Find the value of $x$ in the triangle $ABC$

Question

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

Answer 1

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!

Answer 2

Source: ... (To be completed by this answerer.)

I applied the sine theorem to get $\cos y$ the the cosine theorem to solve for $x$

Answer 3

From the sine theorem you get $cosy=\tfrac{x+2}{2x}$. By half-angle formula in any triangle: $$cos(\frac{\alpha}{2})=\sqrt{\tfrac{s(s-a)}{bc}}$$, where $s$-semiperimeter, $a,b,c$ - corresponding sides of triangle $ABC$. Let $\alpha=2y$, $s=3.5+x; a=x+2;b=x; c=5.$ Then: $$\tfrac{x+2}{2x}=\sqrt{\tfrac{1.5(3.5+x)}{5x}}$$ After squaring both sides and by elementary manipulation you left with quadratic equation: $$x^2+x-20=0$$ So, $x=4$.