A triangle ABC with vertex $C(4,3)$. The bisector and the median line equation drawn from the same vertex are given. Find the vertices A & B. A triangle $\triangle ABC$ one of his vertex is the point $C(4,3)$. 
The bisector line equation is $x+2y-5=0$ and the median line equation is $4x+3y-10=0$ drawn from the same vertex.
Find the coordinates of the vertices A and B.
What i have done:
I started using the fact that the two equations starts on the same vertex, but is not $C$.
So, let $A$ be that vertex, i calculate the intersection point by the system:
$$y_A=-1/2\cdot x_A+5/2$$
$$y_A=-4/3\cdot x_A+10/3$$
Which implies that $A(1,2)$
But i dont know how to find $B$.
 A: The segment connecting $A$ in $(1,2)$ with point $C$ in $(4,3)$ lies on the line $y=\frac{x}{3}+\frac{5}{3}$ and has length $\sqrt{3^2+1^2}=\sqrt{10}$. 
Note that the bisector crosses the $x$-axis in point $(5,0)$. Let us call $D$ this point. The segment $CD$ lies on the line $y=-3x+15$ and has length $\sqrt{3^2+1^2}=\sqrt{10}$. 
Because segments $AC$ and $CD$ have equal length and lie on lines whose slopes are one the negative inverse of the other, we get that $ACD$ is an isosceles right triangle. Thus, the bisector $AD$ forms with $AC$ a $45^o$ degree angle, which means that the angle $CAB$ is right. As a result, the leg $AB$ of the right triangle $ABC$ has slope equal to $CD$, and then lies on the line $y=-3x+5$.
Now let us call $s$ the slope of the segment $BC$ and $E$ its middle point. This segment lies on the line $y=sx+3-4s$, which crosses the median $y=\frac{10-4x}{3}$ in $(\frac{12s+1}{3s+4}, \frac{12-6s}{3s+4})$. So the length of segment $CE$ is 
$$\sqrt{(4-\frac{12s+1}{3s+4})^2+(3-\frac{12-6s}{3s+4})^2} \\ = \frac{\sqrt{15^2+(15s)^2}}{3s+4}= \frac{15\sqrt{s^2+1}}{3s+4}$$
On the other hand, the segment $BC$ crosses the leg $AB$ in $(\frac{4s+2}{s+3}, \frac{9-7s}{s+3})$. So the length of segment $BE$ is 
$$\sqrt{(  \frac{4s+2}{s+3}  -\frac{12s+1}{3s+4})^2+(  \frac{9-7s}{s+3}  -\frac{12-6s}{3s+4})^2 } \\ = \frac{\sqrt{(5-15s)^2+(5s-15s^2)^2}}{(s+3)(3s+4)}=  \frac{5(3s-1)\sqrt{s^2+1}}{(s+3)(3s+4)}$$
Since $BE=CE$ we get
$$\frac{5(3s-1)\sqrt{s^2+1}}{(s+3)(3s+4)}=\frac{15\sqrt{s^2+1}}{3s+4})$$
which reduces to
$$\frac{3s-1}{s+3}=3$$
The last equation has no real solutions, except for $s \rightarrow \infty$. This means that the segment $BC$ is vertical, i.e., it lies on a line of the form $x=k$. Since it includes point $C$, we get that segment $BC$ lies on the line $x=4$. Knowing this, it is not difficult to find its intersection with line $AB$, allowing to conclude that point $B$ is in $(4,-7)$.
