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Question Number 172284 by mnjuly1970 last updated on 25/Jun/22

Answered by mr W last updated on 25/Jun/22

$$sayAC=e,BD=f$$$$\cos B=\frac{a^2+b^2-e^2}{2ab}$$$$\cos D=\frac{c^2+d^2-e^2}{2cd}$$$$\mathrm{\angle }B=\pi -\mathrm{\angle }D$$$$\cos B=-\cos D$$$$\frac{a^2+b^2-e^2}{2ab}=-\frac{c^2+d^2-e^2}{2cd}$$$$\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}=(\frac{1}{ab}+\frac{1}{cd})e^2$$$$\Rightarrow (a^2+b^2)cd+(c^2+d^2)ab=(ab+cd)e^2$$$$\Rightarrow (a^2+b^2+c^2+d^2)abcd+a^2{b}^2{c}^2+b^2{c}^2{d}^2+c^2{d}^2{a}^2+d^2{a}^2{b}^2={(ab+cd)}^2{e}^2$$$$similarly$$$$\Rightarrow (a^2+b^2+c^2+d^2)abcd+a^2{b}^2{c}^2+b^2{c}^2{d}^2+c^2{d}^2{a}^2+d^2{a}^2{b}^2={(bc+ad)}^2{f}^2$$$$$$$$\Rightarrow {(ab+cd)}^2{e}^2={(bc+ad)}^2{f}^2$$$$\Rightarrow (ab+cd)e=(bc+ad)f$$$$\Rightarrow \frac{e}{f}=\frac{bc+ad}{ab+cd}✓$$

Commented by mnjuly1970 last updated on 25/Jun/22

$$thanksalotsirW$$

Commented by Tawa11 last updated on 25/Jun/22

$$\mathrm{Great}\mathrm{sir}$$

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