**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.**

**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.**

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.

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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