Proving bisector using trigonometry ABC is an isosceles triangle with AB = AC.
Point X is an arbitrary point on side BC. Points Y, Z are on the sides
AB, AC, respectively, such that ∠BXY = ∠ZXC. A line parallel to YZ and
passing through B cuts XZ at T.
Prove that AT bisects ∠A.
I'm trying to find a solution to this using trigonometry, but I can't seem to find a nice solution.
My attempt so far:
I let I be the intersection of AT and BC, then I tried to use the sine law on ABT, ABI, and ABI. This gave me a relation on the lengths of AB, BT, and BI. I couldn't manage to relate the length of BI in an equation with the length CI, other than using angles BAT and TAC. But then I don't think that leads to anything, since those angles are the ones that we want to find anyway.
 A: 
First apply sine law in $\triangle BYX$.
$\displaystyle \frac{BY}{\sin \beta} = \frac{BX}{\sin (\alpha + \beta)} \tag1$
Apply sine law in $\triangle BTX$,
$\displaystyle \frac{BT}{\sin \beta} = \frac{BX}{\sin (\beta - \gamma)} \tag2$
From $(1)$ and $(2)$,
$\displaystyle \frac{BY}{BT} = \frac{\sin (\beta - \gamma)}{\sin (\alpha + \beta)} \tag3$
Now applying sine law in $\triangle TZS$,
$\displaystyle \frac{ZS}{\sin (\beta - \gamma)} = \frac{TS}{\sin (\alpha + \beta)} \tag4$
From $(3)$ and $(4)$,
$ \displaystyle \frac{BY}{ZS} = \frac{BT}{TS}$
But we also know that $ \displaystyle \frac{BY}{ZS} = \frac{AB}{AS}$
So in $\triangle BAS$, $ \displaystyle \frac{AB}{AS} = \frac{BT}{TS}$
and so it follows that $AT$ is angle bisector of $\angle A$.
A: Signs in the solution correspond to case $BX < CX$, otherwise some signs need to be changed.
Let $\angle ZXC=\beta$, $\angle TBX=\gamma$. Then $\angle XYZ=\gamma+\beta$, $\angle XZY=\beta-\gamma$.
Sine theorem: $\frac{\sin(\gamma+\beta)}{\sin(\beta-\gamma)}=\frac{XZ}{XY}$.
Similarity of $\triangle BXY$ and $\triangle CXZ$: $\frac{XZ}{XY}=\frac{XC}{XB}$. Therefore $$\frac{XC}{XB}=\frac{\sin(\gamma+\beta)}{\sin(\beta-\gamma)}=\frac{1+\cot \beta \tan \gamma}{1-\cot \beta \tan \gamma}$$
Let $I$ to be the base of perpendicular from $T$ to $BC$. Then $$TI=XI \tan \beta=BI \tan \gamma \Rightarrow XI=BI\cot \beta \tan \gamma$$
$XB=BI-XI$, $XC=CI+XI$, therefore $$\frac{XC}{XB}=\frac{CI+XI}{BI-XI}=\frac{CI+BI\cot \beta \tan \gamma}{BI-BI\cot \beta \tan \gamma}=\frac{CI/BI+\cot \beta \tan \gamma}{1-\cot \beta \tan \gamma}$$
Using previous formula for $XC/XB$ one can get $CI/BI=1$. Therefore I is the middle of BC. Therefore T is on the perpendicular bisector of BC which is bisector of $\angle A$.
