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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×b2+c2−a22bc+b×a2+c2−b22ac=c×a2+b2−c22ab×2abc:a2(b2+c2−a2)+b2(a2+c2−b2)=c2(a2+b2−c2)a2b2+a2c2−a4+a2b2+b2c2−b4=a2c2+b2c2−c42a2b2−a4−b4=−c4(a2−b2)2=c4⇒a2−b2=±c2⇒a2=b2+c2orb2=a2+c2⇒rightangledtriangleinbothcases
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