Converse of Basic Proportionality Theorem

Converse of Basic Proportionality Theorem: 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 AE
---- = ------ then DE || BC
DB EC

Given : A Δ ABC and a line intersecting AB in D and AC in E,
such that AD / DB = AE / EC.

Prove that : DE || BC


Let DE is not parallel to BC. Then there must be another line that is parallel to BC.

Let DF || BC.

Statements
Reasons
1) DF || BC 1) By assumption
2) AD / DB = AF / FC 2) By Basic Proportionality theorem
3) AD / DB = AE /EC 3) Given
4) AF / FC = AE / EC 4) By transitivity (from 2 and 3)
5) (AF/FC) + 1 = (AE/EC) + 1 5) Adding 1 to both side
6) (AF + FC )/FC = (AE + EC)/EC 6) By simplifying
7) AC /FC = AC / EC 7) AC = AF + FC and AC = AE + EC
8) FC = EC 8) As the numerator are same so denominators are equal

This is possible when F and E are same. So DF is the line DE itself.

∴ DF || BC

Examples

1) D and E are respectively the points on the sides AB and AC of a ΔABC such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm, prove that DE || BC.

Solution :

AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm

∴ BD = AB – AD = 5.6 – 1.4 = 4.2 cm

And EC = AC – AE = 7.2 – 1.8 = 5.4 cm

Now, AD / DB =1.4 / 4.2 = 1/3

And AE / EC = 1.8 / 5.4 = 1/3

⇒ AD / DB = AE / EC

DE || BC ( by converse of basic proportionality theorem)

---------------------------------------------------------------
2) State whether PQ || EF.

DP / PE = 3.9 / 3 = 13/10

DQ / QF = 3.6 /2.4 = 3/2

So, DP / PE ≠ DQ / QF

⇒ PQ ∦ EF ( PQ is not parallel to EF)

Similarity in Triangles

• Similarity in Geometry
• Properties of similar triangles
• Basic Proportionality Theorem(Thales theorem)
• Converse of Basic Proportionality Theorem
• Interior Angle Bisector Theorem
• Exterior Angle Bisector Theorem
• Proofs on Basic Proportionality
• Criteria of Similarity of Triangles
• Geometric Mean of Similar Triangles
• Areas of Two Similar Triangles

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