Line (Segment) from the Center to a Chord

Line(Segment) from the Center to a Chord

Let us investigate a line (line segment) drawn from the center of the circle to a chord. BC is a chord of the circle with center at O. D is a point on the chord. What do you notice about the lengths BD and CD when the angle is 90℃.
Move the point D on the chord and look for a situation when the angle is 90℃ at this point what can you say about the lengths BD and CD, are they equal? You will observe that BD = CD. Here we can make the following conjecture:

• The perpendicular from the center of a circle to a chord bisects the chord.

The converse of the above conjecture is also true:

• If a line (line segment) is drawn from the center of a circle to the midpoint of a chord, then the line is perpendicular to the chord.

Let us prove the conjecture 1 above. This proof is based on the congruency of the triangles.

       Given: OD ⊥ BC

       Construction: Join OB and OC.

       Strategy: If we can show that △BDO 
       and △CDO are congruent then the 
       sides BD and CD must be equal.

       To prove: BD = CD

       Proof:

          In △ BDO and △ CDO

          OB = OC (Radii of the circle)

          OD = OD (Common Sides)

         ∠BDO = ∠CDO (given , both 90)

      Hence, △ BDO ≅ △ CDO

      Therefor, BD = CD (Corresponding Parts of Congruent Triangles)