Answer:
\[\left( {{a}^{2}}+{{b}^{2}} \right){{x}^{2}}-2(ac+bd)x+\left( {{c}^{2}}+{{d}^{2}} \right)=0\] For equal roots, \[D=0\] \[{{[-2(ac+bd)]}^{2}}-4({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})=0\] \[4{{(ac+bd)}^{2}}-4({{a}^{2}}{{c}^{2}}+{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}})=0\] \[4[{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}+2abcd-{{a}^{2}}{{c}^{2}}-{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}]=0\] \[-4[{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}-2abcd]=0\] \[-4{{(ad-bc)}^{2}}=0\] \[-4\ne 0\] so, \[{{(ad-bc)}^{2}}=0\] \[ad-bc=0\] \[ad=bc\] \[\frac{a}{b}=\frac{c}{d}\] Hence Proved
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