Triangle Problem

By doing extra construction as shown above we get an isosceles trapezoid $ABCD$ . $AECD$ is a kite and its diagonals are perpendicular. Line segment $AC$ bisects $\angle DAE$ therefore $\angle CAE =10^\circ$

Some angle chasing gives the angles listed.

We can set $AE=EB=BC=1$ .

$EC=2\times sin(10^\circ)=0.3473$

From the law of cosines in $\triangle ACE$ we get $AC=1.2856$

From the law of sines $\angle CAE=arcsin(\frac {CE}{AC}\cdot sin(140^\circ))=\boxed {10^\circ}$

Define the 3 equal lengths each as S, AC = h, EC = f. Clearly, <BEA = 140, < BEC = 80, <AEC = 140. In triangle BEC, by Law of Cosines, F^2 = s^2 + s^2 - 2s^2 cos(20), f = .347296355S. In triangle AEC, by Law of Cosines, h^2 = f^2 + s^2 -2fs cos(140), h = 1.28557515S. In triangle AEC, by the Law of Sines, f/sin(x) = h/sin(140), x = 10. Ed Gray

Since $AE=BE$ , then $\triangle AEB$ is an isosceles triangle. Draw $∠CBD$ , where $∠CBD=20°$ clockwise. Extend $AC$ to $F$ , where $F$ is the intersection of $AC$ and $BD$ . Since $AC⊥BD$ , then $\triangle AFB$ is a right triangle. Note that $∠ABF=3∠ABE$ , then $\triangle AFB$ is a 30-60 right triangle. Therefore, $∠BAF=∠BAE+∠CAE=30°⟹∠CAE=10°$ .