Pythagoras Theorem and it's important questions

Chapter 6 Class 10 Triangles
Serial order wise

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### Transcript

Question 13 D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2 Given : triangle ABC , right angled at C Two points D and E are on the sides CA and CB To prove: AE2 + BD2 = AB2 + DE2 Proof : Let us join the points D with E and B . And point E with A Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 Here , ACE is a right angle triangle AE2 = AC2 + CE2 In right angle triangle DCB (BD)2 = (DC)2 + (BC)2 In right angle triangle ABC (AB)2 = (AC)2 + (BC)2 In right angle triangle DCE (DE)2 = (DC)2 + CE2 Since L.H.S = R.H.S Hence proved

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.