I have the following situation (see pic below). I have two lines $B$, $C$, in the plane, the intersection point $a$, and a point $p$. I need to find the points $b$ and $c$ along $B$ and $C$ such that the line $A$ goes through $p$ and the two segments on each side of $p$ ($pb$ and $pc$) are of equal length, i.e., so that $l_1 = l_2$.


I have tried setting up various equation systems with three unknowns (the lengths $|bc|$, $|ac|$, and $|ab|$), using the Cosine Law but I only get really complicated algebraic equations when trying to solve them.

Any input would be helpful.


3 Answers 3


Reflect $C$ in $p$, obtaining reflected line $C'$. Let $c$ be the intersection of $C'$ and $B$.



I tried using analytic geometry, exploiting the fact that one can always situate $a$ at the origin and $p$ at $(0,1)$ in cartesian coordinates. Then the two lines $B$ and $C$ are simply of the form $y = \alpha_b x$ and $y = \alpha_c x$ and line $A$ is $y = \alpha_a x + 1$; we have solved the problem if we determine $\alpha_a$ in terms of $\alpha_b$ and $\alpha_c$ since we can alwyys rotate and rescale our original problem to fit this pattern.

Well, $$ A \cap B = \left( \frac{1}{\alpha_b - \alpha_a},\frac{\alpha_b}{\alpha_b - \alpha_a} \right) \\ A \cap C = \left( \frac{1}{\alpha_d - \alpha_a},\frac{\alpha_c}{\alpha_d - \alpha_a} \right) $$ Now we write down the expressions for $|pa|^2$ and $|pb|^2$, and cross multiply by the product of the denominators, to get $$ (\alpha_b-\alpha_a)^2 (1+\alpha_a^2) = (\alpha_c-\alpha_a)^2 (1+\alpha_a^2)$$ Then $$(\alpha_b-\alpha_a)^2 = (\alpha_c-\alpha_a)^2 \\ \alpha_a = \frac{\alpha_b+\alpha_c}{2}$$ Well, well, that is a pleasant surprise. THe geometric construction of line $A$ given point $p$ and lines $B$ and $C$ passing through point $a$ is now clear:

  • Construct line $ad \perp pa$; $d$ can be any arbitrary point on this line other than $a$.
  • Construct line $fd \perp ad$. Label the intersection of $df$ with line $B$ as $e$, and the intersection of $fd$ with line $C$ as $f$.
  • Construct point $g$ as the midpoint of line segment $ef$.
  • Draw line $ag$.
  • Construct a line through $p$ which is outside line $ag$, parallel to line $ag$. Label that line $A$ because it is the desired line.

You can ensure equal distances by considering circles around point $P$ and then looking for the intersection points $B$ and $C$ with the lines $c$ and $b$.

(Note: I use upper case letters for points, lower case letters for lines).

Here are some of these:


The interesting question is if we have a circle radius $R$ for which the vectors $u = PB$ and $v = PC$ are antiparallel $u = -v$.

The good news for this image is that there should be one, as the green and light blue cases show, there should be a feasible circle in between.

I do not know a good constructive way how to find that circle, in the above scenario one would try a circle and see if the lines along $PB$ and $PC$ are identical. Or one might look, if the line through $BC$ intersects with $P$.

Another analytical criteria would be that the area of the triangle $PBC$ vanishes.

  • $\begingroup$ Sorry for not mentioning it in the question but I have considered this. It solves the problem but it requires an iterative solution and I was looking for a solution in constant time. Thanks anyway for your input. $\endgroup$
    – mags
    May 12, 2015 at 6:52
  • $\begingroup$ I prepared an analytic solution, but stopped after I saw Steven's solution which looked very simple. It seems to be constructable. $\endgroup$
    – mvw
    May 12, 2015 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.