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Last updated at May 29, 2018 by Teachoo

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Transcript

Ex 6.5, 8 In figure, O is a point in the interior of a triangle ABC, OD ⊥BC, OE ⊥ AC and OF ⊥AB. Show that OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 Given: Triangle ABC and O is a point in the interior of a triangle ABC where, OD ⊥𝐵𝐶,𝑂𝐸⊥𝐴𝐶,𝑂𝐹⊥𝐴𝐵 To prove :- OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 Proof:- Let us join the point O from A , B and C. Using Pythagoras theorem, (Hypotenuse)2 = (Height)2 + (Base)2 In a right angle triangle OAF. (OA)2 = AF2 + OF2 In right angle triangle ODB OB2 = OD2 + BD2 In a right angle triangle OEC (OC)2 = (OE)2 + (EC)2 Adding (1) + (2) + (3) (OA)2 + (OB)2 + (OC)2 = AF2 + OF2 + OD2 + BD2 + OE2 + EC2 OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 Hence proved Ex 6.5,8 In figure, O is a point in the interior of a triangle ABC, OD ⊥BC, OE ⊥AC and OF ⊥AB. Show that (ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2 Using Pythagoras theorem. (Hypotenuse)2 = (Height)2 + (Base)2 In Δ ODB, OB2 = OD2 + BD2 In Δ OFB, OB2 = OF2 + FB2 In Δ OFA, OA2 = OF2 + AF2 In Δ OEA, OA2 = OE2 + AE2 In Δ OEC, OC2 = OE2 + CE2 In Δ ODC, OC2 = OD2 + CD2

Chapter 6 Class 10 Triangles

Example 5
Important

Example 8 Important

Example 10 Important

Example 14 Important

Theorem 6.1 - Basic Proportionality Theorem (BPT) Important

Theorem 6.7 Important

Ex 6.2, 4 Important

Ex 6.2, 5 Important

Ex 6.2, 6 Important

Ex 6.2, 9 Important

Ex 6.3, 11 Important

Ex 6.3, 12 Important

Ex 6.3, 13 Important

Ex 6.3, 14 Important

Ex 6.3, 15 Important

Ex 6.4, 1 Important

Ex 6.4, 3 Important

Ex 6.4, 5 Important

Ex 6.5, 2 Important

Ex 6.5, 3 Important

Ex 6.5, 8 Important You are here

Ex 6.5, 11 Important

Ex 6.5, 12 Important

Ex 6.5, 15 Important

Class 10

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

About the Author

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.