Question

In ΔABC shown below, point A is at (0, 0), point B is at (x2, 0), point C is at (x1, y1), point D is at x sub 1 over 2, y sub 1 over 2, and point E is at the quantity of x sub 1 plus x sub 2 over 2, y sub 1 over 2: Triangle ABC is shown. Point D lies on segment AC and point E lies on segment BC. A segment is drawn between points D and E. Point A is at the origin. Prove that segment DE is parallel to segment AB.

Answer 1

There is two ways to attempt this problem. The first is that you could prove that angle CDE is congruent to angle DAB (of which I'm not totally sure is possible) or option two. Your second choice is to prove that Line segment DE is a flat line because line segment AB is also a flat line.