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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