TITLE: Find the area of ​the shaded region PQGF
QUESTION [6 upvotes]: For reference: Calculate the area of ​​the shaded region.
AQ = 8m; PC = 9 m (Answer:$15m^2$)

My progress:
FP is angle bissector $\triangle AFG$
GQ 1s angle bissector $\triangle FGC$
$\frac{S_{ABP}}{S_{ABQ}}=\frac{AP}{8}\\
\frac{S_{CBQ}}{S_{BCP}}=\frac{CQ}{9}\\
\frac{S_{ABQ}}{S_{BCP}}=\frac{8}{9}\\
S_{FGPQ}=S_{ABC}-S_{BFG}-S_{CGQ}\\
2\alpha+2\theta = 270 \implies \alpha +\theta = 135^o\\
S_{ABC}=BH^2=AH^2=AC^2$
tried to draw some auxiliary lines like the other question but it didn't solve

REPLY [3 votes]: I missed seeing a simpler solution when I looked at the problem a few days back. But as they say, better late than never.

First notice that $\angle APF = \alpha, ~$ given $~\alpha + \theta = 135^\circ$.
Similarly $\angle CQG = \theta$
Now as in $\triangle BFG$, external angle bisectors of $\angle F$ and $\angle G$ meet at a point on the perpendicular bisector of $AC$ and so does the reflection of point $B$ about $AC$ and segment $PQ$ subtends angle of $45^\circ$ at both points, both points must be the same point.
That leads to $\angle BPQ = \alpha~$ and $~\angle BQP = \theta$.
Now notice, $\triangle BAQ \sim \triangle PCB~ \implies \displaystyle \frac{AB}{AQ} = \frac{CP}{BC}$
So, $AB = BC = 6 \sqrt2, AC = 12$. That also leads to $AP = 3, CQ = 4, PQ = 5$.
Now from here, use the fact that $[PFGQ] = [PBQ]$ (shown in my other answer). Alternatively, notice $\triangle PAF \sim \triangle GCQ \sim \triangle BAQ$. That leads to $AF = 2 \sqrt2, BF = 4 \sqrt2, CG = 3 \sqrt2, BG = 3 \sqrt2$
$[PAF] = 3, [GCQ] = 6, [BFG] = 12$. Adding these areas and subtracting from $[ABC] = 36$, we get $[PFGQ] = 15$.