Using converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side

Hint: In order to determine using the converse of basic proportionality theorem, we have to prove that the line joins the midpoint. When a line is drawn parallel to one side of a triangle to cross the other two interesting points, the other two sides are split in the same ratio, which is known as the Thales theorem.

Complete step by step solution:
We have to prove that the Converse of basic proportionality theorem
Statement of basic proportionality theorem (BPT)
According to this theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
We are given as follows
\[\Delta ABC\] in which \[D\] and \[E\] are the midpoints of \[AB\] and \[AC\] respectively such that \[AD = BD\] and \[AE = EC\].
If a line divides any two sides of a triangle in the same ratio then the line must parallel to the third side.
To Prove: \[DE||BC\]
Proof: In the given diagram, \[D\] is the midpoint of \[AB\].
\[\therefore AD = DB\]
\[ \Rightarrow \dfrac{{AD}}{{BD}} = {\text{ }}1\] --------(1)
Also, from the diagram given \[E\] is the midpoint of \[AC\].
\[\therefore AE = EC\]
\[ \Rightarrow \dfrac{{AE}}{{EC}} = 1\] ---------(2)
From equation (1) and (2), we get
\[\dfrac{{AD}}{{BD}} = \dfrac{{AE}}{{EC}}\]
Therefore, \[DE||BC\;\]
As a result, the converse of the Basic Proportionality Theorem is proved.
Hence, the line joining the midpoints of any two sides of a triangle is parallel to the third side.

Note:
This concept has been introduced in similar triangles.  If two triangles are similar to each other then,
1) Corresponding angles of both the triangles are equal
2) Corresponding sides of both the triangles are in proportion to each other
we need to remember that the previously stated theorem, we have to derive the following conclusions:

\[PQ||BC\]if \[P\] and \[Q\] are the midpoints of \[AB\]and \[AC\].
This can be expressed mathematically as follows:
\[PQ||BC\] if \[P\] and \[Q\] are points on \[AB\] and \[AC\] with \[AP = PB = \dfrac{1}{2}(AB)\]and\[AQ = QC = \dfrac{1}{2}(AC)\].
Finally, The converse of the mid-point theorem, which states that the line drawn through the midpoint of a triangular side is also valid, is also true.