SSC (English Medium) Class 9Maharashtra State Board

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Concept: The Mid-point Theorem

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Draw a triangle and mark the mid-points E and F of two sides of the triangle. Join the points E and F in following fig. 

Measure EF and BC. Measure ∠ AEF and ∠ ABC. EF = `1/2` BC and ∠ AEF = ∠ ABC 
so, EF || BC


Theorem : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Observe following Fig.  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. 
In this case, also note that EF = `1/2` ED = `1/2` BC.


Theorem : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.
In following fig.  

Observe that E is the mid-point of AB, line l is passsing through E and is parallel to BC and CM || BA. 
Prove that AF = CF by using the congruence of ∆ AEF and ∆ CDF.


  • Theorem of midpoints of two sides of a triangle
  • Converse of midpoint theroem | Mid-point theorem

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