Last updated at June 23, 2017 by Teachoo

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Ex 6.4, 3 In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that (ππ(π΄π΅πΆ))/(ππ(π·π΅πΆ)) = π΄π/π·π Given: β ABC and β DBC Having common base BC To prove: (ππ β π΄π΅πΆ)/(ππ β π·π΅πΆ)=π΄π/π·π Proof: Since AO and OD are not part of β π΄π΅πΆ and β π·π΅πΆ , we cannot directly use the theorem We know that Area of triangle = 1/2Γπ΅ππ πΓπ΄ππ‘ππ‘π’ππ Lets draw altitude AE [ AE β₯π΅πΆ ] Hence, ar β ABC=1/2Γπ΅πΆΓπ΄πΈ Similarly, for Ξ DBC Let us draw altitude DF (DF β₯ BC) Hence, ar β DBC=1/2Γπ΅πΆΓπ·πΉ Now, taking ratio (ππ βπ΄π΅πΆ)/(ππ βπ·π΅πΆ) = (1/2 Γ π΅πΆ Γ π΄πΈ)/(1/2 Γ π΅πΆ Γ π·πΉ) (ππ βπ΄π΅πΆ)/(ππ βπ·π΅πΆ) = π΄πΈ/π·πΉ Now in β AOE and Ξ DOF β AEO = β DFO β AOE = β DOF So, by using AA similarity criterion β AOE ~ β DOF If two triangle are similar their , corresponding sides are in the same ratio So, π΄πΈ/π·πΉ=π΄π/π·π Putting (2) in (1) (ππππ ππ β π΄π΅πΆ)/(ππππ ππ β π·π΅πΆ)=π΄πΈ/π·πΉ (ππππ ππ β π΄π΅πΆ)/(ππππ ππ β π·π΅πΆ)=π΄π/π·π Hence proved

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 You are here

Ex 6.4, 5 Important

Ex 6.5, 2 Important

Ex 6.5, 3 Important

Ex 6.5, 8 Important

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.