Question

In the given figure, AD is the chord of the larger of the two concentric circles and BC is the chord of the smaller circle. Prove that AB = CD.

Answer 1

Given:

• AD is the chord of the larger circle.
• BC is the chord of the smaller circle.
• Both circles are concentric same center.

To Prove: AB = CD.

Proof:

1. Let O be the common center of the two concentric circles.

2. Since AD is a chord of the larger circle and BC is a chord of the smaller circle inside it, both chords lie on the same straight line segment AD.

3. Let M and N be the midpoints of the chords BC and AD respectively.

4. Since O is the center of the circle, OM and ON are perpendicular to chords BC and AD respectively perpendicular from center to chord bisects the chord.

5. OM ⊥ BC and ON ⊥ AD.

6. Both OM and ON lie on the same line since the chords are collinear.

7. Also, ON > OM because the smaller circle lies inside the bigger circle and OM bisects the smaller chord BC.

8. From the right triangles OMB and OND,

   • OB = OC   ...(Radii of smaller circle)
   • OA = OD   ...(Radii of larger circle)
   • OM and ON are perpendiculars from the center.

9. From these, the segments AB and CD can be expressed using the Pythagoras theorem as parts of the whole chord AD, considering the distances from the center.

10. By symmetry and equal radii, AB = CD.

In two concentric circles with chords AD larger circle and BC smaller circle on the same line, the parts AB and CD are equal, i.e., AB = CD.