Question Number 152647 by mr W last updated on 30/Aug/21
Commented by mr W last updated on 30/Aug/21
$${find}\:\frac{{FG}}{{GB}}=? \\ $$
Answered by mr W last updated on 30/Aug/21
Commented by mr W last updated on 30/Aug/21
$${let}\:{AB}={a},\:{BC}={b} \\ $$$${HA}={HC}={HE}={R} \\ $$$${GE}={GD}={GB}={r} \\ $$$$\frac{{AB}}{{BF}}=\frac{{BF}}{{BC}} \\ $$$$\Rightarrow{BF}^{\mathrm{2}} ={AB}×{BC}={ab}\:\:\:…\left({i}\right) \\ $$$${R}=\frac{{AC}}{\mathrm{2}}=\frac{{a}+{b}}{\mathrm{2}} \\ $$$${HB}={a}−{R}=\frac{{a}−{b}}{\mathrm{2}} \\ $$$${HG}^{\mathrm{2}} ={HE}^{\mathrm{2}} −{GE}^{\mathrm{2}} ={R}^{\mathrm{2}} −{r}^{\mathrm{2}} =\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$${HG}^{\mathrm{2}} ={HB}^{\mathrm{2}} +{BG}^{\mathrm{2}} =\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} =\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{r}^{\mathrm{2}} ={ab}\:\:\:…\left({ii}\right) \\ $$$${BF}^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} \\ $$$$\Rightarrow{BF}=\sqrt{\mathrm{2}}{r}=\sqrt{\mathrm{2}}{GB} \\ $$$$\Rightarrow{FG}+{GB}=\sqrt{\mathrm{2}}{GB} \\ $$$$\Rightarrow\frac{{FG}}{{GB}}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$
Commented by Tawa11 last updated on 30/Aug/21
$$\mathrm{Weldone}\:\mathrm{sir}.\:\mathrm{Thanks}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{more}. \\ $$
Commented by Tawa11 last updated on 30/Aug/21
$$\mathrm{Am}\:\mathrm{learning}\:\mathrm{from}\:\mathrm{all}\:\mathrm{your}\:\mathrm{geometry}\:\mathrm{solution}. \\ $$$$\mathrm{I}\:\mathrm{just}\:\mathrm{start}\:\mathrm{from}\:\mathrm{beginning}. \\ $$