prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side
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The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
You can prove this theorem using the following clue:
Observe the figure in which E and F are mid-points of AB and AC respectively and CD || BA.
∆ AEF ≅ ∆ CDF (ASA Rule)
So, EF = DF and BE = AE = DC. Therefore, BCDE is a parallelogram. This gives EF || BC.
You will see that converse of the above theorem is also true which is stated as below:
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