# Criteria For Similarity of Two Triangles

Table of Contents

## Criteria For Similarity of Two Triangles:

We know that two triangles are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio. That is, if in two triangles ABC and PQR, (i) âˆ A = âˆ P, âˆ B = âˆ Q, âˆ C = âˆ R, and (ii) AB/PQ = BC/QR = CA/RP, then the two triangles are similar. Here, A corresponds to P, B corresponds to Q and C corresponds to R. Symbolically, we write the above relation as ‘â–³ABC ~ â–³PQR’ and read it as ‘â–³ABC is similar to â–³PQR’. Just as in the case of congruence, similar triangles also should be written in the correct correspondence of their vertices. For example, in the above case, it will not be correct to write â–³ABC ~ â–³QRP or â–³ABC ~ â–³RPQ etc. Now a natural question arises: For establishing the similarity of two triangles, should we always look for all the equality relations of corresponding angles and proportionality relations of the corresponding sides of the two triangles? Recall that we found out certain criteria for establishing the congruency of two triangles involving only three pairs of elements of the triangles. Here also, we shall make an attempt to arrive at certain criteria involving less number of elements of triangles for establishing the similarity of two triangles. For this, let us perform some activities.

Let us draw two line segments BC and EF of different lengths (say 2 cm and 5 cm respectively). At B and E, construct angles XBC and PEF of any measure, say 60Â° and at C and F, construct angles YCB and QFE of measure, say 40Â°. Let arms BX and CY intersect at A and arms EP and FQ intersect at D. Thus, we obtain two triangles ABC and DEF. What can you say about âˆ A and âˆ D. Clearly, they are 80Â° each. Thus, in triangles ABC and DEF, we have âˆ A = âˆ D, âˆ B = âˆ E and âˆ C = âˆ F, i.e., corresponding angles are equal. Now, let us measure AB, AC, DE, DF and find the ratios AB/DE, AC/DF and BC/EF. What do you observe?

AB/DE and AC/DF are both equal to 0.4 (or very near to 0.4), i.e., 2/5. Thus, AB/DE = AC/DF = BC/EF (= 2/5), i.e., corresponding sides are in the same ratio and hence the triangles ABC and DEF are similar.