Prove - medial triangle, perpendicular bisector, altitude

The problem is to:

(1) Prove that $\displaystyle \Delta$ABC is the medial triangle for $\displaystyle \Delta$A'B'C'.

(2) Prove that the perpendicular bisector of a side of $\displaystyle \Delta$A'B'C' is an altitude for $\displaystyle \Delta$ABC.

(3) Prove that the three altitudes of any triangle are concurrent.

The given information is:

Given a triangle $\displaystyle \Delta$ABC, draw a line through each vertex that is parallel to the opposite side. The three lines intersect to form a new triangle $\displaystyle \Delta$A'B'C'. The diagram can not be used as the proof.

How do you write this proof out without using the diagram?