I is the incentre of ΔABC. AI when produced meets the circumcircle of ΔABC is D. If ∠BAC = 66° and ∠ACB = 80°. Calculate


(i) ∠DBC      (ii) ∠IBC        (iii) ∠BID

Incentre is the point of intersection of the angle bisectors.

So, <BAD = <DAC = 1/2 <BAD = 1/2 * 66 = 33o

(i) <DBC = <DAC = 33o //angle in the same segment

(ii) In triangle ABC,
<ABC + ACB + BAC = 180 //angle sum property
<ABC + 80 + 66 = 180
<ABC = 180 - 146 = 34o

<IBC = 1/2 ABC = 1/2 * 34 = 17o //since IB is the angle bisector of <ABC

(iii) <BID is the exterior angle of triangle BIA
So, <BID = <IBA + <BAI //exterior angle is equal to the sum of opposite two interior angles

Now, <IBA = 1/2 * ABC = 1/2*34 = 17o
and <BAI = <BAD = 33o

so <BID = 17 + 33 = 50o
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