Rotating Equilateral Triangles

Observe that $CA = CB$ , $CE = CD$ and $\angle BCE = 60 ^ \circ - \angle ACE = \angle ACD$ so CDA is congruent to CEB by SAS.

Therefore, $\angle CDA = \angle CEB = 95 ^ \circ$ and so $\angle ADE = \angle ADC - \angle EDC = 95 ^ \circ - 60 ^ \circ = 35 ^ \circ$ .

Since $C$ is the center of the spiral similarity which maps $\{E,B\}$ to $\{A,D\}$ , then the intersection of lines $BE$ and $AD$ , which we will name as point $K$ , lies on the circumcircles of $ABC$ and $EDC$ .

Thus, since $BEK$ is a straight line, then $\angle KED = 180^\circ - \angle BED = 25^\circ$ .

Similarly, $\angle EKD = 180^\circ - \angle ECD = 120^\circ$ , since $E,K,C,D$ are cyclic.

Thus, in triangle $KED$ , $\angle KDE = 180^\circ - \angle EKD - \angle KED = \boxed{35^\circ}$ .

Also, it is a fun fact to know that $\angle BCE = 45^\circ$ .

Combine the arguments of Lolly Lau, and Manuel Kahayon:

Sides $AC$ and $BC$ are the same length. Likewise, sides $DC$ and $EC$ . Also, $\angle ACB$ and $\angle DCE$ are both 60°. This tells us that by the SAS theorem, that $\triangle ADC$ and $\triangle BEC$ are congruent (rotate the first by 60° CCW on the point $C$ to obtain the second). Thus, $\angle ADC$ is, like $\angle BEC$ , 95°. Since $\angle EDC$ is 60°, the remaining (95° - 60° =) 35° is the measure of $\angle ADE$ .

∆ABC and ∆CDE are both equilateral triangles, so AC=AB=BC and CD=DE=CE

∠ACE = 60-45=15°.

∠ACD=60-15=45°=∠BCE.

BC=AC, CE=CD, and ∠BCE=∠ECD, so according to SAS(side-angle-side), ∆BCE=∆ACD.

∠BEC=95°=∠ADC=60+x

∴ ∠ADE=35°

put sides BC=L, BE=N, EC=M and AD=Z then project perpendicular line( H) from point E to side "L" from geometry of the triangles H=N.(sin 40)=m.(sin 45) so N= M.(sin 45÷sin 40). & L=N.cos 40+M.cos 45 =M.(sin 45÷sin 40).Cos 40+M.cos 45 so L=1.549803828.M. at∆ ADC , Angle ACD=45° so cos 45=1÷√2=. (M^2+1.549803828^2.M^2-Z^2)÷(2.M.1.549M) so Z=1.00062868M at ∆ ACD cos (60+u) and u=requested Angle. cos(60+u)=(M^2+1.00628^2.M^2-1.5498M^2)÷ (2.M.(1.00628.M)) so 60+u=95 so U=35°##########