In triangle ABC, AD is the bisector of angle BAC and I is its incentre .Prove that AI/ID=(AB+AC)/BC Share with your friends Share 7 Vijay Kumar Gupta answered this Consider a traingle ABC in which AD is the bisector of ∠BAC and I isits incentre.To prove: AIID=AB+ACBCWe will first prove that if in △ABC, AD is the bisector of ∠BAC, then BDDC=ABACDraw CE∥DA which meets BA produced at EThe figure is shown below: Proof of BDDC=ABACSince DA∥CE, it implies that ∠2=∠3 interior alternate angles ∠1=∠4 corresponding anglesSince AD is bisector of ∠BACSo ∠1=∠2This further implies that, ∠3=∠4The sides opposite to equal angles are equal in length.So we have AE=ACNow in △BCE, DA∥CESo by basic proportionality theorem. BDDC=ABAE BDDC=ABAC (since AE=AC proved above)Thus, it is proved that, BDDC=ABACAdd 1 to both sides BDDC+1=ABAC+1 BD+DCDC=AB+ACAC BCDC=AB+ACAC ACDC=AB+ACBC .....ilocate the incentre I on the figureHere I lies on AD because AI is also a bisector of ∠BAC Note that the triangles ICA and ICD are similar.The property of similar triangles says that its sides are proportional.This implies that, ACDC=AIID .....iiFrom i and ii, it is proved that, AIID=AB+ACBC -15 View Full Answer
