First Solution. The key observation is the following Lemma.
Lemma. *Segment* $D_1Q$ *is a diameter of circle* $\omega$.
*Proof.* Let $I$ be the center of circle $\omega$, that is, $I$ is the incenter of triangle $ABC$. Extend segment $D_1I$ through $I$ to intersect circle $\omega$ again at $Q'$, and extend segment $AQ'$ through $Q'$ to intersect segment $BC$ at $D'$. We show that $D_2 = D'$, which in turn implies that $Q = Q'$, that is, $D_1Q$ is a diameter of $\omega$.

Let $\ell$ be the line tangent to circle $\omega$ at $Q'$, and let $\ell$ intersect segments $AB$ and $AC$ at $B_1$ and $C_1$, respectively. Then $\omega$ is an excircle of triangle $AB_1C_1$. Let $\mathbf{H}_1$ denote the dilation with center $A$ and ratio $AD'/AQ'$. Since $\ell \perp D_1Q'$ and $BC \perp D_1Q$, $\ell \parallel BC$. Hence, $AB/AB_1 = AC/AC_1 = AD'/AQ'$. Thus, $\mathbf{H}_1(Q') = D'$, $\mathbf{H}_1(B_1) = B$, and $\mathbf{H}_1(C_1) = C$. It also follows that the excircle $\Omega$ of triangle $ABC$ opposite vertex $A$ is tangent to side $BC$ at $D'$.

It is well known that
$$CD_1 = \frac{1}{2}(BC + CA - AB). \qquad (1)$$
We compute $BD'$. Let $X$ and $Y$ denote the points of tangency of circle $\Omega$ with rays $AB$ and $AC$, respectively. Then by equal tangents, $AX = AY$, $BD' = BX$, and $D'C = YC$. Hence,
$$\begin{align*} AX &= AY = \frac{1}{2}(AX + AY) \\ &= \frac{1}{2}(AB + BX + YC + CA) \\ &= \frac{1}{2}(AB + BC + CA). \end{align*}$$
It follows that
$$BD' = BX = AX - AB = \frac{1}{2}(BC + CA - AB). \qquad (2)$$
Combining (1) and (2) yields $BD' = CD_1$. Thus,
$$BD_2 = BD_1 - D_2D_1 = D_2C - D_2D_1 = D_1C = BD',$$
that is, $D' = D_2$, as desired. ■

Now we prove our main result. Let $M_1$ and $M_2$ be the midpoints of segments $BC$ and $CA$, respectively. Then $M_1$ is also the midpoint of segment $D_1D_2$, from which it follows that $IM_1$ is a midline of triangle $D_1QD_2$. Hence,
$$QD_2 = 2IM_1 \qquad (3)$$
and $AD_2 \parallel M_1I$. Similarly, we can prove that $BE_2 \parallel M_2I$.

Let $G$ be the centroid of triangle $ABC$. Thus, segments $AM_1$ and $BM_2$ intersect at $G$. Define transformation $\mathbf{H}_2$ as the dilation with center $G$ and ratio $-1/2$. Then $\mathbf{H}_2(A) = M_1$ and $\mathbf{H}_2(B) = M_2$. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since $AD_2 \parallel M_1I$ and $BE_2 \parallel M_2I$, $\mathbf{H}$ maps lines $AD_2$ and $BE_2$ to lines $M_1I$ and $M_2I$, respectively. It also follows that $\mathbf{H}_2(P) = I$ and that
$$\frac{IM_1}{AP} = \frac{GM_1}{AG} = \frac{1}{2}$$
or
$$AP = 2IM_1. \qquad (4)$$
Combining (3) and (4) yields
$$AQ = AP - QP = 2IM_1 - QP = QD_2 - QP = PD_2,$$
as desired.

Second Solution. From the Lemma, we have
$$\frac{AQ}{AD_2} = \frac{r}{r_a},$$
where $r$ and $r_a$ are the radii of circles $\omega$ and $\Omega$, respectively. Note that
$$r(AB + BC + CA) = 2[ABC]$$
and that
$$\begin{aligned} r_a(AB + AC - BC) &= 2[I_aAB] + 2[I_aAC] - 2[I_aBC] \\ &= 2[I_aBAC] - 2[I_aBC] = 2[ABC], \end{aligned}$$
where $I_a$ is the center of $\Omega$ and $[\mathcal{R}]$ is the area of region $\mathcal{R}$. Thus,
$$\frac{AQ}{AD_2} = \frac{AB + AC - BC}{AB + BC + CA}. \qquad (5)$$

Applying Menelaus's Theorem to triangle $AD_2C$ and line $BE_2$ gives
$$\frac{AP}{PD_2} \cdot \frac{D_2B}{BC} \cdot \frac{CE_2}{E_2A} = 1,$$
or
$$\frac{AP}{PD_2} = \frac{BC \cdot E_2A}{D_2B \cdot CE_2} = \frac{BC \cdot CE_1}{CD_1 \cdot AE_1} \\ = \frac{BC}{AE_1} = \frac{2BC}{AB + AC - BC}.$$
Hence,
$$\frac{AD_2}{PD_2} = 1 + \frac{AP}{PD_2} = \frac{AB + AC + BC}{AB + AC - BC},$$
or
$$\frac{PD_2}{AD_2} = \frac{AB + AC - BC}{AB + AC + BC}. \qquad (6)$$
The desired result now follows from (5) and (6).