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
$$
\frac{\langle k\omega,k\omega+2\rho\rangle}{2(h^\vee+k)}
\,\,\stackrel{(1)}=\,\,
\frac{\| k\omega+\rho\|^2-\|\rho\|^2}{2(h^\vee+k)}
\,\,\stackrel{(2)}=\,\,
\frac{k\langle\omega,\rho\rangle}{h^\vee}
\,\,\stackrel{(3)}=\,\,
\frac{k}{2}\|\omega\|^2.
$$

(1) Easy.
(2) By the lemma below, 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 $\frac{\langle\omega,\rho\rangle}{h^\vee}$, from which the second equality follows.
(3) Again by the lemma, the point $\frac \rho {h^\vee}$ is equidistant to $0$ and to $\omega$. It is on the bisecting hyperplane of the segment $[0,\omega]$, and so $\langle\omega,\frac\rho{h^\vee}\rangle=\langle\omega,\frac\omega2\rangle=\frac12\|\omega\|^2$.<br> QED<br><br>

Lemma: Let $\omega$ be as above. Then $\| \rho - h^\vee\omega\| = \|\rho\|$.

<i>Proof:</i> 
Let $\mathcal A$ be the Weyl alcove. Its isometry group is the automorphism group of the extended Dynkin diagram $\Gamma^e=\Gamma\cup \{\circ\}$.
The vertices of $\Gamma$ whose mark is $1$ are exactly those in the $Aut(\Gamma^e)$ orbit of the extra vertex $\circ$.
Recall that $\rho$ is the unique weight in the interior of $h^\vee \mathcal A$.
Finally, the vertices $0$ and and $h^\vee\omega$ are in the same orbit under the isometry group of $h^\vee \mathcal A$, and therefore equidistant to $\rho$. QED