Theorem – The perpendicular from the center of a circle to a chord bisects the chord

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

Given: A circle with center O, AB is chord of a circle and OC perpendicular from the center O to the chord AB.

i.e.    OC ⊥ AB 

therefore  ∠OCA and ∠OCB, Both angles are 900.  

To prove: AC = CB

Construction: join OA and OB. 

Proof: In △OCA and △ OCB

∠OCA = ∠OCB (Both angles are 900 given)
OA = OB (Radius of circle)
OC = OC (Common side)

Hence, △OCA ≅ △OCB (RHS Congruence rule)

∴ AC = CB …(1) (Corresponding parts of congruent triangles)

                        ∴ AC = CB
 Therefore C is the mid point of chord AB.                         

Hence proved

Related posts:

  1. Theorem – The perpendicular from the center of a circle to a chord bisects the chord
  2. Prove that the angle in a semicircle is a right angle – Theorem – Geometry
  3. Theorem – If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal
  4. The angle subtended by an arc at the centre is double of the angle subtended by it at any point on the remaining part of the circle – Theorem – geometry
  5. Theorem – The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
  6. Theorem – The line drawn through the center of a circle to bisect a chord is perpendicular to the chord
  7. Theorem – Equal chords of a circle subtend equal angles at the center
  8. What is a Vertex in geometry – Definition and Examples
  9. The tangent at any point of a circle is perpendicular to the radius through the point of contact
  10. Theorem – If the diagonals of a Quadrilateral bisect each other then that Quadrilateral is a Parallelogram

Leave a Reply

Your email address will not be published. Required fields are marked *