface of PD⊥.

In RT triangle SAB

You can get SD=SA=SB.

DD 1⊥SD

DD 1⊥SA

DD 1⊥SB

You can get SD1= ad1= BD1.

Similarly, the relationship with CD 1 can be obtained.

It becomes ~ `.