Find the area of ​the non-convex pentagonal region AEIDCA.

Question

For reference:

In triangle $ABC$, the interior angle bisectors $AD$ and $CE$ are drwan which intersect at $I$. If $\angle B = 60^\circ$ and the area of triangular region $AIC$ is $A$, calculate the area of ​​the non-convex pentagonal region $AEIDCA$. (Answer: $2A$)

My progress:

I found many relationships but the main one is missing

$\theta+\alpha=60^\circ$

$\frac{S[ABC]}{S[ABD]}=\frac{AC}{AB}$

$\frac{S[ACE]}{S[BCE}=\frac{AC}{BC}$

$\frac{A}{S[AEI]}=\frac{AC}{AE}$

$\frac{A}{S[CDI]}=\frac{AC}{CD} $

$\triangle ABC: \frac{AB}{BD}=\frac{AC}{CD}$

$\triangle AEC: \frac{AE}{EI}=\frac{AC}{CI}$

$\triangle CDA: \frac{CD}{DI}=\frac{AC}{AI}$

...??

Answer 1

Let $X\in AC$ such that $\angle AIX=\angle XIC=60^\circ$. Observe that $\angle EIA=180^\circ -120^\circ =60^\circ =\angle XIA$, and by definition $\angle IAE=\alpha =\angle IAX$. Therefore, from angle-side-angle criterion we get that the triangles $AIE$ and $AIX$ are congruent. Hence $S[AIX]=S[AIE]$. Analogously, we find that $S[CIX]=S[CID]$.
Finally\begin{align*}S[AEIDC] & =S[AIE]+S[AIX]+S[CID]+S[CIX] \\ & =2\cdot S[AIX]+2\cdot S[CIX] \\ & =2\cdot (S[AIX]+S[CIX]) \\ & =2\cdot S[AIC] \\ & =2\cdot A \end{align*}as wanted.