In the previous section, we stated that two triangles are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).

That is, in Δ ABC and Δ DEF, if

(i) ∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F and

(ii) then the two triangles are similar (see Fig. 10)

Here, you can see that A corresponds to D, B corresponds to E and C corresponds to F. Symbolically, we write the similarity of these two triangles as ‘Δ ABC ~ Δ DEF’ and read it as ‘triangle ABC is similar to triangle DEF’. The symbol ‘~’ stands for ‘is similar to’. Recall that you have used the symbol ‘≅’ for ‘is congruent to’.

It must be noted that as done in the case of congruency of two triangles, the similarity of two triangles should also be expressed symbolically, using correct correspondence of their vertices. For example, for the triangles ABC and DEF of Fig. 10, we cannot write Δ ABC ~ Δ EDF or Δ ABC ~ Δ FED. However, we can write Δ BAC ~ Δ EDF.

Now a natural question arises : For checking the similarity of two triangles, say ABC and DEF, should we always look for all the equality relations of their corresponding angles (∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F) and all the equality relations of the ratios of their corresponding sides Let us examine. You may recall that we have obtained some criteria for congruency of two triangles involving only three pairs of corresponding parts (or elements) of the two triangles. Here also, let us make an attempt to arrive at certain criteria for similarity of two triangles involving relationship between less number of pairs of corresponding parts of the two triangles, instead of all the six pairs of corresponding parts.

**Theorem 3 :** *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.*

This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles.

This theorem can be proved by taking two triangles ABC and DEF such that ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F (see Fig. 11).

Cut DP = AB and DQ = AC and join PQ.

**Remark :** 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.

**Theorem 4 : ** *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.*

This criterion is referred to as the SSS (Side–Side–Side) *similarity criterion for two triangles*. This theorem can be proved by taking two triangles ABC and DEF such that
(see Fig. 12):

Cut DP = AB and DQ = AC and join PQ.

**Remark :** You may recall that either of the two conditions namely, (i) corresponding angles are equal and (ii) corresponding sides are in the same ratio is not sufficient for two polygons to be similar. However, on the basis of Theorems 3 and 4, you can now say that in case of similarity of the two triangles, it is not necessary to check both the conditions as one condition implies the other.

**Theorem 5 :** *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*.

This criterion is referred to as the SAS (Side–Angle–Side) similarity criterion for two triangles.

As before, this theorem can be proved by taking two triangles ABC and DEF such that and ∠ A = ∠ D (see Fig. 13). Cut DP = AB, DQ = AC and join PQ.

**Example 4 :** In Fig. 14, if PQ || RS, prove that Δ POQ ~ Δ SOR.

**Solution :**

**Example 5 :** Observe Fig. 15 and then find ∠ P.

**Solution :** In Δ ABC and Δ PQR,

**Example 6 :** In Fig. 16,

OA . OB = OC . OD.

Show that ∠ A = ∠ C and ∠ B = ∠ D.

**Solution : **

OA . OB = OC . OD(Given)

Also, we have∠ AOD = ∠ COB(Vertically opposite angles)(2)

Therefore, from (1) and (2),Δ AOD ~ Δ COB(SAS similarity criterion)

So,∠ A = ∠ C and ∠ D = ∠ B(Corresponding angles of similar triangles)

**Example 7 :** A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

**Solution :** Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post (see Fig. 17). From the figure, you can see that DE is the shadow of the girl. Let DE be x metres.

Now, BD = 1.2 m × 4 = 4.8 m.

Note that in Δ ABE and Δ CDE,

So, the shadow of the girl after walking for 4 seconds is 1.6 m long.

**Example 8 :** In Fig. 18, CM and RN are respectively the medians of Δ ABC and Δ PQR. If Δ ABC ~ Δ PQR, prove that :

**Solution : **

[**Note :** You can also prove part (iii) by following the same method as used for proving part (i).]