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Question

$\mathrm{In}\mathrm{the}\mathrm{adjoining}\mathrm{figure},\square \mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{square}.△\mathrm{BCE}\mathrm{on}\mathrm{side}\mathrm{BC}\phantom{\rule{0ex}{0ex}}\mathrm{and}△\mathrm{ACF}\mathrm{on}\mathrm{the}\mathrm{diagonal}\mathrm{AC}\mathrm{are}\mathrm{similar}\mathrm{to}\mathrm{each}\mathrm{other}.\text{S}\mathrm{how}\mathrm{that}\phantom{\rule{0ex}{0ex}}\mathrm{A}\left(△\mathrm{BCE}\right)=\frac{1}{2}\mathrm{A}\left(△\mathrm{ACF}\right).$

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Solution

$\mathrm{Given}:\square \mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{square}.\phantom{\rule{0ex}{0ex}}\therefore \mathrm{AC}=\sqrt{2}\mathrm{BC}...\left(1\right)\left[\because \mathrm{Diagonal}\mathrm{of}\mathrm{a}\mathrm{square}=\sqrt{2}×\mathrm{side}\mathrm{of}\text{the}\mathrm{square}\right]\phantom{\rule{0ex}{0ex}}△\mathrm{BCE}~△\mathrm{ACF}\left[\mathrm{Given}\right]\phantom{\rule{0ex}{0ex}}\therefore \frac{\mathrm{A}\left(△\mathrm{BCE}\right)}{\mathrm{A}\left(△\mathrm{ACF}\right)}=\frac{{\left(\mathrm{BC}\right)}^{2}}{{\left(\mathrm{AC}\right)}^{2}}...\left(2\right)\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{A}\left(△\mathrm{BCE}\right)}{\mathrm{A}\left(△\mathrm{ACF}\right)}=\frac{{\mathrm{BC}}^{2}}{{\left(\sqrt{2}\mathrm{BC}\right)}^{2}}\left[\mathrm{From}\left(1\right)&\left(2\right)\right]\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{A}\left(△\mathrm{BCE}\right)}{\mathrm{A}\left(△\mathrm{ACF}\right)}=\frac{{\mathrm{BC}}^{2}}{2{\mathrm{BC}}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{A}\left(△\mathrm{BCE}\right)}{\mathrm{A}\left(△\mathrm{ACF}\right)}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{A}\left(△\mathrm{BCE}\right)=\frac{1}{2}\mathrm{A}\left(△\mathrm{ACF}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$

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