Construct a triangle given side length $b$, the altitude for side $c$, and the angle bisector of $B$.

So far I only found that I can find the angle at $C$ ($\gamma$) by constructing the right triangle with leg $h_c$ and hypotenuse $b$. I played a bit with the angles, but I can't really see how to use the length of the angle bisector, since there aren't many theorems helping me.

I feel like I have to use the inscribed angle theorem and find $B$ as the intersection of 2 loci, but I only found that it lies on side $c$

Any help would be appreciated. Thanks.

  • $\begingroup$ you mean given that $b, n_B , h_c$ $\endgroup$
    – Lion Heart
    Commented Dec 9, 2020 at 17:58
  • $\begingroup$ If $n_B$ means the length of the angle bisector, yes $\endgroup$
    – Airree
    Commented Dec 9, 2020 at 17:59
  • 1
    $\begingroup$ @Airree: Please include your work in finding $\gamma$, so that people don't waste time duplicating your effort. This will help inform readers about the types of tools and techniques you understand. $\endgroup$
    – Blue
    Commented Dec 9, 2020 at 18:33
  • $\begingroup$ @Airree: Are you sure that a ruler/compass construction can be made? $\endgroup$
    – Narasimham
    Commented Dec 10, 2020 at 12:18
  • $\begingroup$ This construction can be carried out using a method called construction by iteration, which uses only ruler and compass. However, this method is neither widely accepted nor used as a method of geometric construction. If you want to see how this method works, yo can find an example at link. If you want a similar answer to your question, let me know. $\endgroup$
    – YNK
    Commented Dec 12, 2020 at 4:12

2 Answers 2


It is not certain if such a construction is possible.

Next best option perhaps is to find important missing sides/angles analytically, to examine their simplicity in order to incorporate them in construction

After no luck with direct construction, a numerical iteration with assumed $ (b,d,h)= (5,4,2) =$ (base,bisector length and altitude) was attempted.

The following generating equation ( derived using standard triangle trig relations) in unknown $a$ gives $a\approx 3.08$.. verified to be ok.

$$\dfrac{d^2}{a}=(\sqrt{b^2-h^2}+\sqrt{a^2-h^2})\cdot (1-\dfrac{b^2}{(\sqrt{b^2-h^2}+a+\sqrt{a^2-h^2})^2})$$

This relation may further guide the sought construction method.

enter image description here

  • $\begingroup$ If you want to cover this problem further, you could perhaps consider the Euler line and the nine-point circle. I suspect that if it can be built it could be perhaps using these two geometric notions. $\endgroup$
    – Piquito
    Commented Dec 10, 2020 at 21:27
  • 1
    $\begingroup$ Thank you. Will try to rope in the centers of a triangle..may be Apollonian circle as well. My thought was like... just as there is the Euler's proof for trisection impossibility going from the generating/combining equation.. I prepared a composite equation ready at first... $\endgroup$
    – Narasimham
    Commented Dec 10, 2020 at 21:42

COMMENT.-Where did you get this problem from? It is very difficult. Analytically you can solve it if you calculate the value of $x$ which gives you the position of the vertex $C = (x, h)$, but $x$ is a solution of the equation

$$k^2(\sqrt{x^2+h^2}+x+b\cos\alpha)^2=(x^2+h^2)(x+b\cos\alpha)(\sqrt{x^2+h^2}+2b)$$ where $k,h,b,\alpha$ are data of your problem (the angle $\alpha$ is determined by the length of the side $b$, $h$ is your altitude and $k$ is the length of the angle bisector).

This equation is of degree eight!


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