Basic Proportionality Theorem 

 A prominent Greek mathematician, Thales, developed the fundamental proportionality theorem, which is also known as a Thales theorem. In his view, the ratio of each two matching sides is always the same for any two equiangular triangles. He gave a theorem of basic proportionality based on this idea (BPT). The basic proportionality theorem states that when the line is drawn parallel to one of the two sides of the triangle in distinct spots, the other two sides are split into the same ratio.

Note: Only equiangular or equilateral triangles that comprise all three angles of 60 degrees are faithful to this theorem.

For instance, in the given figure, line KL is drawn parallel to side QR, such that it joins the other two sides, PQ and QR. As stated by the basic proportionality theorem, it can be inferred that PK/KQ = PL/LR.

Proof of BPT

Construction: Let us design a triangle KLM. A line segment AB is inscribed parallel concerning the side LM, creating two points A on the side KL and B on the side KM of the triangle. In the new triangle, KAB let us draw perpendiculars from vertices A and B such that we have two perpendiculars, AD and BC. Then, connect the point A to M and B to L using dotted lines. The diagram of this construction is given below:

From the above diagram, let us consider the triangles KAB and LAB. Both these triangles share the same base AB and consist of the equal height BC.

(Area of KAB) / (Area of LAB) = (1/2 × KA × BC) / (1/2 × LA × BC)

(Area of KAB) / (Area of LAB) = KA/LA

Again, let us look at the diagram and now consider triangles MAB and KAB. Both these triangles once more share the same base AB and have equal altitude AD.

(Area of KAB) / (Area of MAB) = (1/2 × KB × AD) / (1/2 × BM × AD)

(Area of KAB) / (Area of MAB) = KB/BM

Now the pair of triangles LAB and MAB are in between the same set of parallel lines, AB and LM.

As stated in the properties of the triangle, the area of triangles between two parallel lines are equal; therefore, Area of triangle LAB= Area of triangle MAB

Applying this, we have, (Area of triangle KAB) / (Area of triangle LAB) = (Area of triangle KAB) / (Area of triangle MAB)

KA/AL = KB/BM

Hence proved that the ratio of both sides is equal and proportional.

Special Conclusion:

Mid-Point Theorem: Another significant theorem called the mid-point theorem may also be proved by the above approach. The mid-point theorem asserts that a segment drawn parallel to one side of the triangle splits the other two sides at the midpoints.

The Converse of Basic Proportionality Theorem

The converse of any mathematical statement is the reverse of that statement. The converse of BPT states that if a line segment dissects the two opposite sides of a triangle in an equal ratio, then that line segment is parallel to the third side.

Let us consider the triangle ABC where a line segment PQ divided the side AB and BC into equal proportions. Then according to the converse of BPT, the line PQ is parallel to the third side of the triangle. i.e., BC. PQ||BC.