prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and also half of it.
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Let the ∆ is ABC and EF is the line which passes through the mid points than angle A is common on both ∆ and
AE/AB =AF/AC ( both is 1/2)
Then from SAS similarity
∆ABC≈∆AEF
So,angle ABC=angle AEF
That is only possible when the line is parallel
AE/AB =AF/AC ( both is 1/2)
Then from SAS similarity
∆ABC≈∆AEF
So,angle ABC=angle AEF
That is only possible when the line is parallel
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