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

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.

• you mean given that $b, n_B , h_c$ Dec 9, 2020 at 17:58
• If $n_B$ means the length of the angle bisector, yes Dec 9, 2020 at 17:59
• @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.
– Blue
Dec 9, 2020 at 18:33
• @Airree: Are you sure that a ruler/compass construction can be made? Dec 10, 2020 at 12:18
• 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.
– YNK
Dec 12, 2020 at 4:12

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.

• 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. Dec 10, 2020 at 21:27
• 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... 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!