What is the angle $\angle CTB$ in the following image if we know that $\triangle ABC$ is equilateral ($\lvert AB \rvert = a$), $\lvert AX \rvert = \frac{a}{3}$, $\lvert BY \rvert = \frac{a}{3}$ enter image description here

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!! :)

  • $\begingroup$ if it helps, a little calculation shows that the angle is $90°$, and also that the angle $CTY$ is $60°$ (and btw the distance you guessed to be $a/3$ is actually $a/5$). Perhaps knowing this you will find a purely geometric solution. $\endgroup$ – user8268 May 4 '17 at 18:48
  • $\begingroup$ @user8268 Geogebra already showed it's 90° but it didn't do much good... $\endgroup$ – Collapse May 4 '17 at 18:50
  • $\begingroup$ $CTY = 60^o$ can be seen from rotational symmetry. Draw the third line from $B$ to a point on $AC$ that is $a/3$ away from $C$. The internal triangle must be equilateral. $\endgroup$ – eyeballfrog May 4 '17 at 18:55
  • $\begingroup$ @eyeballfrog hmm I disregarded that earlier but I guess it has to be true, still not sure what to do from there though, it's really bothering me..I'm also pretty sure I managed to prove that the line from X to Y is orthogonal to $BC$ $\endgroup$ – Collapse May 4 '17 at 18:59
  • $\begingroup$ If we call the point I mentioned $Z$ and suppose $CZ$ intersects $AY$ at $S$, then proving that $BS = TS$ would be sufficient. Not sure how to do that though. $\endgroup$ – eyeballfrog May 4 '17 at 19:08

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 $$\angle CTB =180-\angle BTX =180-90=90$$


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