Question:

The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO CO BO DO   Show that ABCD is a trapezium. 6.4 Criteria for Similarity of In the previous section, we stated that two are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). That is, in  ABC and  DEF, if (i)  A =  D,  B =  E,  C =  F and (ii) AB BC CA , DE EF FD   then the two are similar (see Fig. 6.22). Fig. 6.22 Fig. 6.20 Fig. 6.21 86 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 as ‘ ABC ~  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’ in Class IX. It must be noted that as done in the case of congruency of two , the similarity of two should also be expressed symbolically, using correct correspondence of their vertices. For example, for the ABC and DEF of Fig. 6.22, we cannot write  ABC ~  EDF or  ABC ~  FED. However, we can write  BAC ~  EDF. Now a natural question arises : For checking the similarity of two , 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 AB BC CA DE EF FD        ? Let us examine. You may recall that in Class IX, you have obtained some criteria for congruency of two involving only three pairs of corresponding parts (or elements) of the two . Here also, let us make an attempt to arrive at certain criteria for similarity of two involving relationship between less number of pairs of corresponding parts of the two , instead of all the six pairs of corresponding parts. For this, let us perform the following activity: Activity 4 : Draw two line segments BC and EF of two different lengths, say 3 cm and 5 cm respectively. Then, at the points B and C respectively, construct angles PBC and QCB of some measures, say, 60° and 40°. Also, at the points E and F, construct angles REF and SFE of 60° and 40° respectively (see Fig. 6.23). Fig. 6.23 87 Let rays BP and CQ intersect each other at A and rays ER and FS intersect each other at D. In the two ABC and DEF, you can see that  B =  E,  C =  F and  A =  D. That is, corresponding angles of these two are equal. What can you say about their corresponding sides ? Note that BC 3 0.6. EF 5   What about AB DE and CA FD ? On measuring AB, DE, CA and FD, you will find that AB DE and CA FD are also equal to 0.6 (or nearly equal to 0.6, if there is some error in the measurement). Thus, AB BC CA DE EF FD    You can repeat this activity by constructing several pairs of having their corresponding angles equal. Every time, you will find that their corresponding sides are in the same ratio (or proportion). This activity leads us to the following criterion for similarity of two .