Question Number 15761 by Tinkutara last updated on 13/Jun/17
$$\mathrm{If}\:\mathrm{in}\:\mathrm{a}\:\Delta{ABC},\:\frac{\mathrm{2}\:\mathrm{cos}\:{A}}{{a}}\:+\:\frac{\mathrm{cos}\:{B}}{{b}}\:+\:\frac{\mathrm{2}\:\mathrm{cos}\:{C}}{{c}} \\ $$$$=\:\frac{{a}}{{bc}}\:+\:\frac{{b}}{{ac}}\:,\:\mathrm{prove}\:\mathrm{that}\:\angle{A}\:=\:\mathrm{90}°. \\ $$
Answered by ajfour last updated on 13/Jun/17
$$\frac{\mathrm{2cos}\:{A}}{{a}}+\frac{\mathrm{cos}\:{B}}{{b}}+\frac{\mathrm{2cos}\:{C}}{{c}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{abc}} \\ $$$$\Rightarrow\:\:\mathrm{2}{bc}\mathrm{cos}\:{A}+{ac}\mathrm{cos}\:{B}+\mathrm{2}{ab}\mathrm{cos}\:{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${knowing}\:\mathrm{cos}\:{A}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$$\:\:\mathrm{cos}\:{B}=\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}}\:\:,\:\:\mathrm{cos}\:{C}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$${we}\:{get} \\ $$$$\:\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\:\:\:\:\:={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{b}^{\mathrm{2}} −{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\:\Rightarrow\:\:\:\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\:{hence}\:\boldsymbol{{a}}\:{is}\:{the}\:{hypotenuse}\:{and}\:\: \\ $$$$\:\:\:\angle{A}\:=\:\mathrm{90}°\:.\: \\ $$
Commented by Tinkutara last updated on 14/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Jun/17
$$\frac{\mathrm{2}{sosA}}{{a}}+\frac{{cosB}}{{b}}+\frac{\mathrm{2}{cosC}}{{c}}= \\ $$$$\frac{\mathrm{2}{bcosA}+{acosB}}{{ab}}+\frac{\mathrm{2}{cosC}}{{c}}= \\ $$$$\frac{{c}+{bcosA}}{{ab}}+\frac{\mathrm{2}{cosC}}{{c}}=\frac{{c}^{\mathrm{2}} +{bccosA}+\mathrm{2}{abcosC}}{{abc}}= \\ $$$$=\frac{{c}^{\mathrm{2}} +{b}\left({c}.{cosA}+{acosC}\right)+{abcosC}}{{abc}}= \\ $$$$=\frac{{c}^{\mathrm{2}} +{b}^{\mathrm{2}} +{abcosC}}{{abc}}=\frac{\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{\mathrm{2}{abc}}= \\ $$$$=\frac{\mathrm{3}{b}^{\mathrm{2}} +{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{abc}}=\frac{{a}}{{bc}}+\frac{{b}}{{ac}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{abc}}\Rightarrow \\ $$$$\mathrm{3}{b}^{\mathrm{2}} +{a}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \Rightarrow{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={a}^{\mathrm{2}} \:. \\ $$
Commented by Tinkutara last updated on 14/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$