In figure, two equal chords AB and CD intersect each other at E. Prove that arc DA = arc CB.

Given chord AB = chord CD intersect at E.

Construction:

Join AD and BC.

Draw OF ⊥ AB and OG ⊥ CD.

Join OE.

We know that when a perpendicular is drawn from centre to chord, it bisects them.

⇒ AF = FB = and CG = GD =

Given AB = CD,

⇒ AF = FB = CG = GD … (1)

Consider ΔOFE and ΔOGE,

We know that equal chords of a circle are equidistant from each other.

⇒ OF = OG

⇒ OE = OE [Common side]

⇒ ∠OFE = ∠OGE = 90° [Construction]

By RHS congruence rule,

⇒ ΔOFE ≅ ΔOGE

By CPCT,

⇒ FE = GE … (2)

So,

⇒ AE = AF + FE, CE = CG + GE

And EB = FB – FE, ED = GD – GE

From (1) and (2),

⇒ AE = CE … (3)

And EB = ED … (4)

Consider ΔAED and ΔCEB,

⇒ AE = CE [From (3)]

⇒ EB = ED [From (4)]

⇒ ∠AED = ∠CEB [Vertically opposite angles are equal]

By SAS congruence rule,

⇒ ΔAED ≅ ΔCEB

By CPCT,

⇒ AD = CB

∴ Arc AD = Arc CB

Hence proved