This proposition constructs a line segment perpendicular to given line segment from a point on it. The way this is done is very similar to the proof of the previous proposition, and this is because the previous proposition does actually construct a right angle, which is easy to verify. C is not necessarily the midpoint of AB, so instead, D is selected arbitrarily on AC, and E is selected on CB, so that C is the midpoint of DE. An equilateral triangle is constructed of DE resulting in F, and this is connected with C to form the perpendicular line segment. This is true because of SSS congruence.