Alternative methods of solving an angle in an equilateral triangle This question has already been answered here, but the OP said trigonometry was forbidden. I was thinking of different approaches that allow trigonometry, so I decided to post a new question.

An equilateral triangle is given with edges of the length $a$. Let $X\in\overline{AB}$ s.t. $|AX|=\frac{a}{3}$ and let $Y\in\overline{BC}$ s.t. $|BY|=\frac{a}{3}$. Let $T$ be the intersection point of $AY$ and $CX$. Find $\measuredangle{CTB}$.

First of all, $|BX|=2|BY|\;\&\;\measuredangle XBC=60^{\circ}\implies\Delta XBY$ is one half of an equilateral triangle$\implies\measuredangle BYX=90^{\circ}\implies\color{red}{\Delta XYC\;\text{is a right-triangle}}$.
Now, $|XY|^2=\left(\frac{2a}{3}\right)^2-\left(\frac{a}{3}\right)^2=\frac{3a^2}{9}$
$|CX|=\sqrt{|XY|^2+|YC|^2}=\frac{\sqrt{7}a}{3}$
According to the $SAS$ (Side-Angle-Side) theorem, $\Delta ABY\cong\Delta AXC\;\&\;\Delta XBC\cong\Delta AYC$.
Then,
$\measuredangle CXB=\measuredangle AYC\;\&\;\measuredangle YCT=\measuredangle BCX\implies\measuredangle CTY=\measuredangle XBC=60^{\circ}\implies\Delta TYC\sim\Delta BCX$
$$\implies\frac{|XC|}{|YC|}=\frac{|BC|}{|CT|}\implies|CT|=\frac{|BC|\cdot|YC|}{|XC|}=\frac{a\frac{2a}{3}}{\frac{\sqrt{7}a}{3}}=\frac{2a}{\sqrt{7}}$$
In $\Delta XYC$, we have:$\cos(\measuredangle YCX)=\frac{|CY|}{|CX|}=\frac{\frac{2a}{3}}{\frac{\sqrt{7}a}{3}}=\frac{2}{\sqrt{7}}$
In $\Delta BCT$, we have:
$$|BT|=\sqrt{|BC|^2+|CT|^2-2|BC|\dot|CT|\cos(\measuredangle YCX)}=\sqrt{a^2+\frac{4a^2}{7}-2a\cdot\frac{2a}{\sqrt{7}}\cdot\frac{2}{\sqrt{7}}}=\frac{\sqrt{3}a}{\sqrt{7}}$$
In $\Delta YCT$, we have: $\frac{|CY|}{\sin(\measuredangle CTY)}=\frac{|CT|}{\sin(\measuredangle TYC)}\implies \sin(\measuredangle TYC)=\frac{|CT|\sin(\measuredangle CTY)}{|CY|}=\frac{\frac{2a}{\sqrt{7}}\frac{\sqrt{3}}{2}}{\frac{2a}{3}}=\frac{3\sqrt{3}}{2\sqrt{7}}=\sin(\measuredangle BYT)$
In $\Delta BYT$, we have $\frac{|BT|}{\sin(\measuredangle BYT)}=\frac{|BY|}{\sin(\measuredangle YTB)}\implies\measuredangle YTB=\arcsin\frac{|BY|\sin(\measuredangle BYT)}{|BT|}=\arcsin\frac{\frac{a}{3}\frac{3\sqrt{3}}{2\sqrt{7}}}{\frac{\sqrt{3}a}{\sqrt{7}}}=\arcsin\frac{1}{2}\implies\measuredangle YTB=30^{\circ}$
Finally, $\measuredangle CTB=\measuredangle TYB+\measuredangle CTY=90^{\circ}$
Picture:
My question is: is there any shorter way I could solve this via trigonometry or vectors? Thank you in advance!
 A: Here is a proof with vectors. Assume $|{AB}|=|{BC}|=|{CA}|=1$. Then, 
$$\vec {AB}\cdot \vec {AC} = \vec {BC}\cdot \vec {BA}=\vec {CA}\cdot \vec {CB}=\cos60^\circ=\frac12\tag1$$
Given that
$$\frac{|XT|}{|CT|}=\frac{Area_{AXY}}{Area_{ACY}}=\frac{\frac13\cdot\frac13Area_{ABC}}{\frac23Area_{ABC}}=\frac16\implies\frac{|CT|}{|CX|} =\frac67$$
and $\vec {CX} = \frac13 \vec {AB} - \vec {AC}$, we have 
$$\vec {BT} = \vec {CT} - \vec {CB}=\frac67 \vec {CX} - \vec {CB}
= \frac67 \left(\frac13\vec {AB} - \vec {AC} \right)- \vec {CB}$$
Evaluate
$$\vec {BT}\cdot \vec {CX} = \left(\frac27 \vec {AB} - \frac67\vec {AC} - \vec {CB}\right)\cdot \left( \frac13 \vec {AB} - \vec {AC}\right)=0$$
where the doc-products in (1) are used. Thus, $\angle CTB = 90^\circ$.
A: Not a trigonometric solution, but I decided to post it anyway.
It is clear that triangles $CXA$ and $AYB$ are congruent, hence $\angle CXA=\angle AYB$. Hence quadrilateral $BXTY$ is cyclic. The center of this circle is the midpoint $M$ of $XB$ as $MB=MX=\frac 13 a=MY$. Hence $\angle BTY =\frac 12 \angle BMY = 30^\circ$.
