given three distinct lines $g,h,l$ meeting in one point $P$. I want to construct a triangle with vertices on $g,h,l$ such that those lines $g,h,l$ become its angle bisectors.
In general, if we consider a triangle $\Delta(ABC)$ with angles $\alpha,\beta,\gamma$ whose angle bisectors meet at a point $P$, then I recognized the following relations between the angles:
$$\angle (APB)=\frac{1}{2}(\pi+\gamma), \angle (BPC)=\frac{1}{2}(\pi+\alpha), \angle (CPA)=\frac{1}{2}(\pi+\beta).$$
Now suppose we choose points $C,C'\in g$ on different sides of $g$ with respect to $P$ and let $Q\in l,R\in h$ such that $C'$ lies in the interior of the angle $\angle (QPR)$. I started the construction as follows:
- Construct the angle $\angle (QPR)-\pi/2$ on one side of $g$ at the point C. Then we get a intersection point $A$ with the ray $\overrightarrow{PQ}$.
- Construct the angle $\angle (QPR)-\pi/2$ on the other side of $g$ at the point C. Then we get a intersection point $B$ with the ray $\overrightarrow{PR}$.
- Now connect the points $A,B$
But now I'm worried about the angles at $A,B$. I do not know if the angles have the right size and the lines $l,h$ through $A,B$ are indeed angle bisectors? Maybe my construction is wrong. Do you know how to do it right?
Best wishes
Let the angles between the lines be $\phi$, $\psi$ and $\omega$, $\phi+\psi+\omega=2\pi$. Intersection point $P$ is a center of the inscribed circle, so let's draw a circle with arbitrary radius, e.g. $r=1$:
Here we have a system of equations: \begin{align} \tfrac12\alpha+\tfrac12\gamma&=\pi-\phi \\ \tfrac12\beta+\tfrac12\gamma&=\pi-\psi \\ \tfrac12\alpha+\tfrac12\beta&=\pi-\omega \end{align}
which results in \begin{align} \alpha &= \pi-\omega-\phi+\psi \\ \beta &= \pi-\phi-\psi+\omega \\ \gamma &= \pi-\psi-\omega+\phi \end{align}
Now, $|AP|=r/\sin\tfrac\alpha2$, $|BP|=r/\sin\tfrac\beta2$, $|CP|=r/\sin\tfrac\gamma2$.