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konrad509
[tex](x+9)^2=(x+1)^2+(x+5)^2\\ x^2+18x+81=x^2+2x+1+x^2+10x+25\\ x^2-6x-55=0\\ x^2+5x-11x-55=0\\ x(x+5)-11(x+5)=0\\ (x-11)(x+5)=0\\ x=11 \vee x=-5\\\\ |AC|=x+9=11+9=20[/tex]
[tex]This\ question\ is\ illogical.\\\\If\ \Delta ABC\simeq\Delta DE F\ and\\\Delta ABC\ and\ \Delta DE F\ are\ rectangular\ triangle,\ then:\\\\\frac{|AB|}{|BC|}=\frac{|DE|}{|EF|}\\\\If\ m\angle D=60^o\ then\ m\angle A=60^o.\\The\ triangle\ 30^o;60^o;90^o\ has\ the\ relation\ between\ the\ sides\\(look\ at\ the\ picture).[/tex]

[tex]\frac{|DE|\sqrt3}{2}=12\ \ \ \ \ |multiply\ both\ sides\ by\ 2\\|DE|\sqrt3=24\ \ \ \ |multiply\ both\ sides\ by\ \sqrt3\\3|DE|=24\sqrt3\ \ \ \ |divide\ both\ sides\ by\ 3\\|DE|=8\sqrt3\Rightarrow|DF|=2\times8\sqrt3=16\sqrt3\\\\\Delta ABC:\\x+9=2(x+1)\\x+9=2x+2\\x-2x=2-9\\-x=-7\\\boxed{x=7}\\\\|AC|=x+9=7+9=\fbox{16};\\|AB|=x+1=7+1=\fbox8;\\|BC|=x+5=7+5=\fbox{12}\\\\Angles\ of\ \Delta ABC:30^o;\ 60^o;\ 90^o,\ then:[/tex]

[tex]|BC|=\frac{|AC|\sqrt3}{2}\\\\|BC|=\fbox{12}\\\\\frac{|AC|\sqrt3}{2}=\frac{16\sqrt3}{2}=\boxed{8\sqrt2}\\\\12\neq8\sqrt2[/tex]


[tex]If\ use\ Pythagoras\ theorem\ like\ Konrad:\ x=11\\\\|AC|=20\ and\ |AB|=11+1=12\\\\|AC|=2|AB|\to20=2\times12-FALSE...[/tex]
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