Draw CF ⊥ DA ..( F on DA)
Let z= x^2
Applying Pythagoras’s theorem in triangles ABD, BCE and ACF we get
BE^2=x^2-121=z-121
BD^2=z-169
AC^2=CF^2+AF^2 => (BE+BD)^2+4=z
Or 2.BE.BD=z-BE^2-BD^2-4= 286-z
4.BE^2.BD^2=4(z-121)(z-169)=(286-z)^2
Simplify this equation we get 3.z^2-588.z= 0
We get 2 solutions z= 0 and z= 196
Only z=196 is acceptable so x= 14
+2668 Here is the link to the original problem, if anyone is interested in other solutions: https://gogeometry.blogspot.com/2016/03/geometry-problem-1194-equilateral.html
