Proof of this is simple, given the cosine law: $$ c^2 = a^2 + b^2 - 2ab\cos\phi $$ where $a$ , $b$ , and $c$ are the lengths of the sides of the triangle, and angle $\phi$ is the corner angle opposite the side of length $c$ .

Let us define the largest interior angles as angle $\theta$ . Applying this to the parallelogram, we find that \begin{eqnarray*} d_1^2 &=& u^2 + v^2 - 2uv\cos\theta\\ d_2^2 &=& u^2+v^2 - 2uv\cos\left(\pi-\theta\right) \end{eqnarray*}

Knowing that $$ \cos\left(\pi-\theta\right) = - \cos\theta $$ we can add the two expressions together, and find ourselves with \begin{eqnarray*} d_1^2+d_2^2 &=& 2u^2 + 2v^2 - 2uv\cos\theta + 2uv\cos\theta\\ d_1^2+d_2^2 &=& 2u^2 + 2v^2 \end{eqnarray*}which is the theorem we set out to prove.