# 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? – Wojowu 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

## 2 Answers

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. – Rory Daulton Jun 29 '15 at 0:58
• @RoryDaulton Yes, of course, thanks for spotting that. – Wojowu Jun 29 '15 at 6:45
• @RoryDaulton Is it unique? Would more solutions appear when Geogebra leaves vertices free? – Narasimham Nov 8 '17 at 17:39
• @Narasimham: IIRC I chose 3 points then formed the triangle and the rest, including the 3 "starting lines," from those. I then checked the values of the angles and saw that they disagreed with the 1st version of this answer. If I were to choose the 3 lines first I would choose 1 point on 1 of the lines, then the rest would result from that--so there is 1 degree of freedom. (I believe--I can't check that right now.) – Rory Daulton Nov 8 '17 at 19:03

I was attempting to address the general question you posed earlier (by Mathematica) with an assumption of existence of common angular bisectors of given the three lines for variable triangles schematically hand sketched here:

which would perhaps be also interesting. Whether they would be a set of parallel similar triangles of parallel sides needs yet to be confirmed.