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Similarity of Triangles

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Similarity of Triangles - Lesson Summary

In general, there are several objects which have something common between them. Observing them closely, let see that some of them have same shape but may have different or same size.

For example, if the photographs a person developed from same negative, they all look same in all respect except for their size. Such objects are called similar objects. 

Two line segments of different sizes, two circles of different radii, two squares of different sizes, two rectangles of different dimensions - come under similar figures. One smaller circle can be got by shrinking a larger circle. One bigger square can be got by stretching a smaller square.

Then, what about the similarity of triangles? Is it true to say any two given triangles are similar? The answer is NO. This is true only when the  triangles are equilateral.

For all other triangles, we have the following statement which lays down the condition for the similarity of two triangles.

“Two triangles are said to be similar if their corresponding angles are equal and their sides are proportional”

The symbol ~ for the similarity of two triangles. Write ΔABC ~ ΔA'B'C' for the similarity of ΔABC and ΔA'B'C'. Further, we follow that the vertices A corresponds to A', B corresponds to B' and C corresponds to C', the angles ∠A = ∠A', ∠B = ∠B' and ∠C = ∠C' and   AB A'B' =  BC B'C' =  AC A'C'. This ratio is called Scale factor.

Basic proportionality theorem or Thales’ theorem:
If a line is drawn parallel to one side of a triangle and it intersects the other two sides in two distinct points then it divides the two sides in the same ratio.

In the ΔABC , if DE || BC, then  AD DB =  AE EC.

Given: In ΔABC, DE is parallel to BC.

To prove:   AD DB  =   AE EC 

Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB

Proof: Consider ΔADE and ΔBDE.

ar(ΔADE) = ½ x AD x EN

ar(ΔBDE) = ½ x DB x EN

ar( ∆ ADE) ar( ∆ BDE)    =  1 2 x AD x DM 1 2 x DB x DM         = AD DB  ...........(1)

Consider ΔADE and ΔDEC

ar(ΔADE) = ½ x AE x DM

ar(ΔDEC) = ½ x EC x DM

ar( ∆ ADE) ar( ∆ DEC)    =   1 2 x AE x DM 1 2 x EC x DM     = AE EC  ..........(2)

Consider ΔBDE and ΔCED

ar(ΔBDE) = ar(ΔCED) .........(3)

[Since ΔBDE and ΔCED are on the same base, DE, and between the same parallels, BC and DE.]

∴ AD DB  =  AE EC   [From (1), (2) and (3)]

Converse of Basic Proportionality Theorem:
If a line divides any two sides of a triangle in the same ratio, the line must be parallel to the third side.

In theΔABC, if D and E are two points on AB and AC respectively such that, AD DB =  AE EC, then DE||BC

Given: In ΔABC, D and E are two points on AB and AC, respectively, such that,  AD DB  = AE EC

To prove: DE∥BC

Proof: In ΔABC,

   AD DB  = AE EC   (Given) ............. (1)

Since DE' ∥ BC, by basic proportionality Theorem, We have

  AD DB  = AE' E'C .........................(2)

From (1) and (2),

  AE EC  = AE' E'C .........................(3)

  AE EC + 1 = AE' E'C + 1

  AE+EC EC  = AE'+E'C E'C     

∴    AC EC  = AC E'C

⇒ EC = E'C

∴ DE ∥ BC


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