# Basic Proportionality Theorem (BPT) Proof and Examples

**Basic Proportionality Theorem** (BPT) is also called **Thales Theorem**. Because Thales, who introduced the study of geometry in Greece, made an important fact related to similar triangles, **"Similar triangles always have the same ratio of the lengths of any two corresponding sides."** was proved.

In this article, along with the Basic Proportionality Theorem, we will study the proportionality theorem statement, the basic proportionality theorem proof.

**Theorem1:** Basic Proportionality Theorem Statement, Basic Proportionality Theorem Proof, Thales Theorem Proof, Thales Theorem Statement-

**"A line drawn parallel to one side of a triangle intersects the other two sides, those points divide those sides in equal proportion."**

This concept has been introduced in similar triangles. If two triangles are similar to each other then,

- Corresponding angles of both the triangles are equal
- Corresponding sides of both the triangles are in proportion to each other

Thus two triangles ΔABC and ΔPQR are similar if,

- ∠A=∠P, ∠B=∠Q and ∠C=∠R
- AB/PQ, BC/QR, AC/PR

## Proof of Basic Proportionality Theorem

Let us now try to prove the basic proportionality theorem statement

Consider a triangle ΔABC, as shown in the given figure. In this triangle, we draw a line PQ parallel to the side BC of ΔABC and intersect the sides AB and AC in P and Q, respectively.

According to the basic proportionality theorem as stated above, we need to prove:

AP/PB = AQ/QC

### Construction of Basic Proportionality Theorem

Join the vertex B of ΔABC to Q and the vertex C to P to form the lines BQ and CP and then drop a perpendicular QN to the side AB and also draw PM⊥AC as shown in the figure.

### Proof

Now the area of ∆APQ = 1/2 × AP × QN (Since, area of a triangle= 1/2× Base × Height)

Similarly, area of ∆PBQ= 1/2 × PB × QN

area of ∆APQ = 1/2 × AQ × PM

Also,area of ∆QCP = 1/2 × QC × PM ………… (1)

Now, if we find the ratio of the area of triangles ∆APQand ∆PBQ, we have

According to the property of triangles, the triangles drawn between the same parallel lines and on the same base have equal areas.

Therefore, we can say that ∆PBQ and QCP have the same area.

area of ∆PBQ = area of ∆QCP …..(3)

Therefore, from the equations (1), (2) and (3) we can say that,

AP/PB = AQ/QC

Also, ∆ABC and ∆APQ fulfil the conditions for similar triangles, as stated above. Thus, we can say that ∆ABC ~∆APQ.

The MidPoint theorem is a special case of the basic proportionality theorem.

According to the mid-point theorem, a line drawn joining the midpoints of the two sides of a triangle is parallel to the third side.

Consider an ∆ABC.

### Conclusion

We arrive at the following conclusions from the above theorem:

If P and Q are the mid-points of AB and AC, then PQ || BC. We can state this mathematically as follows:

If P and Q are points on AB and AC such that AP = PB = 1/2 (AB) and AQ = QC = 1/2 (AC), then PQ || BC.

Also, the converse of mid-point theorem is also true which states that the line drawn through the mid-point of a side of a triangle that is parallel to another side, bisects the third side of the triangle.

Hence, the basic proportionality theorem is proved.