Theory:

Theorem IX
The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
 
A9.png
 
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]
 
EF=12ED
 
EF=12BC [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.
 
A9.png
 
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.