Theorem 8.9
The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Given : ABCD is a triangle where
E and F are mid points of
AB and AC respectively
To Prove : EF BC
Construction : Through C draw a line segment parallel to AB
& extend EF to meet this line at D.
Proof : Since EB DC
with transversal ED.
AEF = CDF
In AEF and CDF
AEF = CDF
AF = CF
AFE = CFD
AEF CDF
So, EA = DC
But, EA = EB
Hence, EB = DC
Now,
In EBCD,
EB DC & EB = DC
Thus, one pair of opposite sides is equal and parallel.
Hence EBCD is a parallelogram.
Since opposite sides of parallelogram are parallel.
So, ED BC
i.e. EF BC
Hence, proved.

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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