Answer
Proof given below.
Work Step by Step
The diagonal is ${\bf u}+{\bf v}$
The cosine of the angle $\theta_{1}$, between ${\bf u}+{\bf v}$ and ${\bf u}$ is
$\displaystyle \cos\theta_{1}=\frac{({\bf u+v})\cdot{\bf u}}{|{\bf u+v}||{\bf u}|}$
The cosine of the angle $\theta_{2}$, between ${\bf u}+{\bf v}$ and ${\bf v}$ is
$\displaystyle \cos\theta_{2}=\frac{({\bf u+v})\cdot{\bf v}}{|{\bf u+v}||{\bf v}|}$
The denominators in these expressions are equal, because $|{\bf u}|=|{\bf v}|$.
If we prove that the numerators are equal, because cosine is one-to-one on $[0,\pi]$, it would follow that $\theta_{1}=\theta_{2}.$
We will apply Properties of the Dot Product (boxed on p. 709)
$({\bf u+v})\cdot{\bf u}={\bf u}\cdot({\bf u+v}) \quad$ ... property 1
=${\bf u}\cdot{\bf u} +{\bf u}\cdot{\bf v}\quad$ ... property 3
= $|{\bf u}|^{2} +{\bf u}\cdot{\bf v}\quad$ ... property 4
= $|{\bf v}|^{2} +{\bf u}\cdot{\bf v}\quad$ ... given in the problem
=${\bf v}\cdot{\bf v} +{\bf u}\cdot{\bf v}\quad$ ... property $4$
$={\bf v}\cdot({\bf u+v}) \quad$ ... property $3$
$=({\bf u+v})\cdot{\bf v} \quad$ ... property 1
The numerators are equal as well.
Thus, $\theta_{1}=\theta_{2}.$
