Euclid II-1. (pg 51)
- Author:
- Emily Crum
Explanation:
Create circle ABC. Connect AB. Draw a perpendicular bisector of AB such that D is the midpoint of AB and EC is perpendicular to AB and D lies on EC. Then create the midpoint of EC, labeled pt F.
F can either be the center of the circle or not. If it is, end proof.
If it is not, let G be the center. Join GA, GD, and GB.
AD=DB because D is the midpoint. GD=GD, and AG=BG because they are both radii. Thus by SSS, triangle ADG is congruent to triangle BDG. Thus angle ADG= angle BDG. Because these two angles form a straight line (AB) and equal each other, they must both be right angles.
But, angle ADF is also right because of the given conditions earlier where EC is perpendicular to AB. Thus angle ADF= angle ADG. This is a contradiction. Thus G is NOT the center of the circle and F therefore is.