Let $M$ the intersection of the bisector of $\angle ACD$ with $AD$, and $N$ the intersection of the bisector of $\angle ACB$ with $AB$ and let $K=AC\cap NM$. 1- $\frac{CD}{AC}=\frac{MD}{AM}$ and $\frac{CB}{AC}=\frac{NB}{AN}$, using $BC=CD$ we get $\frac{NB}{AN}=\frac{MD}{AM}$ and then, $NM||BD$ 2- easy angle chasing to proof that $MC$ bisects $\angle KMD$ and $NC$ bisects $\angle KNB$ 3- Conclude that KNBC and KMDC are kites and that $\astrosun I_1$, $\astrosun I_2$ are their incircles.

Let the external tangent of $\astrosun I_1$ , $\astrosun I_2$ closer to $A$ intersect $AC$ at $P$. This solution is based on the observation that $P$ is the Incenter of $\triangle ACD$. Since easily $BC=CD$ by simple angle chasing, $P$ is the incenter iff $PC=BC=CD$ . This follows from property of Incenter and false positioning. For the proof we use false positioning. Let $Q$ be the point on $AC$ such that $QC=BC=CD$. Construct points $E$ and $F$ on $AB$,$AD$ such that $EQ$,$FQ$ is tagent to $\astrosun I_1$ , $\astrosun I_2$ respectively. Then Quadrilateral $QCBE$ has an incircle, therefore $QC+BE=CB+EQ$ , $BE=EQ$. $QCBE$ is a kite, similarly for $QCDF$. Finally, $\angle EQF= \angle EQC+ \angle FQC= \angle ABC+\angle ADC= 180$. So the other tangents (besides $AC$) from $Q$ to the two incircles, $EQ$ and $FQ$ is the same line. Therefore $Q$ belongs to a common external tangent. To complete the proof, $BD$ is parallel to the tangent at $C$ of $\astrosun ABCD$ from simple angle chasing. The conclusion follows readily from that fact that $EQCB$ and $FQCD$ are kites.

Apply complex numbers with $(ABCD)$ as the unit circle, such that $A=a^2,B=b^2,C=c^2,D=d^2,i_1=-(ab+bd+da),i_2=-(ad+dc+ca)$. Then $\frac{(i_1-i_2)^2}{(b^2-c^2)(a^2-d^2)}=\frac{(dc+ca-ab-bd)^2}{(b+c)(b-c)(a+d)(a-d)}=\frac{(a+d)(b-c)}{(b+c)(a-d)}\in\mathbb{R}$, as desired.

@hector explain your 2 statement

Let $CI_2$ and $C_1$ meet $AD$ and $AB$ at $F$ and $E$ and Let $S$ be reflection of $B$ across $CI_1$. Note that $\frac{BE}{EA} = \frac{BC}{CA} = \frac{DC}{CA} = \frac{DF}{FA} \implies EF || BD$. Note that $\angle BES = 2\angle BEC = \angle BCA + 2\angle BAC = \angle 180 - \angle ABD = \angle 180 - \angle AEF$ so $S$ lies on $EF$ so $EF$ is reflection of $AB$ across $CI_1$ so $EF$ is tangent to $\astrosun I_1$. with same approach we have $EF$ is reflection of $AD$ across $CI_2$ so $EF$ is tangent to $\astrosun I_2$. we're Done.

$ABCD$ is a cyclic quadrilateral, and $\angle BAC = \angle DAC$. $\odot I_1$ and $\odot I_2$ are the incircles of $\triangle ABD$ and $\triangle ADC$ respectively. Prove that one of the common external tangents of $\odot I_1$ and $\odot I_2$ is parallel to $BD$.