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Question-221369

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Question Number 221369 by mr W last updated on 01/Jun/25
Commented by mr W last updated on 01/Jun/25
no
$${no} \\ $$
Commented by fantastic last updated on 01/Jun/25
∠FDC=90^0 ??
$$\angle{FDC}=\mathrm{90}^{\mathrm{0}} ?? \\ $$
Answered by vnm last updated on 07/Jun/25
let EF=1, q=EB+BF=ED+DF ∡FED=α+ϕ, ∡EFD=α−ϕ FD=((sin(α+ϕ))/(sin2α)), ED=((sin(α−ϕ))/(sin2α)) q=((sin(α+ϕ)+sin(α−ϕ))/(sin2α))=((2sinαcosϕ)/(2sinαcosα))=((cosϕ)/(cosα)) cosα=((cosϕ)/q), α=arccos((cosϕ)/q) ∡FED=arccos((cosϕ)/q)+ϕ ∡EFD=arccos((cosϕ)/q)−ϕ similarly: ∡EFB=arccos((cosθ)/q)+θ ∡FEB=arccos((cosθ)/q)−θ ϕ,θ − some parameters, they can be of any sign. BC−AB=(FC−FB)−(q−(FB+AE))= FC−FB−q+FB+AE=FC+AE−q similarly: CD−AD=EC+AF−q to prove that BC−AB=CD−AD, it is sufficient to prove that FC+AE=EC+AF. let f(x)=arccos((cosx)/q)+x ∡FED=f(ϕ), ∡EFD=f(−ϕ) ∡EFB=f(θ), ∡FEB=f(−θ) EC=((sin∡EFB)/(sin(∡EFB+∡FED)))=((sinf(θ))/(sin(f(θ)+f(ϕ)))) AF=((sin∡FEB)/(sin(∡FEB+∡EFD)))=((sinf(−θ))/(sin(f(−θ)+f(−ϕ)))) FC=((sin∡FED)/(sin(∡FED+∡EFB)))=((sinf(ϕ))/(sin(f(ϕ)+f(θ)))) AE=((sin∡EFD)/(sin(∡EFD+∡FEB)))=((sinf(−ϕ))/(sin(f(−ϕ)+f(−θ)))) we have to prove that ((sinf(ϕ))/(sin(f(ϕ)+f(θ))))+((sinf(−ϕ))/(sin(f(−ϕ)+f(−θ))))= ((sinf(θ))/(sin(f(θ)+f(ϕ))))+((sinf(−θ))/(sin(f(−θ)+f(−ϕ)))), −(π/2)<ϕ,θ<(π/2). Let F(x,y)= sinf(x)sin(f(−x)+f(−y))+ sinf(−x)sin(f(x)+f(y)). We need to simplify it and check that it is a symmetric function of x,_ y. F(x,y)=((2cosxcosy)/q^3 )(q^2 cos(x+y)−cosxcosy+(√((q^2 −cos^2 x)(q^2 −cos^2 y)))) BC−AB=CD−AD= ((F(ϕ,θ))/(sin(f(ϕ)+f(θ))sin(f(−ϕ)+f(−θ))))−q
$$ \\ $$$$\mathrm{let}\:{EF}=\mathrm{1},\:{q}={EB}+{BF}={ED}+{DF} \\ $$$$\measuredangle{FED}=\alpha+\varphi,\:\measuredangle{EFD}=\alpha−\varphi \\ $$$${FD}=\frac{\mathrm{sin}\left(\alpha+\varphi\right)}{\mathrm{sin2}\alpha},\:\:{ED}=\frac{\mathrm{sin}\left(\alpha−\varphi\right)}{\mathrm{sin2}\alpha} \\ $$$${q}=\frac{\mathrm{sin}\left(\alpha+\varphi\right)+\mathrm{sin}\left(\alpha−\varphi\right)}{\mathrm{sin2}\alpha}=\frac{\mathrm{2sin}\alpha\mathrm{cos}\varphi}{\mathrm{2sin}\alpha\mathrm{cos}\alpha}=\frac{\mathrm{cos}\varphi}{\mathrm{cos}\alpha} \\ $$$$\mathrm{cos}\alpha=\frac{\mathrm{cos}\varphi}{{q}},\:\:\alpha=\mathrm{arccos}\frac{\mathrm{cos}\varphi}{{q}} \\ $$$$\measuredangle{FED}=\mathrm{arccos}\frac{\mathrm{cos}\varphi}{{q}}+\varphi \\ $$$$\measuredangle{EFD}=\mathrm{arccos}\frac{\mathrm{cos}\varphi}{{q}}−\varphi \\ $$$$\mathrm{similarly}: \\ $$$$\measuredangle{EFB}=\mathrm{arccos}\frac{\mathrm{cos}\theta}{{q}}+\theta \\ $$$$\measuredangle{FEB}=\mathrm{arccos}\frac{\mathrm{cos}\theta}{{q}}−\theta\: \\ $$$$\varphi,\theta\:−\:\mathrm{some}\:\mathrm{parameters},\:\mathrm{they}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{of}\:\mathrm{any}\:\mathrm{sign}. \\ $$$${BC}−{AB}=\left({FC}−{FB}\right)−\left({q}−\left({FB}+{AE}\right)\right)= \\ $$$${FC}−{FB}−{q}+{FB}+{AE}={FC}+{AE}−{q} \\ $$$$\mathrm{similarly}: \\ $$$${CD}−{AD}={EC}+{AF}−{q} \\ $$$$\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\:{BC}−{AB}={CD}−{AD},\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{sufficient}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\:{FC}+{AE}={EC}+{AF}. \\ $$$$\mathrm{let}\:{f}\left({x}\right)=\mathrm{arccos}\frac{\mathrm{cos}{x}}{{q}}+{x} \\ $$$$\measuredangle{FED}={f}\left(\varphi\right),\:\:\measuredangle{EFD}={f}\left(−\varphi\right) \\ $$$$\measuredangle{EFB}={f}\left(\theta\right),\:\:\measuredangle{FEB}={f}\left(−\theta\right) \\ $$$${EC}=\frac{\mathrm{sin}\measuredangle{EFB}}{\mathrm{sin}\left(\measuredangle{EFB}+\measuredangle{FED}\right)}=\frac{\mathrm{sin}{f}\left(\theta\right)}{\mathrm{sin}\left({f}\left(\theta\right)+{f}\left(\varphi\right)\right)} \\ $$$${AF}=\frac{\mathrm{sin}\measuredangle{FEB}}{\mathrm{sin}\left(\measuredangle{FEB}+\measuredangle{EFD}\right)}=\frac{\mathrm{sin}{f}\left(−\theta\right)}{\mathrm{sin}\left({f}\left(−\theta\right)+{f}\left(−\varphi\right)\right)} \\ $$$${FC}=\frac{\mathrm{sin}\measuredangle{FED}}{\mathrm{sin}\left(\measuredangle{FED}+\measuredangle{EFB}\right)}=\frac{\mathrm{sin}{f}\left(\varphi\right)}{\mathrm{sin}\left({f}\left(\varphi\right)+{f}\left(\theta\right)\right)} \\ $$$${AE}=\frac{\mathrm{sin}\measuredangle{EFD}}{\mathrm{sin}\left(\measuredangle{EFD}+\measuredangle{FEB}\right)}=\frac{\mathrm{sin}{f}\left(−\varphi\right)}{\mathrm{sin}\left({f}\left(−\varphi\right)+{f}\left(−\theta\right)\right)} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{sin}{f}\left(\varphi\right)}{\mathrm{sin}\left({f}\left(\varphi\right)+{f}\left(\theta\right)\right)}+\frac{\mathrm{sin}{f}\left(−\varphi\right)}{\mathrm{sin}\left({f}\left(−\varphi\right)+{f}\left(−\theta\right)\right)}= \\ $$$$\frac{\mathrm{sin}{f}\left(\theta\right)}{\mathrm{sin}\left({f}\left(\theta\right)+{f}\left(\varphi\right)\right)}+\frac{\mathrm{sin}{f}\left(−\theta\right)}{\mathrm{sin}\left({f}\left(−\theta\right)+{f}\left(−\varphi\right)\right)}, \\ $$$$−\frac{\pi}{\mathrm{2}}<\varphi,\theta<\frac{\pi}{\mathrm{2}}. \\ $$$$\mathrm{Let}\:{F}\left({x},{y}\right)= \\ $$$$\mathrm{sin}{f}\left({x}\right)\mathrm{sin}\left({f}\left(−{x}\right)+{f}\left(−{y}\right)\right)+ \\ $$$$\mathrm{sin}{f}\left(−{x}\right)\mathrm{sin}\left({f}\left({x}\right)+{f}\left({y}\right)\right). \\ $$$$\mathrm{We}\:\mathrm{need}\:\mathrm{to}\:\mathrm{simplify}\:\mathrm{it}\:\mathrm{and}\:\mathrm{check}\:\mathrm{that} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{symmetric}\:\mathrm{function}\:\mathrm{of}\:{x},_{} {y}. \\ $$$${F}\left({x},{y}\right)=\frac{\mathrm{2cos}{x}\mathrm{cos}{y}}{{q}^{\mathrm{3}} }\left({q}^{\mathrm{2}} \mathrm{cos}\left({x}+{y}\right)−\mathrm{cos}{x}\mathrm{cos}{y}+\sqrt{\left({q}^{\mathrm{2}} −\mathrm{cos}^{\mathrm{2}} {x}\right)\left({q}^{\mathrm{2}} −\mathrm{cos}^{\mathrm{2}} {y}\right)}\right) \\ $$$${BC}−{AB}={CD}−{AD}= \\ $$$$\frac{{F}\left(\varphi,\theta\right)}{\mathrm{sin}\left({f}\left(\varphi\right)+{f}\left(\theta\right)\right)\mathrm{sin}\left({f}\left(−\varphi\right)+{f}\left(−\theta\right)\right)}−{q} \\ $$

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