Ex 7.1, 7 - Chapter 7 Class 9 Triangles (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 7 Class 9 Triangles
Example 3
Important
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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
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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
Transcript
Ex7.1, 7 AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB (See the given figure). Show that DAP EBP (ii) AD = BE Given: P is the mid point of AB, So, AP = BP BAD = ABE EPA = DPB To prove: (i) DAP EBP (ii) AD = BE Proof: Since EPA = DPB We add DPE both sides EPA + DPE = DPB + DPE DPA = EPB In DAP and EBP, DAP = EBP AP = BP DPA = EPB DAP EBP AD = BE
