How can i solve this tough question of class 9th ncert of Circles of math of exercise 10.6 of question no.4 . Give me the best and simple way for solving this question. 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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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. Q.4
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Consider the diagram
Here AD = CE
We know, any exterior angle of a triangle is equal to the sum of interior opposite angles.
So,
∠DAE = ∠ABC+∠AEC (in ΔBAE) ——————-(i)
DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.
So,
∠DAE = (½)∠DOE ——————-(ii)
Similarly, ∠AEC = (½)∠AOC ——————-(iii)
Now, from equation (i), (ii), and (iii) we get,
(½)∠DOE = ∠ABC+(½)∠AOC
Or, ∠ABC = (½)[∠DOE-∠AOC] (hence proved).