how to find triangle for given two bisector and one vertex. Suppose $ (1,2)$ be given vertex A of a triangle. Let $y =x $ and $ y = 0$  be two bisectors of other two vertices B and C. How to find equation of the line joining B and C?
 A: I will give you a purely mathematically approach to this. You should draw out each of these steps to see how they translate into shapes.
I will call the vertex with bisector $y=0$ vertex $C$. The vertex with bisector $y=x$ is $B$. Noe that the line $AC$ has the form
$$AC=y=\frac{-2}{c-1}(x-c)$$
Reflecting this line over the bisector $y=x$ will yield the line $BC$:
$$BC=\frac{c-1}{-2}x+c$$
Note that this line must contain the point $(0,c)$. Setting $x=c$ yield
$$0=c^2-3c$$
So $c=0$ or $c=3$. You can verify that the only valid solution is $c=3$.
Note that the line $BC$ must also go through vertex $B$. The only step now is to set $x=b$ in $BC$. You should see at this point that $b$ must be negative to satisfy the bisector $y=0$:
$$b=\frac{c-1}{-2}b+c$$
which gives us 
$$b=\frac{2c}{c+1}=\frac{6}{4}=\frac{3}{2}$$
Uh oh... Turns out $b$ is positive. You can verify the triangle defined by these $A,B,C$ do not satisfy the constraints of the problem. There is no such triangle!
A: 
It is not hard to prove that $\phi=90-\frac{\beta}2$, $\psi=90-\frac{\gamma}2$, $\omega=90-\frac{\alpha}2$, are all acute angles for any triangle.
But in your problem one of the angles $\phi, \psi, \omega$ is obtuse, when you connect A to the origin. So there is no such triangle that satisfy your conditions.
