Friday, November 15, 2019

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)

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