# 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?

• Have you tired drawing this situation on the $x-y$ plane and seeing what results? Nov 9, 2017 at 20:59
• i tried by using reflection of A about two bisectors...i got solution but it not satisfy triangle property....i have doubt that whether solution always possible or not. Nov 9, 2017 at 21:02
• You're on the right track using reflection. Also think about what these bisectors are telling you. Having a bisector of $y=0$ means the vertex C must lay on the x-axis and be of the form (c,0). Similarly, the bisector $y=x$ means the vertex B must be of the form (b,b). Can you go from here? If not, I can write up a formal answer. Nov 9, 2017 at 21:24

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!

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.