Question :

PROVE that in two concentric circles the chord of the lager circle, which touches the SMALLER circle, is bisected by it at the point of contact .

ANSWER

Given : -

2 concentric circles with 'O' as the common circle for both the circles.

AB is the chord of the lager circle.

Required to prove : -

AC = BC

Congruency criteria used : -

Side,Side,Angle (S,S,A) congruency criteria

Construction : -

Before solving this question we need to perform some bit of constructions !

1. Join O to C . 'C' is a point of contact of smaller circle with the chord AB

2. Join A to O and B to O

3. While joining O to C MAKE sure it is perpendicular to AB (chord)

PROOF : -

Consider ∆AOC & ∆BOC

In ∆AOC & ∆BOC

→ OC = OC (side)

[ Reason : Common side ]

→ OA = OB (side)

[ Reason : In a circle, all radii are equal ]

→ ∠ACO = ∠BCO (angle)

[ Reason : AB is perpendicular to OC ]

From the above we can conclude that;

By using SSA congruency criteria

∆ AOC ≅ ∆BOC

This implies;

AC = BC ( side )

[ Reason : Corresponding Parts of Congruent Triangles (CPCT)]

Therefore,

The Chord AB is bisected by the point 'C' which is the point of the contact .

HENCE Proved !