Question

In the following figure, AB and CD are two chords of a circle intersecting each other at point E. Prove that ∠AEC = `1/2` (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre).

Answer 1

Given: In a figure, two chords AB and CD intersecting each other at point E.

To prove: ∠AEC = `1/2`  ...[Angle subtended by arc C × A at centre + angle subtended by arc DYB at the centre]

Construction: Extend the line DO and BO at the points l and H on the circle. Also, join AC.

Proof: We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

∴ ∠1 = 2∠6   ...(i)

And ∠3 = 2∠7  ...(ii)

In ΔAOC, OC = OA  ...[Both are the radius of circle]

∠OCA = ∠4  ...[Angles opposite to equal sides are equal]

Also, ∠AOC + ∠OCA + ∠4 = 180°  ...[By angle sum property of triangle]

⇒ ∠AOC + ∠4 + ∠4 = 180°

⇒ ∠AOC = 180° – 2∠4   ...(iii)  

Now, in ΔAEC, ∠AEC + ∠ECA + ∠CAE = 180°  ...[By angle property sum of a triangle]

⇒ ∠AEC = 180° – (∠ECA + ∠CAE)

⇒ ∠AEC = 180° – [(∠ECO + ∠OCA) + ∠CAO + ∠OAE]

= 180° – (∠6 + ∠4 + ∠4 + ∠5)  ...[In ΔOCD, ∠6 = ∠ECO angles opposites to equal sides are equal]

= 180° – (2∠4 + ∠5 + ∠6)

= 180° – (180° – ∠AOC + ∠7 + ∠6)  ...[From equation (iii) and in ΔAOB, ∠5 = ∠7, as (angles opposite to equal sides are equal)]

= `∠AOC - (∠3)/2 - (∠1)/2`  ...[From equations (i) and (ii)]

=  `∠AOC - (∠1)/2 - (∠2)/2 - (∠3)/2 + (∠2)/2`  ...`["Adding and subtracting" (∠2)/2]` 

= `∠AOC - 1/2 (∠1 + ∠2 + ∠3) + (∠8)/2`  ...[∵ ∠2 = ∠8 vertically opposite angles]

= `∠AOC - (∠AOC)/2 + (∠DOB)/2`

⇒ `∠AEC = 1/2(∠AOC + ∠DOB)`

= `1/2` ...[Angle subtended by arc CXA at the centre + angle subtended by arc DYB at the centre]