Question

If two chords AB and BC of a circle subtends equal angle to the diameter through their point of intersection , prove that the chords are equal. ​

Answer 1

Answer:

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Answer 2

Given: AB and CD are two chords of a circle with centre O, intersecting at point E. PQ is a diameter through E, such that ∠AEQ = ∠DEQ.

To prove: AB = CD

Construction: Draw OL perpendicular AB and OM perpendicular CD

Proof: ∠LOE + ∠LEO + ∠OLE = 180° (Angle sum property of a triangle)

⇒ ∠LOE + ∠LEO = 90° ...(i)

Similarly ∠MOE + ∠MEO + ∠OME = 180°

⇒ ∠MOE + ∠MEO + 90° = 180°

∠MOE + ∠MEO = 90° ...(ii)

From (i) and (ii) we get

∠LOE + ∠LEO = ∠MOE + ∠MEO ...(iii)

Also, ∠LEO = ∠MEO (Given) ...(iv)

From (i) and (iv) we get

∠LOE = ∠MOE

Now in triangles OLE and OME

∠LEO = ∠MEO (Given)

∴ ∠LOE = ∠MOE (Proved above)

EO = EO (Common)

∴ △OLE ≅ △OME (ASA congruence criterion)

∴ OL = OM (CPCT)

Thus, chords AB and CD are equidistant from the centre of the circle. Since, chords of a circle which are equidistant from the centre are equal.

∴ AB = CD