Question

In the given figure, AB and CD are two chords of a circle, intersecting each other at a point E.
Prove that ∠AEC = $\frac{1}{2}$(angle subtended by arc CXA at the centre + angle subtended by arc DYB at the centre).

Answer 1

Join AD


We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc AXC subtends ∠AOC at the centre and ∠ADC at D on the circle.

∴ ∠AOC = 2∠ADC
⇒ $\angle ADC=\frac{1}{2}\left(\angle AOC\right)$ ...(1)

Also, arc DYB subtends ∠DOB at the centre and ∠DAB at A on the circle.

∴ ∠DOB = 2∠DAB
⇒ $\angle DAB=\frac{1}{2}\left(\angle DOB\right)$ ...(2)

Now, in ∆ADE,
∠AEC = ∠ADC + ∠DAB (Exterior angle)
⇒ ∠AEC = $\frac{1}{2}\left(\angle AOC+\angle DOB\right)$ (from (1) and (2))

Hence, ∠AEC = $\frac{1}{2}$(angle subtended by arc CXA at the centre + angle subtended by arc DYB at the centre).