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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 Ī” ABC such that šµš·/š¶š· = š“šµ/š“š¶ ā‹… Prove that AD is the bisector of āˆ  BAC. Given : A Ī”ABC Where šµš·/š¶š· = š“šµ/š“š¶ To Prove: AD is the bisector of āˆ BAC i.e. āˆ BAD = āˆ DAC Construction : Produce BA to E such that AE = AC. Now join CE. Proof: In Ī” AEC, AE = AC āˆ“ āˆ  AEC = āˆ  ACE Also, šµš·/š¶š· = š“šµ/š“š¶ Now, by construction AC = AE šµš·/š¶š· = š“šµ/š“šø In Ī” BEC Now, we have a line AD which divides the two sides BE and BC of āˆ† BEC in the same ratio, Thus, by Converse of Basic proportionality Theorem,DA āˆ„ CE Now, DA āˆ„ CE and CA is the transversal āˆ  BAD = āˆ  AEC āˆ  DAC = āˆ  ACE Now, From (1) āˆ  AEC = āˆ  ACE Put āˆ AEC = āˆ BAD and āˆ ACE = āˆ DAC from equations (2) and (3) āˆ  BAD = āˆ  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 āˆ  BAC Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.