+1234 Can you solve this?
In quadrilateral $BCED$, sides $\overline{BD}$ and $\overline{CE}$ are extended past $B$ and $C$, respectively, to meet at point $A$. If $BD = 18$, $BC = 18$, $CE = 6$, $AC = 14$ and $AB = 10$, then what is $DE$?
+129852 
Law of Cosines
BC^2 = BA^2 + AC^2 -2 ( BA * AC) cos (DAE)
18^2 = 10^2 + 14^2 - 2 ( 10 * 14) cos (DAE)
[ 18^2 - 10^2 - 14^2 ] / [ - 2 (10 * 14) ] = cos (DAE)
-1/10 = cos (DAE)
Again
DE^2 = AD^2 + AE^2 - 2(AD * AE) cos (DAE)
DE^2 = 28^2 + 20^2 - 2(28 * 20) (-1/10)
DE^2 = 1296
DE = sqrt [1296] = 36
