Let's consider a circle and a chord AB. To draw the perpendicular bisector CD, we place the compass tip on A and draw two arcs. With the same compass setting, we repeat the process with the compass placed on B.
Doing a similar process, we draw the perpendicular bisector of CD. Let O be the intersection point.
Our mission is to prove that CD passes through the center of the circle. To do it, we will assume that the center of the circle is not on CD. Let X be the center of the circle.
By construction, E is the midpoint of AB and then AE≅ EB. Also, because all radii of a circle are congruent, we have that XA≅XB.
Additionally, the Reflexive Property of Congruence gives us that XE≅XE. Therefore, by the Side-Side-Side Congruence Theorem we get that △ AEX ≅ △ BEX.
△ AEX ≅ △ BEX
⇓
∠AEX ≅ ∠BEX
Notice that these two angles also form a linear pair. Therefore, they must be right angles. In consequence, XE⊥AB implying that XE is the perpendicular bisector of AB.
CD is the perpendicular bisector ofAB
XE is the perpendicular bisector ofAB
The last fact contradicts the uniqueness of the perpendicular bisector of a segment. Therefore, the initial assumption is false. This means that the center of the circle is on CD.
b
From Part A, X is on CD. By construction C and D lie on ⊙ X, which means that CD is a diameter. Also, by construction O is the midpoint of CD.
Since the midpoint of a diameter is the center of the circle, we have that O is the midpoint of the circle. Therefore, point O is point X.
