<s>[INST] Find the value of $x$ in the triangle $ABC$ A triangle $ABC$ has angle $A=2y$ and angle $C=y$. Furthermore, $AB=x,BC=x+2$, and $AC=5$. Solve for $x$.
I tried this question whole day but could not find solution of it . Please help me
 [/INST] Let $AD$ be a bisector of $\Delta ABC$ and $AD=y$.
Thus, $AD=DC=y$ and $BD=x+2-y$.
Now, since $\Delta{ABD}\sim\Delta{CBA}$, we obtain
$$\frac{y}{5}=\frac{x}{x+2}=\frac{x+2-y}{x},$$ which gives
$xy=5x-2y$ and $x^2=(x+2)^2-xy-2y$, which is $xy=4x+4-2y$ 
and from here $4x-2y+4=5x-2y$ or $x=4$ and we are done!
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