What is the angle $\angle CTB$ in the following image if we know that $\triangle ABC$ is equilateral ($\lvert AB \rvert = a$), $\lvert AX \rvert = \frac{a}{3}$, $\lvert BY \rvert = \frac{a}{3}$
Problem here is that I can't use trigonometry or anything complex really, I have to keep it simple. My idea was that maybe the intersection of $B$ and $AC$ is actually $\frac{a}{3}$ away from A but I'm not even sure that's true. I've also tried rotating it $60°$ but that yielded no results either and I'm kind of lost. Maybe I should try finding some congruent/similar triangles but I'm not sure in which direction I'm supposed to go. I've also thought about using the info I've been given( that the triangle is equilateral) but I'm still not quite certain what to do.
Thanks in advance!! :)
Triangles $ABY$ and $CAX$ are congruent. Therefore $$\angle BYT=\angle BYA=\angle AXC=\angle AXT=180-\angle BXT$$ hence quad $BYTX$ is cyclic. Therefore $$\angle BTX=\angle BYX$$ However $BX=2\, BY$ and $\angle XBY=60$ yielding that $\angle BYX =90$. Consequently $$\angle BTX=\angle BYX=90$$ so $$\angle CTB =180-\angle BTX =180-90=90$$