Question Number 43265 by Cheyboy last updated on 09/Sep/18
Commented by ajfour last updated on 09/Sep/18
Commented by ajfour last updated on 09/Sep/18
![inclination of ED with horizontal = ∠AED−φ = 90°−((180°−(θ+φ))/2)−φ = ((𝛉−𝛗)/2) let AE = AD = r Let ∠APG = 𝛂 , ∠APF =𝛃 AC=2Rsin φ ⇒ AT=((AE+AC)/2) ⇒ AT = Rsin φ+(r/2) = AM+TM ⇒ TM = (r/2) = GP sin ∠GPM ⇒ (r/2)=Rsin (𝛂−𝛗) ....(i) Similarly on the other side SN = (r/2) = Rsin (𝛃−𝛉) ...(ii) (i) and (ii) ⇒ 𝛃−𝛂 = 𝛉−𝛗 ...(I ) inclination of GF with horizontal is = ∠ of GP with horizontal − ∠PGF = 90°−α−((180−(α+β))/2) = ((α+β)/2)−α = ((β−α)/2) = ((𝛉−𝛗)/2) [ See (I ) ] Hence ED // GF .](https://www.tinkutara.com/question/Q43303.png)
$${inclination}\:{of}\:{ED}\:{with}\:{horizontal} \\ $$$$\:\:\:\:=\:\angle{AED}−\phi\:=\:\mathrm{90}°−\frac{\mathrm{180}°−\left(\theta+\phi\right)}{\mathrm{2}}−\phi \\ $$$$\:\:\:=\:\frac{\boldsymbol{\theta}−\boldsymbol{\phi}}{\mathrm{2}}\: \\ $$$${let}\:{AE}\:=\:{AD}\:=\:\boldsymbol{{r}} \\ $$$${Let}\:\angle{APG}\:=\:\boldsymbol{\alpha}\:\:,\:\:\angle{APF}\:=\boldsymbol{\beta} \\ $$$${AC}=\mathrm{2}{R}\mathrm{sin}\:\phi\:\:\Rightarrow\:{AT}=\frac{{AE}+{AC}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{AT}\:=\:{R}\mathrm{sin}\:\phi+\frac{{r}}{\mathrm{2}}\:\:=\:{AM}+{TM} \\ $$$$\Rightarrow\:{TM}\:=\:\frac{\boldsymbol{{r}}}{\mathrm{2}}\:=\:{GP}\:\mathrm{sin}\:\angle{GPM} \\ $$$$\Rightarrow\:\:\:\frac{\boldsymbol{{r}}}{\mathrm{2}}=\boldsymbol{{R}}\mathrm{sin}\:\left(\boldsymbol{\alpha}−\boldsymbol{\phi}\right)\:\:\:….\left({i}\right) \\ $$$${Similarly}\:{on}\:{the}\:{other}\:{side} \\ $$$$\:\:\:\:{SN}\:=\:\frac{\boldsymbol{{r}}}{\mathrm{2}}\:=\:\boldsymbol{{R}}\mathrm{sin}\:\left(\boldsymbol{\beta}−\boldsymbol{\theta}\right)\:\:…\left({ii}\right) \\ $$$$\left({i}\right)\:{and}\:\left({ii}\right)\:\Rightarrow\: \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\beta}−\boldsymbol{\alpha}\:=\:\boldsymbol{\theta}−\boldsymbol{\phi}\:\:\:\:\:\:\:\:\:…\left({I}\:\right) \\ $$$${inclination}\:{of}\:{GF}\:{with}\:{horizontal} \\ $$$${is}\:\:=\:\angle\:{of}\:{GP}\:{with}\:{horizontal} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\angle{PGF} \\ $$$$\:\:\:\:\:=\:\mathrm{90}°−\alpha−\frac{\mathrm{180}−\left(\alpha+\beta\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\:\frac{\alpha+\beta}{\mathrm{2}}−\alpha\:\:=\:\frac{\beta−\alpha}{\mathrm{2}}\:=\:\frac{\boldsymbol{\theta}−\boldsymbol{\phi}}{\mathrm{2}}\: \\ $$$$\:\:\left[\:{See}\:\left({I}\:\right)\:\right]\:\:{Hence}\:\:\boldsymbol{{ED}}\://\:\boldsymbol{{GF}}\:. \\ $$$$ \\ $$
Commented by Cheyboy last updated on 09/Sep/18
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir},\mathrm{Really}\:\mathrm{appreciated} \\ $$