The problem is to:
(1) Prove that ABC is the medial triangle for A'B'C'.
(2) Prove that the perpendicular bisector of a side of A'B'C' is an altitude for ABC.
(3) Prove that the three altitudes of any triangle are concurrent.
The given information is:
Given a triangle ABC, draw a line through each vertex that is parallel to the opposite side. The three lines intersect to form a new triangle A'B'C'. The diagram can not be used as the proof.
How do you write this proof out without using the diagram?