Home/ Questions/Q 550863
Next
In Process
Virat
  • 1
Asked: 2026-10-03T06:00:47+00:00 In:

A more general midpoint theorem and its converse. In a quadrilateral ABCD, suppose AB ∥ DC. Recall that such ABCD is called a trapezium. Let E be the midpoint of AD. A line drawn through E intersects side BC at F. (i) If EF ∥ AB, then show that F is the midpoint of BC. Conclude that EF = (AB + CD)/2. (ii) If F is the midpoint of BC, then show that EF ∥ AB. There are at least two ways to solve (ii). Do it directly by using the midpoint M of BD and showing that EM and FM are the same lines. Or use (i) along with the same idea used in the second proof of the converse of the Midpoint Theorem. Which way do you think is simpler?

  • 1

(i) Diagonal BD meets EF at M. By Theorem 7, M and F bisect BD and BC. Adding segment lengths gives EF = (AB + CD)/2. (ii) Both methods work; using parallel line uniqueness is simpler.

Class 9 Ganita Manjari Part 2 chapter 12 question answer
Class 9 Ganita Manjari Part 2 chapter 12 Quadrilaterals solutions

) if f is the midpoint of bcalong with the same idea used in the second proofcbse class 9 maths ganita manjari part 2 solutionsclass 9 maths ganita manjari part 2 chapter 12 question answergeneral midpoint theorem and its conversemidpoint m of bd and showing that em and fm
Share

    1 Answer

    1. (i) Draw diagonal BD meeting line EF at M. In triangle DAB, line EM through midpoint E parallel to AB bisects BD at M. In triangle BDC, MF ∥ DC bisects BC at F, yielding EF = EM + MF = AB/2 + CD/2 = (AB + CD)/2. (ii) Midpoints of AD and BC join to form EF. Using the uniqueness of a parallel line through E combined with part (i) is simpler than collinearity proofs.

       

      For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)

      https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/

      • 0
    Leave an answer

    Leave an answer

    Browse