Suppose $\omega_1$ is tangent to $AB$, $AD$ at $P_1$, $Q_1$ respectively and $\omega_2$ is tangent to $CD$, $BC$ at $P_2$, $Q_2$ respectively.

Lemma. Let a circle $\omega$ be tangent to the half-lines $BA$, $BC$ at $P$, $Q$ respectively. $\omega$ is tangent to $\omega_1$ if and only if $\sqrt{BP} = \sqrt{AB} \pm \sqrt{AP_1}$ for some choice of the sign.

Proof. $PP_1$ is the common external tangent of $\omega$ and $\omega_1$. So $\omega$, $\omega_1$ are tangent if and only if $PP_1 = 2\sqrt{rr_1}$, where $r$, $r_1$ are the radii of $\omega$, $\omega_1$ respectively. If we let $\angle ABC = 2\alpha$, then $r = BP \cdot \tan\alpha$ and $r_1 = AP \cdot \cot\alpha$. So $2\sqrt{rr_1} = 2\sqrt{BP \cdot AP_1}$. So $\omega$, $\omega_1$ are tangent if and only if $AB - AP_1 - BP = \pm 2\sqrt{BP \cdot AP_1}$ and the claim follows. $\square$

By assumption, there is a circle $\omega$ tangent to the lines $DA$, $DC$ at $Q$, $P$ respectively and externally tangent to $\omega_1$, $\omega_2$. It is easily seen that $\omega$ should be inside the angle $\angle ADC$. So, by the lemma we have:
$$\sqrt{DQ} = \sqrt{AD} \pm \sqrt{AQ_1} \\ \sqrt{DP} = \sqrt{CD} \pm \sqrt{CP_2}$$
for some choice of the signs. We have $DP = DQ$, $AQ_1 = AP_1$, $CP_2 = CQ_2$, $AD = BC$ and $CD = AB$. So by the above equations we get
$$\sqrt{BC} \pm \sqrt{AP_1} = \sqrt{AB} \pm \sqrt{CQ_2} \Rightarrow \sqrt{BC} \pm \sqrt{CQ_2} = \sqrt{AB} \pm \sqrt{AP_1}.$$

Let $\omega'$ be a circle tangent to the half-lines $BA$, $BC$ at $P'$, $Q'$ respectively, such that both sides of this equation are equal to $\sqrt{BP'}$ (note that the value is positive). So $\omega'$ is tangent to $\omega_1$, $\omega_2$ by the lemma and the assertion is proved. $\square$