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Maths / Triangles / Basic Proportionality Theorem and its Converse

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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 $large Delta$ DEC 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 :.

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

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 $large x> 0,x=1$

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Basic Proportionality Theorem and its Converse

₹17200 ₹12000 +
Taxes