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 : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side, and is equal to half of it.
- Converse of midpoint theroem : The straight line drawn through the mid-point of one side of a triangle parallel to another, bisects the third side.
Shaalaa.com | Mid-point theorem
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