It turns out that it's easier to prove the following generalization:

Let $\mathfrak g$ be a simple Lie algebra (not necessarily simply laced), 
let $\omega$ be a fundamental weight whose Dynkin mark is 1, and let $k$ be any number.
Then we have

$$
\frac{\langle k\omega,k\omega+2\rho\rangle}{2(h^\vee+k)}
=
\frac{k}{2}\|\omega\|^2.
$$

(The original question is the case $k=1$)
  
<hr>
<i>Proof:</i>
We'll show that
$$
\frac{\langle k\omega,k\omega+2\rho\rangle}{2(h^\vee+k)}
=
\frac{\| k\omega+\rho\|^2-\|\rho\|^2}{2(h^\vee+k)}
=
\frac{k\langle\omega,\rho\rangle}{h^\vee}
=
\frac{k}{2}\|\omega\|^2.
$$

(1) The first equality is easy.<br>
(2) Let $\mathcal A$ be the Weyl alcove. Recall that $\rho$ is the unique weight in the interior of $h^\vee \mathcal A$.
The vertices $0$ and and $h^\vee\omega$ are in the same $Aut(\mathcal A)$-orbit, and are therefore equidistant to $\rho$.
It follows that the numerator $\| k\omega+\rho\|^2-\|\rho\|^2$ vanishes when $k=-h^\vee$. The function $k\mapsto \frac{\| k\omega+\rho\|^2-\|\rho\|^2}{2(h^\vee+k)}$ is therefore linear. One easily computes that its derivative at zero is $\frac{\langle\omega,\rho\rangle}{h^\vee}$, and the second equality follows.<br>
(3) For the third equality one notes that, by symmetry reasons, the vector $\frac \rho {h^\vee}$ is on the bisecting hyperplane of the segment $[0,\omega]$. Therefore $\langle\omega,\frac\rho{h^\vee}\rangle=\langle\omega,\frac\omega2\rangle=\frac12\|\omega\|^2$.<br> QED