Class 10 - Basic Proportionality Theorem

The Basic Proportionality Theorem (BPT), also known as Thales’ theorem, is a key concept in geometry that explains how a line drawn parallel to one side of a triangle divides the other two sides proportionally. It plays an important role in understanding the similarity of triangles and solving ratio-based problems. In this guide, you’ll explore the statement, converse and proof of the basic proportionality theorem, along with clear formulas and practical examples to build a strong foundation in geometry.

Table of Contents

• What is Basic Proportionality Theorem
• Converse of Basic Proportionality Theorem
• Solved Examples on Basic Proportionality Theorem
• Practice Questions on Basic Proportionality Theorem
• Common Mistakes to Avoid

What is Basic Proportionality Theorem 

Let ABC be a triangle. Draw a line DE that is parallel to side BC, where D lies on AB and E lies on AC. The theorem says that the line DE divides AB and AC in exactly the same ratio.

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

Proof: Let ABC be a triangle. Draw a line DE that is parallel to side BC, where D lies on AB and E lies on AC. 

To prove that: AD/DB = AE/EC

Construction: Join vertices B to E and C to D to form line segments BE and CD. Then draw DM perpendicular to AC, and EN perpendicular to AB.

DM ⊥ AC and  EN  ⊥ AB

 

Consider the triangles ADE and BDE

Area of △ADE = (1/2) × b × h =  (1/2) × AD × EN    -------- (1)

Area of △BDE = (1/2) × b × h = (1/2) × DB × EN  --------- (2)

When we take their ratio of (1) and (2)

ar(△ADE) / ar(△BDE) = AD / DB -------- (3)

Now consider the triangles ADE and CDE
Area of △ADE = (1/2) × b × h =  (1/2) × AE × DM    -------- (4)

Area of △CDE = (1/2) × b × h = (1/2) × EC × DM  --------- (5)

When we take their ratio of (5) and (4)

ar(△ADE) / ar(△CDE) = AE / EC --------- (6)

△BDE and △CDE lie on the same base DE and between the same parallel lines DE and BC.
∴ ar(△BDE) = ar(△CDE) (∵triangles  on same base and lie between the same pair of parallel lines are equal in area). --------- (7)

Hence, ar(△ADE) / ar(△BDE) = ar(△ADE) / ar(△CDE) 

∴  AD/DB = AE/EC ( from (3) and (6))
Hence proved.

• Corollary: DE ∥ BC ⟹ AD/AB = AE/AC
  
  

Converse of Basic Proportionality Theorem

Converse states the reverse of what the original theorem says.

Statement: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Proof: Given: In △ABC, line DE divides AB and AC such that 

AD/DB = AE/EC     ---------- (1)

To prove: DE ∥ BC.

Assume DE is not parallel to BC.

Construction: Draw a line DE’ parallel to BC.

 

Hence, by similar triangles,
AD/DB = AE’/E’C     ---------- (2)
From eq. (1) and (2), we get AE/EC = AE’/E’C
Adding 1 on both the sides.
AE/EC + 1 = AE’/E’C + 1
(AE + EC)/EC = (AE' + E'C)/E'C
AC/EC = AC/E’C
So, EC = E’C

This is possible only when E and E’ coincide.
But, DE’ || BC
DE ||BC.
∴ Hence, proved.

Solved Examples on Basic Proportionality Theorem

Example 1: In △ABC, DE ∥ BC. If AD = 4 cm, DB = 6 cm, and AE = 5 cm, find EC.
Solution: Since DE ∥ BC, by BPT:
AD/DB = AE/EC
4/6 = 5/EC
EC = (5 × 6) / 4 = 30/4 = 7.5 cm
∴ EC =  7.5 cm

Example 2: In △ABC, D and E are points on AB and AC respectively. AD = 2 cm, DB = 4 cm, AE = 3 cm, EC = 6 cm. Show that DE ∥ BC.
Solution: Given AD = 2 cm, DB = 4 cm, AE = 3 cm, EC = 6 cm.
AD/DB = 2/4 = 1/2
AE/EC = 3/6 = 1/2
Since AD/DB = AE/EC = 1/2, by the converse of BPT, DE ∥ BC.

Example 3: In △PQR, ST ∥ QR. PS = x cm, SQ = 3 cm, PT = (x + 1) cm, and TR = 4.5 cm. Find the value of x.
Solution: Given ST ∥ QR, PS = x  cm, SQ = 3 cm, PT =  (x + 1) cm, TR = 4.5 cm.
By BPT (since ST ∥ QR): PS/SQ = PT/TR
x/3 = (x+1)/4.5 
4.5x = 3x + 3
1.5x = 3 
x = 2

Practice Questions on Basic Proportionality Theorem

1.  In △ABC, DE ∥ BC and AD/DB = AE/EC. Also, ∠ADE = ∠ACB. Prove that △ABC is isosceles.

2. In triangle MNO, a line KL is drawn parallel to side NO, intersecting MN at K and MO at L. If MK = 2 cm, KN = 3 cm, and KL = 4 cm, find the length of NO.

3. In triangle ABC, line DE is parallel to BC. If AD = 5 cm, AE = 7.5 cm, and EC = 6 cm, find the length of DB

4. In △ABC, D and E are points on AB and AC, respectively. AD = 2 cm, DB = 4 cm, AE = 3 cm, EC = 6 cm. Prove that DE ∥ BC.

Common  Mistakes to Avoid

• Mixing up the ratio: AD/DB = AE/EC is correct. Writing AD/AB = DB/EC is wrong. Make sure you are comparing corresponding parts

• Forgetting that the line must intersect both sides in distinct points: BPT applies when the line cuts through the interior of the triangle and not at a vertex.

• Applying BPT when the line is not parallel: The theorem only holds when the line is parallel to one side. If that condition is not given or proved, you cannot use BPT directly.

• Confusing BPT with the similarity criteria: BPT tells you about proportional sides when a parallel line is drawn. It does not by itself establish that two triangles are similar; you need additional conditions for that.