Last updated at May 29, 2018 by Teachoo

Transcript

Ex7.1, 8 In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that: ΔAMC ≅ ΔBMD Given: ∠ ACB = 90° M is the mid-point of AB So, AM = BM Also, DM = CM To prove: ΔAMC ≅ ΔBMD Proof: Lines CD & AB intersect… So, ∠AMC = ∠BMD In ΔAMC and ΔBMD, AM = BM ∠AMC = ∠BMD CM = DM ∴ ΔAMC ≅ ΔBMD Ex7.1, 8 In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that: (ii) ∠DBC is a right angle. From part 1, ΔAMC ≅ ΔBMD ∴ ∠ACM = ∠BDM But ∠ACM and ∠BDM are alternate interior angles for lines AC & BD If a transversal intersects two lines such that pair of alternate interior angles is equal, then lines are parallel. So, BD || AC Now, Since DB || AC and Considering BC as transversal , ⇒ ∠DBC + ∠ACB = 180° ⇒ ∠DBC + 90° = 180° ⇒ ∠DBC = 180° – 90° ⇒ ∠DBC = 90° Hence, ∠ DBC is a right angle Hence proved Ex 7.1, 8 In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that: (iii) ΔDBC ≅ ΔACB From part 1, ΔAMC ≅ ΔBMD AC = BD In ΔDBC and ΔACB, DB = AC ∠DBC = ∠ACB BC = CB ∴ ΔDBC ≅ ΔACB Ex7.1, 8 In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that: (iv) CM = 1/2 AB From part 3, ΔDBC ≅ ΔACB ∴ DC = AB 1/2 DC = 1/2 AB CM = 1/2 AB Hence proved

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.