A) Have the same area
B) Are similar
C) Are congruent
D) None of these
Correct Answer: B
Solution :
Let the complex number \[a,b,c\]and \[u,v,w\] represent the vertices \[A,B,C\]and \[D,E,F\] of the two triangle \[ABC\] and \[DEF\] respectively. Put \[b-a={{r}_{1}}{{e}^{i{{\theta }_{1}}}}\] \[c-a={{r}_{2}}{{e}^{i{{\theta }_{2}}}}\] \[v-u={{\rho }_{1}}{{e}^{i{{\varphi }_{1}}}},w-u={{\rho }_{2}}{{e}^{i{{\varphi }_{2}}}}\]and \[r=\lambda {{e}^{i\alpha }}\] Substituting these values in the given relations \[c-a=r(b-a)\]and \[w-u=(v-u)r,\] we have \[{{r}_{2}}{{e}^{i{{\theta }_{2}}}}=\lambda {{e}^{i\alpha }}{{r}_{1}}{{e}^{i{{\theta }_{1}}}}=\lambda {{r}_{1}}{{e}^{i(\alpha +{{\theta }_{1}})}}\] .......(i) and \[{{\rho }_{2}}{{e}^{i{{\varphi }_{2}}}}={{\rho }_{1}}{{e}^{i{{\varphi }_{1}}}}\lambda {{e}^{i\alpha }}=(\lambda {{\rho }_{1}}){{e}^{i({{\varphi }_{1}}+\alpha )}}\] .......(ii) Equating moduli and arguments of the complex numbers on both sides (i), we get \[{{r}_{2}}=\lambda {{r}_{1}},{{\theta }_{2}}=\alpha +{{\theta }_{1}}\] i.e., \[AC=\lambda AB\]and \[\angle CAB={{\theta }_{2}}-{{\theta }_{1}}=\alpha \] Similarly from (ii), we shall get \[DF=\lambda DE\] and \[\angle FDE={{\varphi }_{2}}-{{\varphi }_{1}}=\alpha \] Thus we get \[\frac{AC}{DF}=\frac{AB}{DE}\]and \[\angle CAB=\angle FDE\] Hence the triangle \[ABC\] and \[DEF\] are similar.
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