EXAMPLE PROBLEMS ON BASIC PROPORTIONALITY THEOREM

Basic Proportionality Theorem :

If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

Example 1 :

In a triangle ABC,D and E are points on the sides AB and AC respectively such that DE is parallel to BC.

(i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, then find AC.

Solution :

In the given triangle ABC the sides DE is parallel to the side BC. By using “Thales theorem”, we get

AD/DB = AE/EC

6/9 = 8/EC

EC = (9  8)/6

EC = (3  4)

EC = 12 cm

Example 2 :

In triangle ABC, if AD = 8 cm, AB = 12 cm and AE = 12 cm, then find CE.

Solution :

In the given triangle ABC the sides DE is parallel to the side BC. By using “Thales theorem”, we get

AD/DB = AE/EC

AB = AD + DB

12 = 8 + DB

DB = 4 cm

AD/DB = AE/EC

8/4 = 12/EC

EC = (12  4)/8

EC = 48/8

EC = 6 cm

Example 3 :

In triangle ABC, if AD = 4 x – 3, BD = 3 x – 1, AE = 8 x – 7 and EC = 5 x – 3, then find the value of x.

Solution :

In the given triangle ABC the sides DE is parallel to the side BC. By using “Thales theorem”, we get

AD/DB = AE/EC

(4x – 3)/(3x – 1) = (8x – 7)/(5x – 3)

(4x – 3)(5x – 3) = (8x – 7)(3x – 1)

20x2 – 12x – 15 x + 9 = 24x2 – 8x – 21x + 7

 20x2 – 24x + 9 =  8x2 – 29x + 7

20x2 - 24x2 – 27x + 29x + 9 – 7 = 0

-4x2 + 2 x + 2 = 0

Multiply by -2.

2x2 – x - 1 = 0

(x – 1)(2 x + 1) = 0

x - 1 = 0

x = 1

2x + 1 = 0

2x = -1

x = -1/2

So the value of x is 1.

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