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 $k=1$)
Remark: The right left hand side $\frac{\langle k\omega,k\omega+2\rho\rangle}{2(h^\vee+k)}
$ is the minimal energy of the positive energy representation of the affine Lie algebra $\hat{\mathfrak g}$ of level $k$ and highest weight $k\omega$.
Proof: 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 checks that it vanishes at zero and that its derivative at zero is $\frac{\langle\omega,\rho\rangle}{h^\vee}$.
(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 $\langle\omega,\frac\rho{h^\vee}\rangle=\langle\omega,\frac\omega2\rangle=\frac12\|\omega\|^2$.
QED
Lemma: Let $\omega$ be as above. Then $\| \rho - h^\vee\omega\| = \|\rho\|$.
Proof: 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 Dynkin 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 $h^\vee \mathcal A$. To finish the argument, note that 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