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Transcript

Theorem 6.7: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then right triangle on both sides of the perpendicular are similar to the whole triangle and to each other Given: ∆ABC right angled at B & perpendicular from B intersecting AC at D. (i.e. BD ⊥ AC) To Prove: ∆ADB ~ ∆ABC ∆BDC ~ ∆ABC & ∆ADB ~ ∆BDC Theorem 6.7: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then right triangle on both sides of the perpendicular are similar to the whole triangle and to each other Given: ∆ABC right angled at B & perpendicular from B intersecting AC at D. (i.e. BD ⊥ AC) To Prove: ∆ADB ~ ∆ABC ∆BDC ~ ∆ABC & ∆ADB ~ ∆BDC Proof: In ∆ADB & ∆ABC ∠ A = ∠ A ∠ ADB = ∠ ABC ∆ADB ~ ∆ABC Similarly, In ∆BDC & ∆ABC ∠ C = ∠ C ∠ BDC = ∠ ABC ∆BDC ~ ∆ABC From (1) and (2) ∆ADB ~ ∆ABC & ∆BDC ~ ∆ABC If one triangle is similar to another triangle, and second triangle is similar to the third triangle, then first and third triangle are similar ∴ ∆ADB ~ ∆BDC Hence Proved Rough This is same as a = b, b = c then a = c

  1. Chapter 6 Class 10 Triangles
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo