Question Number 82878 by mr W last updated on 25/Feb/20
Commented by behi83417@gmail.com last updated on 25/Feb/20
$$\mathrm{Beautiful}\:\mathrm{quistion}.\mathrm{waiting}\:\mathrm{for}\:\mathrm{answer}... \\ $$
Commented by ajfour last updated on 25/Feb/20
$$\mathrm{so}\:\mathrm{please}\:\mathrm{answer}\:\mathrm{Behi}\:\mathrm{Sir}.. \\ $$
Commented by behi83417@gmail.com last updated on 25/Feb/20
$$\mathrm{dear}\:\mathrm{Ajfour}\:\mathrm{sir}!\:\mathrm{i}\:\mathrm{am}\:\mathrm{waiting}\:\mathrm{for}\:\mathrm{answer} \\ $$$$\mathrm{of}\:\mathrm{my}\:\mathrm{master}:\:\mathrm{mrW}.\left(\mathrm{also}\:\mathrm{your}\:\mathrm{answer}\right. \\ $$$$\left.\mathrm{too},\mathrm{most}\:\mathrm{welcome}.\right) \\ $$
Answered by mr W last updated on 26/Feb/20
Commented by mr W last updated on 26/Feb/20
![ΔAFG: semi perimeter=u incircle radius =n excircle radius =a u=((p+q+s)/2) n=(Δ_(AFG) /u) ⇒(1/n)=(u/Δ_(AFG) ) a=(Δ_(AFG) /(u−s)) ⇒(1/a)=((u−s)/Δ_(AFG) ) ΔAGE: semi perimeter=v incircle radius =b excircle radius =m v=((q+r+t)/2) similarly ⇒(1/b)=(v/Δ_(AGE) ) ⇒(1/m)=((v−t)/Δ_(AGE) ) (1/a)+(1/b)=((u−s)/Δ_(AFG) )+(v/Δ_(AGE) )=(1/Δ_(AFG) )(u−s+(Δ_(AFG) /Δ_(AGE) )v) ⇒(1/a)+(1/b)=(1/Δ_(AFG) )(u−s+(s/t)v) (1/n)+(1/m)=(u/Δ_(AFG) )+((v−t)/Δ_(AGE) )=(1/Δ_(AFG) )[u+(Δ_(AFG) /Δ_(AGE) )(v−t)] (1/n)+(1/m)=(1/Δ_(AFG) )[u+(s/t)(v−t)] ⇒(1/n)+(1/m)=(1/Δ_(AFG) )(u+(s/t)v−s) ⇒(1/a)+(1/b)=(1/m)+(1/n)](Q83013.png)
$$\Delta{AFG}: \\ $$$${semi}\:{perimeter}={u} \\ $$$${incircle}\:{radius}\:={n} \\ $$$${excircle}\:{radius}\:={a} \\ $$$${u}=\frac{{p}+{q}+{s}}{\mathrm{2}} \\ $$$${n}=\frac{\Delta_{{AFG}} }{{u}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{n}}=\frac{{u}}{\Delta_{{AFG}} } \\ $$$${a}=\frac{\Delta_{{AFG}} }{{u}−{s}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}=\frac{{u}−{s}}{\Delta_{{AFG}} } \\ $$$$ \\ $$$$\Delta{AGE}: \\ $$$${semi}\:{perimeter}={v} \\ $$$${incircle}\:{radius}\:={b} \\ $$$${excircle}\:{radius}\:={m} \\ $$$${v}=\frac{{q}+{r}+{t}}{\mathrm{2}} \\ $$$${similarly} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{b}}=\frac{{v}}{\Delta_{{AGE}} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{{m}}=\frac{{v}−{t}}{\Delta_{{AGE}} } \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{{u}−{s}}{\Delta_{{AFG}} }+\frac{{v}}{\Delta_{{AGE}} }=\frac{\mathrm{1}}{\Delta_{{AFG}} }\left({u}−{s}+\frac{\Delta_{{AFG}} }{\Delta_{{AGE}} }{v}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{\Delta_{{AFG}} }\left({u}−{s}+\frac{{s}}{{t}}{v}\right) \\ $$$$\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{m}}=\frac{{u}}{\Delta_{{AFG}} }+\frac{{v}−{t}}{\Delta_{{AGE}} }=\frac{\mathrm{1}}{\Delta_{{AFG}} }\left[{u}+\frac{\Delta_{{AFG}} }{\Delta_{{AGE}} }\left({v}−{t}\right)\right] \\ $$$$\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{m}}=\frac{\mathrm{1}}{\Delta_{{AFG}} }\left[{u}+\frac{{s}}{{t}}\left({v}−{t}\right)\right] \\ $$$$\Rightarrow\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{m}}=\frac{\mathrm{1}}{\Delta_{{AFG}} }\left({u}+\frac{{s}}{{t}}{v}−{s}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{{m}}+\frac{\mathrm{1}}{{n}} \\ $$
Commented by ajfour last updated on 28/Feb/20
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{what}\:\mathrm{a}\:\mathrm{presentation}! \\ $$