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  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Ex 6.6, 9 In Fig. 6.63, D is a point on side BC of Ξ” ABC such that 𝐡𝐷/𝐢𝐷 = 𝐴𝐡/𝐴𝐢 β‹… Prove that AD is the bisector of ∠ BAC. Given : A Ξ”ABC Where 𝐡𝐷/𝐢𝐷 = 𝐴𝐡/𝐴𝐢 To Prove: AD is the bisector of ∠BAC i.e. ∠BAD = ∠DAC Construction : Produce BA to E such that AE = AC. Now join CE. Proof: In Ξ” AEC, AE = AC ∴ ∠ AEC = ∠ ACE Also, 𝐡𝐷/𝐢𝐷 = 𝐴𝐡/𝐴𝐢 Now, by construction AC = AE 𝐡𝐷/𝐢𝐷 = 𝐴𝐡/𝐴𝐸 In Ξ” BEC Now, we have a line AD which divides the two sides BE and BC of βˆ† BEC in the same ratio, Thus, by Converse of Basic proportionality Theorem,DA βˆ₯ CE Now, DA βˆ₯ CE and CA is the transversal ∠ BAD = ∠ AEC ∠ DAC = ∠ ACE Now, From (1) ∠ AEC = ∠ ACE Put ∠AEC = ∠BAD and ∠ACE = ∠DAC from equations (2) and (3) ∠ BAD = ∠ DAC Converse of Basic proportionality Theorem If a line divides any two sides of a triangle in the same ratio, then the line is Parallel to the third side. Thus, AD is the bisector of ∠ BAC Hence proved

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