In the tetrahedral p-abc, D, e and F are the midpoint of AB, BC and AC, respectively

In the tetrahedral p-abc, D, e and F are the midpoint of AB, BC and AC, respectively

① If you want to make the vertical plane of plane ABC through point P in this regular tetrahedron, then the vertical plane through point P must pass through the midpoint of triangle ABC, that is, the intersection of CD, BF and AE ~!
Then I will help you solve the projection problem you asked: a tetrahedron is an individual composed of four equilateral triangles. According to your question, PAB, PBC, PAC and ABC are congruent triangles. The projection of point P on ABC is not on De, but on the intersection of CD, BF and AE

For tetrahedron, the lines of three groups of relative edges are made respectively, and the three lines intersect at one point;
RT, how to prove it

I can only dictate that you draw a picture by yourself
Let four vertices of the tetrahedron be ABCD and six midpoint efghij be on edge ad DC BC AB BD AC respectively
To prove that eg, ij and HF intersect at one point, let eg and HF intersect at m, connect eh, Hg, gf and Fe is obviously a parallelogram, connect DJ, bj and EF, Hg intersect at s t respectively
In triangle JBD, s is the midpoint of DJ, t is the midpoint of BJ, I is the midpoint of BD, ST is a median line of triangle JBD, it must pass the midpoint of ij, and st and ij are equally divided with each other, efgh is a parallelogram, s is the midpoint of EF, t is the midpoint of Hg, obviously so m is the key point of St, that is, M is the midpoint of ij, so the midpoint of ij eg HF three lines coincides, that is, the three lines intersect at one point

There is a regular tetrahedron. The cross section passing through one edge of the regular tetrahedron and the center of the circumscribed sphere intersects the other edge at point A. It is proved that a is the midpoint of this edge

The straight line passing through the upper fixed point and the center of circumscribed circle extends to the point m where the bottom surface intersects
Then on the bottom, connect the upper point of the triangle with m to one side
Because it is a regular tetrahedron, all triangles are equilateral
Three lines in one, so the line on the bottom intersects at the midpoint over there