Similar Triangles

Two or more figures may have the same shape but not necessarily same size then those figures are said to be similar.

However, if two or more figures have the same shape and same size then those figures are said to be congruent.

Note:

  1. The congruent figures are similar but, the converse is not true, i.e. similar figures need not be congruent.

  2. All regular polygons of the same number of sides are similar like, all equilateral triangles, all squares, etc. are similar.

  3. All circles are always similar irrespective of their diameter or radius.

  4. The similarity of figures is denoted by the ‘~’ symbol.

Similarity of Two Polygons

Two polygons of the same number of sides are similar, if 

  1. their corresponding angles are equal and 

  2. their corresponding sides are in the same ratio (or proportion).

Similarity of Two Triangles

Since, we know that triangle is the smallest three-sided polygon. So, we can state the same conditions for the similarity of two triangles. That is: Two triangles are similar, if 

  1. their corresponding angles are equal and 

  2. their corresponding sides are in the same ratio (or proportion).

In geometry, correspondence means that a particular part of one polygon relates exactly to a similarly positioned part of another polygon. Even if two triangles are oriented differently from each other, if we rotate them to orient in the same way and see that their angles are alike, we can say those angles are corresponding angles of the two triangles.

Example:

In the given figure, two triangles ΔABC and ΔPQR are similar only if,

i)       \[\angle A{\text{ }} = {\text{ }}\angle P,{\text{ }}\angle B{\text{ }} = {\text{ }}\angle Q{\text{ }}and{\text{ }}\angle C{\text{ }} = {\text{ }}\angle R\]

 ii)       \[\frac{{AB}}{{PQ}} = \frac{{BC}}{{QR}} = \frac{{AC}}{{PR}}\]

Hence if the above-mentioned conditions are satisfied then we can say that ΔABC ~ ΔPQR

Note that if corresponding angles of two triangles are equal, then they are known as equiangular triangles. 

A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which is as follows: The ratio of any two corresponding sides in two equiangular triangles is always the same. It is believed that he had used a result called the Basic Proportionality Theorem (now known as the Thales Theorem) for the same.

Basic Proportionality Theorem (or Thales Theorem) 

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

Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.

To prove:  \[\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\]

Construction: Let us join BE and CD and then draw DM ⊥ AC and EN ⊥ AB.

Proof:   Since, area of Δ ADE = (\[\frac{1}{2}\]× base × height) = \[\frac{1}{2}\] × AD × EN.

Area of Δ ADE is denoted as ar(ADE).

So, ar(ADE) = \[\frac{1}{2}\]× AD × EN

Similarly, ar(BDE) = \[\frac{1}{2}\] × DB × EN,

                 ar(ADE) = \[\frac{1}{2}\] × AE × DM and ar(DEC) = \[\frac{1}{2}\]× EC × DM.

Therefore, \[\frac{{ar\left( {ADE} \right)}}{{ar\left( {BDE} \right)}} = \frac{{\frac{1}{2} \times AD \times EN}}{{\frac{1}{2} \times DB \times EN}} = \frac{{AD}}{{DB}}\]                                         (1) 

                                                                                                                  

And           \[\frac{{ar\left( {ADE} \right)}}{{ar\left( {DEC} \right)}} = \frac{{\frac{1}{2} \times AE \times DM}}{{\frac{1}{2} \times EC \times DM}} = \frac{{AE}}{{EC}}\]                                        (2)

 Since, Δ BDE and DEC are on the same base DE and between the same parallels BC and DE. 

So, ar(BDE) = ar(DEC)                                                                   (3)

Therefore, from (1), (2) and (3), 

we obtain:   \[\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\]           (Proved)

Similarity Criterion of a Triangle

There are three similarity criteria of triangles which are used to solve the problems based on similar triangles. They are as following:

  1. AA (or AAA) or Angle-Angle similarity criterion

  2. SSS or Side-Side-Side similarity criterion

  3. SAS or Side-Angle-Side similarity criterion

AA (or AAA) or Angle-Angle similarity criterion

Statement: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

Also, if two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal. 

Therefore, AAA similarity criterion can also be stated as follows:

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. This may be referred to as the AA similarity criterion for two triangles.

SSS or Side-Side-Side Similarity Criterion

Statement: If in two triangles, sides of one triangle are proportional to (or in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

SAS or Side-Angle-Side Similarity Criterion

Statement: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Areas of Similar Triangles

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. i.e.,

For given two triangles ABC and PQR such that Δ ABC ~ Δ PQR     

Then, \[\frac{{ar\left( {ABC} \right)}}{{ar\left( {PQR} \right)}} = {\left( {\frac{{AB}}{{PQ}}} \right)^2} = {\left( {\frac{{BC}}{{QR}}} \right)^2} = {\left( {\frac{{AC}}{{PR}}} \right)^2}\]

Solved Examples:

Q.1. Give Examples of Pairs of (i) Similar Figures (ii) Non-Similar Figures.

Ans. (i) similar figures: all equilateral triangles, all circles.

         (ii) non-similar figures: a circle and a square, a triangle and a trapezium.

Q.2. In the Figure Given Below, DE || BC. Find EC.

Ans. Let EC = x cm.

        It is given that DE || BC.

By using Basic proportionality theorem, we obtain

   \[\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\]

\[\frac{{1.5}}{3} = \frac{1}{x}\]

\[X = \frac{3}{{1.5}}\]

x = 2

So, EC = 2 cm.

Q.3. Let Δ ABC ~ Δ DEF and their Areas be 64 cm2 and 121 cm2 Respectively. If EF = 15.4 cm then Find BC.

Ans.  It is given that Δ ABC ~ Δ DEF

So, \[\frac{{ar\left( {ABC} \right)}}{{ar\left( {DEF} \right)}} = {\left( {\frac{{AB}}{{DE}}} \right)^2} = {\left( {\frac{{BC}}{{EF}}} \right)^2} = {\left( {\frac{{AC}}{{DF}}} \right)^2}\]

Given that, 

EF = 15.4 cm

ar(ABC) = 64 cm2

ar(DEF) = 121 cm2


So, \[\frac{{ar\left( {ABC} \right)}}{{ar\left( {DEF} \right)}}\]=\[{\left( {\frac{{BC}}{{EF}}} \right)^2}\]


  \[\left( {\frac{{64{\text{ }}cm2}}{{121{\text{ }}cm2}}} \right) = \frac{{BC2}}{{15.4{\text{ }}cm}}\]

\[\frac{{BC}}{{15.4{\text{ }}cm}} = \frac{8}{{11}}{\text{ }}cm\]

\[BC = \frac{{8 \times 15.4}}{{11}}\]= 11.2 cm