# Construct the triangle with given angle bisectors

given three lines $\ell_1,\ell_2, \ell_3$ which intersect in one point $P$. How can one construct a triangle such that the given lines become its angle bisectors?

So far I tried to find conditions on how the three given lines have to intersect such that the desired triangle exists. There are altogether six rays $\omega_1^1, \omega_1^2; \omega_2^1,\omega_2^2;\omega_3^1,\omega_3^2$ - any line $\ell_j$ determines the two rays $\omega_j^1,\omega_j^2$- emanating from the common point $P$.

Any of the three vertices $V_1,V_2,V_3$ of the desired triangle has to lie on exactly one line, i.e. $V_j\in \ell_j=\omega_j^1\cup \omega_j^2$. Hence we have either $V_j\in \omega_j^1$ or $V_j\in \omega_j^2$.

Therefore the triangle should exist if and only if it is possible to choose three rays $\tilde{\omega}_1\in \lbrace \omega_1^1, \omega_1^2\rbrace$, $\tilde{\omega}_2\in \lbrace \omega_1^2, \omega_1^2\rbrace$, $\tilde{\omega}_3\in \lbrace \omega_1^3, \omega_1^3\rbrace$ such that any three of the rays $\tilde{\omega_1}, \tilde{\omega_2}, \tilde{\omega_3}$ form an angle less than $\pi$.

But I do not know how to construct the triangle?!

I started to draw a circle around $P$ with any radius. Now I can choose a Point $V_1$ on the ray $\tilde{\omega_1}$ outside the circle. And now I can draw the tangents through $V_1$ to the circle. I think those tangents will meet the other rays at some points $V_2,V_3$ (Why?). Now we can draw the line $V_2,V_3$. But this line needs to be tangent to the circle and I'm not sure about that. Will this construction work, or is it done in a different way?

Best regards

• Hint: Can you figure out what are the angles of the triangle given these lines? If so, how does that help you? Jun 28 '15 at 8:29
• No I do not know :-(. But I realized that my construction does not work.
– asd
Jun 28 '15 at 10:10

Let $I$ be the point of intersection of the lines (renamed it so that it's clear that we want it to be the incenter) and let $A,B,C$ be the points we want to construct. Let $\alpha,\beta,\gamma$ be angles of the triangle, and let $\delta,\epsilon,\zeta$ be angles $BIC,CIA,AIB$ respectively (see the picture).
We can easily find angles $\delta,\epsilon,\zeta$ given the three lines, so now the aim is to express $\alpha,\beta,\gamma$ in terms of these. Let's try to find value of $\delta$. We have $\delta=180°-\beta/2-\gamma/2=180°-(\beta+\gamma)/2=180°-(180°-\alpha)/2=90°+\frac{\alpha}{2}$, so that $\alpha=2\delta-180°$. Similarly we can find values of $\beta,\gamma$ given $\epsilon,\zeta$. Now if we know the angles of triangle $ABC$ we can easily construct it on the lines, I'll leave it for you to do.
• Shouldn't that be $\delta=90°+\frac{\alpha}2$, so that $\alpha=2\delta-180°$? (Or perhaps $\frac\alpha2=\delta-90°$ would be more useful.) That seems to be the correct development of your algebra, and it also agrees with my Geogebra testing. Jun 29 '15 at 0:58