+478 In quadrilateral $BCED$, sides $\overline{BD}$ and $\overline{CE}$ are extended past $B$ and $C$, respectively, to meet at point $A$. If $BD = 8$, $BC = 3$, $CE = 1$, $AC = 19$ and $AB = 13$, then what is $DE$?
+15079 \(If\ BD = 8,\ BC = 3,\ CE = 1,\ AC = 19\ and\ AB = 13,\ then\ what\ is\ DE?\)
\(f_{AB}(x)=\sqrt{13^2-x^2}\\ f_{BC}(x)=\sqrt{3^2-(x-19)^2}\\ 169-x^2=9-x^2+38x-361\\ 38x=169-9+361=521\\ x_B=13.7105\\ y_B=\color{red }\sqrt{-18.9...}\)
The triangle ABC, with sides 3, 19, 13 is not possible.
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