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20/07/2021 4:10 pm
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Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
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20/07/2021 4:14 pm
Consider the diagram
Here AD = CE
We know, any exterior angle of a triangle is equal to the sum of interior opposite angles.
∠DAE = ∠ABC+∠AEC (in ΔBAE) ..........(i)
DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.
∠DAE = (1/2)∠DOE ...........(ii)
Similarly, ∠AEC = (1/2)∠AOC .........(iii)
Now, from equation (i), (ii), and (iii) we get,
(1/2)∠DOE = ∠ABC + (1/2)∠AOC
∠ABC = (1/2)[∠DOE-∠AOC] (hence proved).