#### text

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

**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

**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.

#### description

**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.