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Question Number 152647 by mr W last updated on 30/Aug/21

Commented by mr W last updated on 30/Aug/21

find ((FG)/(GB))=?

$${find}\:\frac{{FG}}{{GB}}=? \\ $$

Answered by mr W last updated on 30/Aug/21

Commented by mr W last updated on 30/Aug/21

let AB=a, BC=b HA=HC=HE=R GE=GD=GB=r ((AB)/(BF))=((BF)/(BC)) ⇒BF^2 =AB×BC=ab ...(i) R=((AC)/2)=((a+b)/2) HB=a−R=((a−b)/2) HG^2 =HE^2 −GE^2 =R^2 −r^2 =(((a+b)/2))^2 −r^2 HG^2 =HB^2 +BG^2 =(((a−b)/2))^2 +r^2 ⇒(((a+b)/2))^2 −r^2 =(((a−b)/2))^2 +r^2 ⇒(((a+b)/2))^2 −(((a−b)/2))^2 =2r^2 ⇒2r^2 =ab ...(ii) BF^2 =2r^2 ⇒BF=(√2)r=(√2)GB ⇒FG+GB=(√2)GB ⇒((FG)/(GB))=(√2)−1

$${let}\:{AB}={a},\:{BC}={b} \\ $$$${HA}={HC}={HE}={R} \\ $$$${GE}={GD}={GB}={r} \\ $$$$\frac{{AB}}{{BF}}=\frac{{BF}}{{BC}} \\ $$$$\Rightarrow{BF}^{\mathrm{2}} ={AB}×{BC}={ab}\:\:\:...\left({i}\right) \\ $$$${R}=\frac{{AC}}{\mathrm{2}}=\frac{{a}+{b}}{\mathrm{2}} \\ $$$${HB}={a}−{R}=\frac{{a}−{b}}{\mathrm{2}} \\ $$$${HG}^{\mathrm{2}} ={HE}^{\mathrm{2}} −{GE}^{\mathrm{2}} ={R}^{\mathrm{2}} −{r}^{\mathrm{2}} =\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$${HG}^{\mathrm{2}} ={HB}^{\mathrm{2}} +{BG}^{\mathrm{2}} =\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} =\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{r}^{\mathrm{2}} ={ab}\:\:\:...\left({ii}\right) \\ $$$${BF}^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} \\ $$$$\Rightarrow{BF}=\sqrt{\mathrm{2}}{r}=\sqrt{\mathrm{2}}{GB} \\ $$$$\Rightarrow{FG}+{GB}=\sqrt{\mathrm{2}}{GB} \\ $$$$\Rightarrow\frac{{FG}}{{GB}}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$

Commented by Tawa11 last updated on 30/Aug/21

Weldone sir. Thanks. God bless you more.

$$\mathrm{Weldone}\:\mathrm{sir}.\:\mathrm{Thanks}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{more}. \\ $$

Commented by Tawa11 last updated on 30/Aug/21

Am learning from all your geometry solution. I just start from beginning.

$$\mathrm{Am}\:\mathrm{learning}\:\mathrm{from}\:\mathrm{all}\:\mathrm{your}\:\mathrm{geometry}\:\mathrm{solution}. \\ $$$$\mathrm{I}\:\mathrm{just}\:\mathrm{start}\:\mathrm{from}\:\mathrm{beginning}. \\ $$

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