#2**+1 **

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

Guest May 6, 2022

#4**+1 **

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

BuilderBoi May 6, 2022