"Question 4 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.
Class 9 - Math - Circles Page 186"
Answers
The angle subtended by an Arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
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We know that and exterior angle of a triangle is equal to the sum of the interior opposite angles.
In ∆BDC,
∠ADC= ∠DBC+∠DCB......(i)
Also we know that angle at the centre is twice of the angle at a point on the remaining part of circle.
∠DCE= ½ ∠DOE
∠DCB= ½ ∠DOE
[∠DCE= ∠DCB]
∠ADC= ½ ∠AOC
On putting the values of ∠ADC and ∠DCB in eq i,
½ ∠AOC= ∠ABC+1/2 ∠DOE
[∠DBC=∠ABC]
∠ABC= ½(∠AOC - ∠DOE)
Hence, ∠ABC is equal to have the difference of the angle subtended by chords AC and DE at the centre.
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Hope this will help you....
