Question 2:

This can be done easily with some properties of parallel lines and transversals.

Looking at the diagram, let the ratio of ED : DC be p : (1-p)... p can be thought of as a "factor" of the length EC. and 1-p is the complement factor as ED + DC = EC.

By properties of parallel lines (AE//BD), AB : BC = p : (1-p)

Once again by the properties of parallel lines (BF//CE), AF : FE = p : (1-p)

BUT... by the same argument, using FD//AC, we know that AF : FE = CD : DE

Therefore (1-p) : p = p : (1-p), and this can only occur if the ratio 1-p = p that is p = 1/2 (this means that ED : DC = AB : BC = AF : FE = 1/2 : 1/2 = 1 : 1) meaning that B, D, F are indeed midpoints of the 3 corresponding sides.