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Question-40960




Question Number 40960 by behi83417@gmail.com last updated on 30/Jul/18
Commented by MrW3 last updated on 30/Jul/18
a×((b^2 +c^2 −a^2 )/(2bc))+b×((a^2 +c^2 −b^2 )/(2ac))=c×((a^2 +b^2 −c^2 )/(2ab))  ×2abc:  a^2 (b^2 +c^2 −a^2 )+b^2 (a^2 +c^2 −b^2 )=c^2 (a^2 +b^2 −c^2 )  a^2 b^2 +a^2 c^2 −a^4 +a^2 b^2 +b^2 c^2 −b^4 =a^2 c^2 +b^2 c^2 −c^4   2a^2 b^2 −a^4 −b^4 =−c^4   (a^2 −b^2 )^2 =c^4   ⇒a^2 −b^2 =±c^2   ⇒a^2 =b^2 +c^2  or b^2 =a^2 +c^2   ⇒right angled triangle in both cases
$${a}×\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}+{b}×\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}}={c}×\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$×\mathrm{2}{abc}: \\ $$$${a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)={c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right) \\ $$$${a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} −{a}^{\mathrm{4}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} ={a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{c}^{\mathrm{4}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{a}^{\mathrm{4}} −{b}^{\mathrm{4}} =−{c}^{\mathrm{4}} \\ $$$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} ={c}^{\mathrm{4}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\pm{c}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:{or}\:{b}^{\mathrm{2}} ={a}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\Rightarrow{right}\:{angled}\:{triangle}\:{in}\:{both}\:{cases} \\ $$

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