E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and EF = 1/2 (AB + CD) [Hint: Join BE and produce it to meet CD produced at G.]
Solution:
Given, ABCD is a trapezium in which AB || CD
E and F are the midpoints of the non-parallel sides AD and BC
We have to prove that EF || AB and EF = 1/2 (AB + CD)
Join BE and extend it to meet CD produced at G
Draw BD which intersects EF at O.
Consider triangle GCB,
E and F are the midpoints of BG and BC.
The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”
By Midpoint theorem,
EF || GC
Given, AB || GC or CD || AB
So, EF || AB
Considering triangle ABD,
AB || EO
E is the midpoint of AD
The converse of the midpoint theorem states that if a line is drawn through the midpoint of one side of a triangle, and parallel to the other side, it bisects the third side.
By converse of midpoint theorem,
O is the midpoint of BD
EO = 1/2 AB -------------- (1)
Considering triangle BDC,
OF || CD
O is the midpoint of BD
By converse of midpoint theorem,
OF = 1/2 CD ------------- (2)
Adding (1) and (2),
EO + OF = 1/2 AB + 1/2 CD
From the figure,
EF = EO + OF
Therefore, EF = 1/2 (AB + CD)
✦ Try This: Can a quadrilateral ABCD be a parallelogram if DC = 8 cm, AD = 4 cm and
BC = 4.4 cm?
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 8
NCERT Exemplar Class 9 Maths Exercise 8.4 Problem 12
E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and EF = 1/2 (AB + CD). [Hint: Join BE and produce it to meet CD produced at G.
Summary:
E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. It is proven that EF || AB and EF = 1/2 (AB + CD)
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