Question Number 27605 by ajfour last updated on 10/Jan/18
Answered by mrW2 last updated on 13/Jan/18
![((DB)/(DC))=((AB)/(AC))=(c/b) ⇒DB=((ac)/(b+c)) ⇒DC=((ab)/(b+c)) AD=((2(√(s(s−a)bc)))/(b+c)) ((AE)/(AB))=((AD)/(DB))=((2(√(s(s−a)bc)))/(b+c))×((b+c)/(ac))=((2(√(s(s−a)bc)))/(ac)) ⇒AE=((2c(√(s(s−a)bc)))/(2(√(s(a−a)bc))+ac)) similarly ⇒AF=((2b(√(s(s−a)bc)))/(2(√(s(a−a)bc))+ab)) a^2 =b^2 +c^2 −2bc×cos A ⇒cos A=((b^2 +c^2 −a^2 )/(2bc)) EF^2 =AE^2 +AF^2 −2×AE×AF×cos A =((4c^2 s(s−a)bc)/(4s(s−a)bc+a^2 c^2 +2ac(√(s(s−a)bc))))+((4b^2 s(s−a)bc)/(4s(s−a)bc+a^2 b^2 +2ab(√(s(s−a)bc))))−2×((4bcs(s−a)bc)/(4s(s−a)bc+a^2 bc+2a(b+c)(√(s(s−a)bc))))×((b^2 +c^2 −a^2 )/(2bc)) =4s(s−a)bc[(c^2 /(4s(s−a)bc+a^2 c^2 +2ac(√(s(s−a)bc))))+(b^2 /(4s(s−a)bc+a^2 b^2 +2ab(√(s(s−a)bc))))−((b^2 +c^2 −a^2 )/(4s(s−a)bc+a^2 bc+2a(b+c)(√(s(s−a)bc))))] =4s(s−a)bc[(c^2 /(4s(s−a)bc+a^2 c^2 +2ac(√(s(s−a)bc))))+(b^2 /(4s(s−a)bc+a^2 b^2 +2ab(√(s(s−a)bc))))−((b^2 +c^2 −a^2 )/(4s(s−a)bc+a^2 bc+2a(b+c)(√(s(s−a)bc))))] ⇒EF=2(√(s(s−a)bc[(c^2 /(4s(s−a)bc+a^2 c^2 +2ac(√(s(s−a)bc))))+(b^2 /(4s(s−a)bc+a^2 b^2 +2ab(√(s(s−a)bc))))−((b^2 +c^2 −a^2 )/(4s(s−a)bc+a^2 bc+2a(b+c)(√(s(s−a)bc))))]))](Q27709.png)
$$\frac{{DB}}{{DC}}=\frac{{AB}}{{AC}}=\frac{{c}}{{b}} \\ $$$$\Rightarrow{DB}=\frac{{ac}}{{b}+{c}} \\ $$$$\Rightarrow{DC}=\frac{{ab}}{{b}+{c}} \\ $$$${AD}=\frac{\mathrm{2}\sqrt{{s}\left({s}−{a}\right){bc}}}{{b}+{c}} \\ $$$$\frac{{AE}}{{AB}}=\frac{{AD}}{{DB}}=\frac{\mathrm{2}\sqrt{{s}\left({s}−{a}\right){bc}}}{{b}+{c}}×\frac{{b}+{c}}{{ac}}=\frac{\mathrm{2}\sqrt{{s}\left({s}−{a}\right){bc}}}{{ac}} \\ $$$$\Rightarrow{AE}=\frac{\mathrm{2}{c}\sqrt{{s}\left({s}−{a}\right){bc}}}{\mathrm{2}\sqrt{{s}\left({a}−{a}\right){bc}}+{ac}} \\ $$$$ \\ $$$${similarly} \\ $$$$\Rightarrow{AF}=\frac{\mathrm{2}{b}\sqrt{{s}\left({s}−{a}\right){bc}}}{\mathrm{2}\sqrt{{s}\left({a}−{a}\right){bc}}+{ab}} \\ $$$$ \\ $$$${a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}×\mathrm{cos}\:{A} \\ $$$$\Rightarrow\mathrm{cos}\:{A}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$${EF}^{\mathrm{2}} ={AE}^{\mathrm{2}} +{AF}^{\mathrm{2}} −\mathrm{2}×{AE}×{AF}×\mathrm{cos}\:{A} \\ $$$$=\frac{\mathrm{4}{c}^{\mathrm{2}} {s}\left({s}−{a}\right){bc}}{\mathrm{4}{s}\left({s}−{a}\right){bc}+{a}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{2}{ac}\sqrt{{s}\left({s}−{a}\right){bc}}}+\frac{\mathrm{4}{b}^{\mathrm{2}} {s}\left({s}−{a}\right){bc}}{\mathrm{4}{s}\left({s}−{a}\right){bc}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{2}{ab}\sqrt{{s}\left({s}−{a}\right){bc}}}−\mathrm{2}×\frac{\mathrm{4}{bcs}\left({s}−{a}\right){bc}}{\mathrm{4}{s}\left({s}−{a}\right){bc}+{a}^{\mathrm{2}} {bc}+\mathrm{2}{a}\left({b}+{c}\right)\sqrt{{s}\left({s}−{a}\right){bc}}}×\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$$=\mathrm{4}{s}\left({s}−{a}\right){bc}\left[\frac{{c}^{\mathrm{2}} }{\mathrm{4}{s}\left({s}−{a}\right){bc}+{a}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{2}{ac}\sqrt{{s}\left({s}−{a}\right){bc}}}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}{s}\left({s}−{a}\right){bc}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{2}{ab}\sqrt{{s}\left({s}−{a}\right){bc}}}−\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{s}\left({s}−{a}\right){bc}+{a}^{\mathrm{2}} {bc}+\mathrm{2}{a}\left({b}+{c}\right)\sqrt{{s}\left({s}−{a}\right){bc}}}\right] \\ $$$$=\mathrm{4}{s}\left({s}−{a}\right){bc}\left[\frac{{c}^{\mathrm{2}} }{\mathrm{4}{s}\left({s}−{a}\right){bc}+{a}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{2}{ac}\sqrt{{s}\left({s}−{a}\right){bc}}}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}{s}\left({s}−{a}\right){bc}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{2}{ab}\sqrt{{s}\left({s}−{a}\right){bc}}}−\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{s}\left({s}−{a}\right){bc}+{a}^{\mathrm{2}} {bc}+\mathrm{2}{a}\left({b}+{c}\right)\sqrt{{s}\left({s}−{a}\right){bc}}}\right] \\ $$$$ \\ $$$$\Rightarrow{EF}=\mathrm{2}\sqrt{{s}\left({s}−{a}\right){bc}\left[\frac{{c}^{\mathrm{2}} }{\mathrm{4}{s}\left({s}−{a}\right){bc}+{a}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{2}{ac}\sqrt{{s}\left({s}−{a}\right){bc}}}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}{s}\left({s}−{a}\right){bc}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{2}{ab}\sqrt{{s}\left({s}−{a}\right){bc}}}−\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{s}\left({s}−{a}\right){bc}+{a}^{\mathrm{2}} {bc}+\mathrm{2}{a}\left({b}+{c}\right)\sqrt{{s}\left({s}−{a}\right){bc}}}\right]} \\ $$
Commented by ajfour last updated on 13/Jan/18
$${so}\:{lengthy}\:{and}\:{you}\:{could}\:{do}\:{it}\:{sir}! \\ $$$${thank}\:{you}\:{immensely}. \\ $$