Question

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

Answer 1

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Answer 2

Prove that ∠AEC =1/2

Step-by-step explanation:

Draw the line AC and BC ,

As we know that the ∠ subtended by 'an arc of a circle at its center is twice the angle subtended by the same arc at a point on the circumference'

This gives,

∠AOC = 2∠ABC (...1)

Similarly,

∠BOD = 2∠BCD (...2)

By adding both the equations

∠AOC + ∠BOD = 2∠ABC + 2∠BCD

Taking 2 as common,

∠AOC + ∠BOD = 2(∠ABC+∠BCD)

We can write it as

∠AOC + ∠BOD = 2(∠EBC+∠BCE)

This gives,

∠AOC + ∠BOD = 2(180°  −∠CEB)

It can be written as

∠AOC + ∠BOD = 2(180°  −(180°  −∠AEC))

So, we get

∠AOC + ∠BOD = 2∠AEC

By dividing the equation by 2 ,

∠AEC=1/2(∠AOC+∠BOD or angle subtended by arc CXA at the center + angle subtended by arc DYB at the center)

Hence, it is proved that ∠AEC=1/2 (angle subtended by arc CXA. At the center + angle subtended by arc DYB at the center).

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