Question

In a circle with centre o chords ab and cd intersect inside the circumference at e prove that angle aoc = angle bod = 2angle aec

Answer 1

A circle with centre O is shown in figure . where chords AB and CD intersect inside the Circumference at E .
To prove : ∠AOC + ∠BOD = 2∠AEC

arc AC subtends the ∠AOC on the centre and ∠ABC on the point B .
we know, angle subtended on the centre of circle is double the angle subtended on the other part of circle by the same arc
so, ∠AOC = 2∠ABC ------(1)

similarly we can see that , arc BD subtends the ∠BOD on the centre and ∠BCD on the point C.
so, ∠BOD = 2∠BCD ---------(2)

now, add equations (1) and (2),
∠AOC + ∠BOD = 2(∠ABC + ∠BCD) -----(3)
now see the ∆BCE,
∠AEC is the exterior angle of ∆BCE ,
that's why,
∠AEC = (∠ABC + ∠BCD) now use it in equation (3)

hence, ∠AOC + ∠BOD = ∠AEC

Answer 2

A circle with centre O is shown in figure . where chords AB and CD intersect inside the Circumference at E .

To prove : ∠AOC + ∠BOD = 2∠AEC

arc AC subtends the ∠AOC on the centre and ∠ABC on the point B .

we know, angle subtended on the centre of circle is double the angle subtended on the other part of circle by the same arc

so, ∠AOC = 2∠ABC ------(1)

similarly we can see that , arc BD subtends the ∠BOD on the centre and ∠BCD on the point C.

so, ∠BOD = 2∠BCD ---------(2)

now, add equations (1) and (2),

∠AOC + ∠BOD = 2(∠ABC + ∠BCD) -----(3)

now see the ∆BCE,

∠AEC is the exterior angle of ∆BCE ,

that's why,

∠AEC = (∠ABC + ∠BCD) now use it in equation (3)

hence, ∠AOC + ∠BOD = ∠AEC

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