Question Number 172284 by mnjuly1970 last updated on 25/Jun/22
Answered by mr W last updated on 25/Jun/22
$${say}\:{AC}={e},\:{BD}={f} \\ $$$$\mathrm{cos}\:{B}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{e}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$\mathrm{cos}\:{D}=\frac{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} −{e}^{\mathrm{2}} }{\mathrm{2}{cd}} \\ $$$$\angle{B}=\pi−\angle{D} \\ $$$$\mathrm{cos}\:{B}=−\mathrm{cos}\:{D} \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{e}^{\mathrm{2}} }{\mathrm{2}{ab}}=−\frac{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} −{e}^{\mathrm{2}} }{\mathrm{2}{cd}} \\ $$$$\frac{{a}}{{b}}+\frac{{b}}{{a}}+\frac{{c}}{{d}}+\frac{{d}}{{c}}=\left(\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{cd}}\right){e}^{\mathrm{2}} \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){cd}+\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right){ab}=\left({ab}+{cd}\right){e}^{\mathrm{2}} \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right){abcd}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} {d}^{\mathrm{2}} +{c}^{\mathrm{2}} {d}^{\mathrm{2}} {a}^{\mathrm{2}} +{d}^{\mathrm{2}} {a}^{\mathrm{2}} {b}^{\mathrm{2}} =\left({ab}+{cd}\right)^{\mathrm{2}} {e}^{\mathrm{2}} \\ $$$${similarly} \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right){abcd}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} {d}^{\mathrm{2}} +{c}^{\mathrm{2}} {d}^{\mathrm{2}} {a}^{\mathrm{2}} +{d}^{\mathrm{2}} {a}^{\mathrm{2}} {b}^{\mathrm{2}} =\left({bc}+{ad}\right)^{\mathrm{2}} {f}^{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\left({ab}+{cd}\right)^{\mathrm{2}} {e}^{\mathrm{2}} =\left({bc}+{ad}\right)^{\mathrm{2}} {f}^{\mathrm{2}} \\ $$$$\Rightarrow\left({ab}+{cd}\right){e}=\left({bc}+{ad}\right){f} \\ $$$$\Rightarrow\frac{{e}}{{f}}=\frac{{bc}+{ad}}{{ab}+{cd}}\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 25/Jun/22
$$\:\:\:{thanks}\:{alot}\:\:{sir}\:{W} \\ $$
Commented by Tawa11 last updated on 25/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$