Question

proof for converse of basic proportionality theorem?

Answer 1

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.

StatementsReasons1) DF || BC1) By assumption2) AD / DB = AF / FC2) By Basic Proportionality theorem3) AD / DB = AE /EC3) Given4) AF / FC = AE / EC4) By transitivity (from 2 and 3)5) (AF/FC) + 1 = (AE/EC) + 15) Adding 1 to both side6) (AF + FC )/FC = (AE + EC)/EC6) By simplifying7) AC /FC = AC / EC7) AC = AF + FC and AC = AE + EC8) FC = EC8) 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) 

Answer 2

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