Point inside a tetrahedron joined to corners creates how many new internal planes? When a point inside of a tetrahedron (a solid with four triangular surfaces) is connected by
straight lines to its corners, how many (new) internal planes are created with these lines?
How do we visualize this?
Are some tools available online, for creating such a geometry?
 A: Earlier the tetrahedron had 4 points in space.It has a total of 4 planes. After adding an internal point we now have 5 points in the space. If we take any three points we can form a plane. so total planes we can form now with 5 points is 5c3 = 10.
So the number of new planes=10-4=6
A: I do not know about nice tools for visualizing this, but imagine following:
You get four lines each from the point inside to the corner points. Every set of three of those lines are linerl independent. Otherwise the point would be on the surface of the tetrahedron. Therefore each pair of two lines spans a new distinct plane. 
So if you number the lines, the first line creates three new planes (with each of the remaining lines) the second line creates two new planes (each with line three of four) and line three creates one new plane with line four. And we've already created all planes that contain line four, so we have a total of 6 new planes.
A: Another way of thinking would be in the following manner. A tetrahedron is a solid with 4 points in space of which none of the 3 are linearly independent. If we add a point inside the tetrahedron we form 4 new mini tetrahedrons. Each tetrahedron has 4 faces. So in total if we cut all these 4 tetrahedrons then we have 16 faces. 
Now the initial tetrahedron has already 4 faces. Remaining we have 12 faces. Also any three internal tetrahedrons share 2 sides in common. So out of 4 internal tetrahedrons we have total 6 faces in common. so finally we can say that adding a new point inside the tetrahedron creates 6 new faces
A: The Tetrahedron has 4 triangular surfaces with 4 vertices/corners ( say A,B,C and D) as can be seen here. http://www.sjsu.edu/faculty/wa…
Now if you take a point inside a tetrahedron (suppose O) and connect it with any two of its corners which are nothing but vertices( suppose A and B), you will get 1 internal plane as OAB.
So we can see from here that, no of new internal planes = no of different pair of corners or vertices
Similarly you can take any other 2 corners like (A,C) or (A,D) or (B,C) or (B,D) or (C,D),
hence total possible pair of corners are 6. Therefore 6 new internal planes possible.
We could also calculate the possible corners by using combinations formula,
which is nCr, i.e. no of ways to select a combination of r things from a given set of n things.
here n = 4 ( as total 4 vertices, A,B,C and D)
and r =2 ( as we need two corners at a time)
Thus, 4C2 = 6.
