Basic Proportionality Theorem and its Converse


 
 
Concept Explanation
 

Basic Proportionality Theorem and its Converse

Basic Proportionality Theorem ( Thales Theorem ) and its converse

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

Given: A triangle ABC and  DE || BC

large To Prove: frac{AD}{DE}=frac{AE}{EC}

Construction: Join BE and CD and draw DM large perp AC and EN large perp AB

Proof:

large area;of;(Delta ADE)= frac{1}{2} ;X ;AD ;X ;EN            .......(1)

large area;of;(Delta BDE)= frac{1}{2} ;X ;DB ;X ;EN            ......(2)

large Also;area;of;(Delta ADE)= frac{1}{2} ;X ;AE ;X ;DM    .....(3)

large area;of;(Delta DEC)= frac{1}{2} ;X ;EC ;X ;DM               .....(4)

Using Eq 1, 2, 3 and 4

large frac{ar(ADE)}{ar(BDE)}= frac{frac{1}{2}X AD X EN}{frac{1}{2}X DB X EN}=frac{AD}{DB}                   ........(5)

large frac{ar(ADE)}{ar(DEC)}= frac{frac{1}{2}X AE X DM}{frac{1}{2}X EC X DM}=frac{AE}{EC}                   ........(6)

Now large Delta BDE and large DeltaDEC are on the same base DE and between the same parallel  BC || DE

So ar(BDE) = ar( DEC)

Using the relation in Eq 5 and 6 we get

large frac{AD}{DB}=frac{AE}{EC}

Hence Proved

Converse of the basic proportionality theorem is also true,

Theorem   If a line divides any twosides of a triangle in the same ratio, the line is parallel to the third side.

Given : a triangle ABC  and That is, if in fig. large frac{AD}{DB}=frac{AE}{EC},

To Prove :  large DEparallel BC

Construction: Draw DF || BC

Proof: If DF || BC

then large frac{AD}{DB}= frac{AF}{FC}                            ...... (1)

But we are given large frac{AD}{DB}=frac{AE}{EC}         ...............(2)

Then from 1 and 2 we get

large frac{AF}{FC}= frac{AE}{EC}

This is possible only when F and E coincide.

This implies DE || BC

Example  :.

large dpi{120} large If ; DEparallel BC; and;   large AD=4x-3, AE=8x-7,BD=3x-1large and;CE=5x-3,; Find ;x ;when; x>0 

Solution   

             large As;DEparallel BC,

           large frac{AD}{DB}=frac{AE}{EC}        [ Basic Proportionality Theorem]

large Rightarrow       large frac{4x-1}{3x-1}=frac{8x-7}{5x-3}               

large Rightarrow        large (4x-3)(5x-3)=(3x-1)(8x-7)

large Rightarrow       large 20x^{2}-15x-12x+9=24x^{2}-8x-21x+7

large Rightarrow       large 4x^{2}-2x-2=0;;or;;2x^{2}-x-1=0

large Rightarrow       large 2x^{2}-2x+x-1=0

large Rightarrow       large 2x(x-1)+(x-1)=0

large Rightarrow       large (2x+1)(x-1)=0;;;Rightarrow ;;x=-frac{1}{2},1

As          large x> 0,x=1

Sample Questions
(More Questions for each concept available in Login)
Question : 1

large Delta ABC sim Delta DEF.   If DE = 2 AB and BC = 3cm then EF is equal to ____________

Right Option : C
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Question : 2

In a given triangle, DE || BC and large frac{AD}{DB}=frac{3}{5,}  large AC=5.6cm, find AE.

Right Option : B
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Explanation
Question : 3

In the figure, D and E are two points on the sides AB and AC respectively of large Delta ABC. If dpi{110} DE | BC,  AD = 2 cm, AB = 6 cm and AC = 9 cm, find AE.

Right Option : B
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Explanation
 
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