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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×b2+c2a22bc+b×a2+c2b22ac=c×a2+b2c22ab×2abc:a2(b2+c2a2)+b2(a2+c2b2)=c2(a2+b2c2)a2b2+a2c2a4+a2b2+b2c2b4=a2c2+b2c2c42a2b2a4b4=c4(a2b2)2=c4a2b2=±c2a2=b2+c2orb2=a2+c2rightangledtriangleinbothcases

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