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I'm trying to arrive to the shortest expression possible for finding the diagonal in a convex (and non-cyclic) quadrilateral, knowing its four sides lengths and the other diagonal.

My best try matches the second proposed solution in this other question: How can I find the diagonal of a quadrilateral? proposed by Yves Daoust, using the law of cosines:

$$\mu_0=\arccos\left(\frac{MQ^2+MP^2-PQ^2}{2MQ\cdot MP}\right)\\ \mu_1=\arccos\left(\frac{MP^2+MN^2-NP^2}{2MP\cdot MN}\right)\\ NQ=\sqrt{MQ^2+MN^2-2MQ\cdot MN\cos\left(\mu_0+\mu_1\right)}.$$ where MN, NP, PQ, and MQ are the sides of the convex quadrilateral, MP is the known diagonal, and NQ is the unknown diagonal.

However, that's a still relatively large expression. I tried to simplify taking advantage that $$\arccos\alpha \pm \arccos\beta = \arccos\left(\alpha\beta \mp \sqrt{(1-\alpha^2)(1-\beta^2)}\right)$$ but Mathematica turns this into quite a messy expression which, if I subtract from the original, won't result zero. So, maybe I'm doing something wrong.

Anyway, is there any other approach which would give a shorter expression for the diagonal?

Thanks!

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cesss
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1 Answers1

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I think this alternate approach could be even messier but works.

enter image description here

$ABCD$ is a quadrilateral with its diagonal $BD$ and all its sides known. Then, we have the following facts:_

  1. By Heron’s formula, $[⊿ABD]$ and $[⊿CBD]$ are known. Therefore, $[ABCD]$ is known.

  2. In $⊿ABD, h = \dfrac {2[⊿ABD]}{AB}$, is known.

  3. By cosine law, $\alpha$ is also known.

Through $C, CX$ is drawn parallel to $DB$ cutting $AB$ extended at $X$. Then $\beta = \alpha$.

After joining $DX$, it should be understood that $[ADX] = [ABCD]$. This means $AX = \dfrac {2[⊿ABCD]}{h}$, another know quantity. This further means $BX$ is known.

$⊿BCX$ with $BX, \beta$ and $BC$ known is solvable. This implies $CX$ is known.

Finally, $AC$ is found by applying cosine law to $⊿AXC$.

Mick
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