CBSE Class 10 Mathematics chapter 6 - Triangle - Criteria for Similarity of Triangles

Criteria for Similarity of Two Triangles

Two triangles are said to be similar triangles, If: 

• Their corresponding angles are equal.
• Their corresponding sides are in the same proportion/ratio.

Let's take two triangles ABC and DEF as fallow:

• In ΔABC and ΔDEF
• ∠A = ∠D
• ∠B = ∠E
• ∠C = ∠F and
• $$\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}$$
• then the two triangles are similar.
• The similarity of two triangles can be represented symbolically as '∆ABC ~ ∆DEF'.
• The symbol '∼' is used to denote similarity, and it stands for 'is similar to'.
• So, '∆ABC ~ ∆DEF' is read as 'triangle ABC is similar to triangle DEF'.

Criteria for Similarity of Triangles

There are five criteria for similarity of triangles which is used to determining the similarity of triangles are:

• AAA Similarity (Angle-Angle-Angle Similarity)
• AA Similarity (Angle-Angle Similarity)
• SAS Similarity (Side-Angle-Side Similarity)
• SSS Similarity (Side-Side-Side Similarity)

Let us discuss in detail.

AAA Similarity (Angle-Angle-Angle Similarity):

• According to this criteria "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."

Proof:

Let's take two triangle ∆ABC and ∆DEF such that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F as shown in below in the figure.

Now, we cut DP = AB and DQ = AC and Join PQ

Because of ∆ ABC and ∆ DPQ are congruent, so we can write

∆ ABC ≅ ∆ DPQ

which gives ∠B = ∠P = ∠E and the line PQ is parallel to EF.

By using the basic proportionality theorem, we can write

$$\frac{DP}{PE}=\frac{DQ}{QF}$$

$$i.e., \frac{AB}{DE}=\frac{AC}{DF}................... (i)$$

Similarly,

$$\frac{DF}{DE}=\frac{BC}{EF}...................... (ii)$$

From (1) and (2), we can write:

$$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$$

Hence proved that the two triangles ABC and DEF are similar.

AA Similarity (Angle-Angle Similarity)

• We know that— 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 and they are similar triangle.
• So, It is known as AA Similarity (Angle-Angle Similarity) criteria.
• Symbolically it can be represent as : If ∠A = ∠D and ∠B = ∠E, then ∆ABC ~ ∆DEF.

Example: Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm,CA = 8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm

So, we have $$\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}$$

Let us check them

$$\frac{AB}{DE}=\frac{3}{4.5}=\frac{30}{45}=\frac{2}{3}$$

$$\frac{BC}{EF}=\frac{6}{9}=\frac{2}{3}$$

$$\frac{CA}{FD}=\frac{8}{12}=\frac{2}{3}$$

Each equal to $$\frac{2}{3}$$

When we measure ∠A, ∠B, ∠C, ∠D, ∠E and ∠F

∠A = ∠D

∠B = ∠E

∠C = ∠F

Because the corresponding angles of the two triangles are equal.

SAS Similarity (Side-Angle-Side Similarity)

According to this criteria "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."

Proof:

To prove this theorem, takes two triangles ABC and DEF

By basic proportionality theorem, we get

$$\frac{AB}{DE}=\frac{AC}{DF}$$

and ∠A = ∠D

In ∆DEF, the line PQ is parallel to EF.

So, ∆ ABC ≅ ∆ DPQ.

Hence, we can say that ∠A = ∠D, ∠B=∠P and ∠C= ∠Q

which means that the triangle ABC is similar to the triangle DEF.

Its proved that ∆ ABC ~ ∆ DEF.

SSS Similarity (Side-Side-Side Similarity)

According to this criteria "If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar."

Proof:

To prove this theorem, takes two triangles ABC and DEF

such that $$\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}$$

We have to prove that ∠A = ∠D, ∠B=∠E and ∠C= ∠F and ∆ ABC ~ ∆ DEF.

Now, we draw DP = AB and DQ = AC and Join PQ

as per given $$\frac{AB}{DE}=\frac{CA}{FD}$$

and by construction DP = AB and DQ = AC

$$\frac{DP}{DE}=\frac{DQ}{DF}$$

$$\frac{DE}{DP}=\frac{DF}{DQ}$$

subtract 1 from both sides

$$\frac{DE}{DP}-1=\frac{DF}{DQ}-1$$

$$\frac{DE-DP}{DP}=\frac{DF-DQ}{DQ}$$

$$\frac{PE}{DP}=\frac{QF}{DQ}$$

$$\frac{DP}{PE}=\frac{DQ}{QF}$$

We know that "If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side." (Theorem 6.2)

Therefore, PQ || EF

So, ∠P = ∠E ( because of corresponding angles) ............ (i)

and ∠Q = ∠F ( because of corresponding angles) .......... (ii)

⇒ ∆ ABC ~ ∆ DEF  (by AA similarity)

We know that corresponding sides of similar triangles proportional, therefore

$$\frac{DP}{DE}=\frac{DQ}{DF}=\frac{PQ}{EF}............... (iii)$$

also given, $$\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}$$

and by construction AB = DP and AC = DQ

thus $$\frac{DP}{DE}=\frac{BC}{EF}=\frac{DQ}{FD}............ (iv)$$

from equation (ii) and (iii)

thus $$\frac{BC}{EF}=\frac{PQ}{EF}$$

therefore, BC = PQ ................. (v)

in triangle ABC and DPQ

AB = DP ( by construction)

AC = DQ (by construction)

BC = PQ ( by equation (v))

⇒∆ ABC ≅ ∆ DPQ (by SSS congruency)

therefore

∠B = ∠P (by CPCT)

∠C = ∠Q (by CPCT)

∠A = ∠D (by CPCT)

But from equation (i) & (ii)

∠P = ∠E

∠Q = ∠F

therefore

∠B = ∠P = ∠E ................ (vi)

and ∠C = ∠Q = ∠F ........... (vii)

Therefore, in triangle ABC and DEF

by equation (vi) and (vii)

∠B = ∠E and ∠C = ∠F

Therefore  ∆ ABC ~ ∆ DEF (by AA similarity criteria)

Hence proved.