Diagonals in a parallelogram

The lengths of the diagonals in a parallelogram can be computed using the law of cosines for the constructed triangle including the sides $a$ and $b$ and the angle $\alpha$ or $\beta$.

The diagonal $f$, from the upper left corner to the lower right corner of the parallelogram, uses therefore the following formula$f = \sqrt{a^2 + b^2 - 2ab \cos\left(\alpha\right)}.$The other diagonal $e$, from the lower left corner to the upper right corner of the parallelogram, includes in the constructed triangle the angle $\beta$. The angles in a parallelogram have the property that$\beta = 180 \degree - \alpha \left(= \pi - \alpha\right).$So we get the following formula for computing the length of the diagonal $e$\begin{aligned} e &= \sqrt{a^2+b^2 -2ab \cos \left(\beta\right)}\\ &= \sqrt{a^2+b^2 -2ab \cos \left(\pi -\alpha\right)} \\ &= \sqrt{a^2+b^2 +2ab \cos \left(\alpha\right)}. \end{aligned}

Both diagonals intersect at half of their length, i.e. they bisect each other.