Last updated at May 29, 2018 by Teachoo

Transcript

Ex 7.2, 6 ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that ∠BCD is a right angle. Given: AB = AC Also, AD = AB i.e. AC = AB = AD To prove: ∠BCD = 90° Proof: In ΔABC, AB = AC ⇒ ∠ACB = ∠ABC In ΔACD, AC = AD ∠ADC = ∠ACD In ΔBCD, ∠ABC + ∠BCD + ∠BDC = 180° ∠ACB + ∠ BCD + ∠ACD = 180° (∠ACB +∠ACD) + ∠ BCD = 180° (∠ BCD) + ∠ BCD = 180° 2∠ BCD = 180° ∠ BCD = (180°)/2 ∠ BCD = 90° Hence proved

Chapter 7 Class 9 Triangles

Class 9

Important Questions for Exam - Class 9

- Chapter 1 Class 9 Number Systems
- Chapter 2 Class 9 Polynomials
- Chapter 3 Class 9 Coordinate Geometry
- Chapter 4 Class 9 Linear Equations in Two Variables
- Chapter 5 Class 9 Introduction to Euclid's Geometry
- Chapter 6 Class 9 Lines and Angles
- Chapter 7 Class 9 Triangles
- Chapter 8 Class 9 Quadrilaterals
- Chapter 9 Class 9 Areas of parallelograms and Triangles
- Chapter 10 Class 9 Circles
- Chapter 11 Class 9 Constructions
- Chapter 12 Class 9 Herons Formula
- Chapter 13 Class 9 Surface Areas and Volumes
- Chapter 14 Class 9 Statistics
- Chapter 15 Class 9 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.