### Theory:

Theorem IX
The line segment joining the mid-points of two sides of a triangle is parallel to the third side. Given: $$ABC$$ is a triangle, where $$E$$ is the mid-point of $$AB$$ and $$F$$ is the mid-point of $$AC$$.

To prove: $$EF \ || \ BC$$

Construction: Draw a line segment through $$C$$ parallel to $$AB$$ and extend $$EF$$ to meet this line at $$D$$.

Proof: Since $$E$$ is the mid-point of $$AB$$, $$AE = EB$$ - - - - (I)

Since $$F$$ is the mid-point of $$AC$$, $$AF = FC$$ - - - - (II)

By construction $$AB \ || \ CD$$.

$$\Rightarrow AE \ || \ CD$$ and $$ED$$ is a transversal.

$$\angle AEF = \angle CDF$$ [alternate interior angles] - - - - - (III)

In $$\Delta AEF$$ and $$\Delta CDF$$:

$$\angle AFE = \angle CFD$$ [vertically opposite angles]

$$\angle AEF = \angle CDF$$ [(from (III)]

$$AF = FC$$ [from (II)]

Therefore, $$\Delta AEF \cong \Delta CDF$$ [by $$AAS$$ congruence rule].

$$\Rightarrow AE = CD$$ [by CPCT] - - - - - (IV)

From equations (I) and (IV), we can say that:

$$BE = CD$$

In quadrilateral $$EBCD$$:

$$BE \ || \ CD$$ and $$BE = CD$$

One pair of opposite side is equal and parallel.

Thus, $$EBCD$$ is a parallelogram.

So, $$ED \ || \ BC$$

$$\Rightarrow EF \ || \ BC$$

Hence, proved.

Note: We proved that $$\Delta AEF \cong \Delta CDF$$.

$$EF = FD$$ [by CPCT]

$\mathit{EF}=\frac{1}{2}\mathit{ED}$

$\mathit{EF}=\frac{1}{2}\mathit{BC}$ [Since $$EBCD$$ is a parallelogram, $$ED = BC$$].

Thus, we can conclude that the parallel mid-line segment is half of the third side of the triangle.

Important!
The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.
Theorem X
The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side. Given: $$ABC$$ is a triangle, where $$E$$ is the mid-point of $$AB$$. Also, $$EF \ || \ BC$$.

Construction: Draw a line segment through $$C$$ parallel to $$AB$$ and extend $$EF$$ to meet this line at $$D$$.

To prove: $$F$$ is the mid-point of $$AC$$. That is, $$AF = FC$$

Proof: Since $$E$$ is the mid-point of $$AB$$, $$AE = EB$$ - - - - (I)

In quadrilateral $$EBCD$$:

$$EF \ || \ BC$$ [since $$ED \ || \ BC$$]

$$EB \ || \ DC$$ [by construction]

Both pairs of opposite sides are parallel.

Thus, $$EBCD$$ is a parallelogram.

We know that "opposite sides of a parallelogram are equal".

$$EB = DC$$ - - - - (II)

On comparing equation (I) and (II), we get, $$AE = DC$$ - - - - (III).

Since $$AE \ || \ DC$$ with transversal $$ED$$ by construction.

$$\angle AEF = \angle CDF$$ [alternate interior angles] - - - -  (IV)

In $$\Delta AEF$$ and $$\Delta CDF$$:

$$\angle AEF = \angle CDF$$ [from (IV)]

$$\angle AFE = \angle CFD$$ [vertically opposite angles]

$$AE = DC$$ [from (III)]

Therefore, $$\Delta AEF \cong \Delta CDF$$ [by $$AAS$$ congruence criterion].

$$\Rightarrow AF = CF$$ [by CPCT]

Thus, $$F$$ is the mid-point of $$AC$$.

Hence, proved.