We’ve shown that a diameter that bisects a chord is perpendicular to that chord. Now let’s prove the opposite: that a diameter that is perpendicular to a chord bisects that chord.
In the circle O, the diameter AB is perpendicular to a chord CD. Prove that it bisects the chord.
We will show the two line segments , CE and ED, are equal, using triangle congruency.
We can connect the chord’s endpoints with the center, with lines that are radii of the circle, and thus equal, and use the fact that AB⊥CD to prove congruency of these 2 right triangles using the Hypotenuse -Leg postulate.
(1) OE=OE // Common side, reflexive property of equality
(2) OC=OD=r // Radii of a circle are all equal
(3) AB⊥CD // Given
(4) ΔOEC≅ ΔOED // Hypotenuse-Leg
(5) CE=ED // (4), Corresponding sides in congruent triangles
And as a side benefit, we have also proven that the perpendicular diameter bisects the arc subtended by the chord, since:
(6) ∠COB ≅ ∠DOB // (4), Corresponding angles in congruent triangles
(7) ArcCB≅ArcBD //(6)