Basic Proportionality Theorem (BPT) | Triangles | Class 10 Mathematics

01 · THE THEOREM

Theorem 6.1 (Thales Theorem)

The Basic Proportionality Theorem (BPT) states:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

NCERT PROOF DIAGRAM

Triangle ABC with DE ∥ BC & Proof Constructions

Given: DE ∥ BC
Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.
To Prove: AD / DB = AE / EC

02 · PROOF

Proof of BPT

Using the area of triangles formula Area = 1/2 × Base × Height:

Step 1

Draw perpendiculars DM ⊥ AC and EN ⊥ AB. Join BE and CD.

Step 2

area(△ADE) / area(△BDE) = (1/2 × AD × EN) / (1/2 × DB × EN) = AD / DB.

Step 3

area(△ADE) / area(△DEC) = (1/2 × AE × DM) / (1/2 × EC × DM) = AE / EC.

Step 4

Since △BDE and △DEC lie on the same base DE and between same parallels DE ∥ BC, area(△BDE) = area(△DEC).

Therefore, AD / DB = AE / EC (Proved)

03 · CONVERSE

Converse of Basic Proportionality Theorem (Theorem 6.2)

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Theorem Statement

If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
If AD / DB = AE / EC, then DE ∥ BC.

04 · QUICK SUMMARY

Key Takeaways

BPT: DE ∥ BC ⟹ AD / DB = AE / EC.

Corollary: AD / AB = AE / AC & DB / AB = EC / AC.

Converse of BPT: AD / DB = AE / EC ⟹ DE ∥ BC.