I stumbled into an Irregular convex quadrilateral in one of my projects and I cannot figure out if the diagonal |BD| can be found.
Problem:
I have an Irregular convex quadrilateral ABCD defined by the picture as [1]: https://i.sstatic.net/PWLL2.png
where I know all sides |AB|, |BC|, |CD| and |DA| and angle $\angle ADC$ (inside angle of point D = $\delta$). I need to find the length of diagonal |BD|.
Attempts:
I have found the length of |AC| diagonal by the law of cosines:
$$ |AC|^2 = |DA|^2 + |CD|^2 – 2\cdot |DA|\cdot |CD| \cdot \cos(\delta) $$
but I cannot seem to figure out the |BD| diagonal. I tried using the area of quadrilateral or law of sines but to no avail…
Does anybody have any ideas on how to proceed?
Thank you!
Best Answer
You found $AC$ using the cosine rule. With another version of that same law it is possible to find any angle given all three sides of a triangle. You now know all three sides of $\triangle ABC$ and $\triangle ADC$. With that you can derive $\angle DAC$ and $\angle CAB$, hence $\angle DAB$.
Now, consider $\triangle DAB$. You have sides $AD$ and $AB$, and $\angle DAB$. Use the cosine rule once more to solve for $BD$.