Find the area of ​the shaded region PQGF 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

 A: Unbelievable setup! I had great joy of simply angle chasing and discovering and I wish to share with you.
$\triangle BFG$ is the reference triangle. $FP,GQ$ are given its external angle bisectors; let them meet at excenter $E$. Drop perpendiculars from $E$ onto $AB,BC$ and complete the square $ABCE$. Lets observe that $AEC$ is exact copy of $ABC$, so $\angle PEQ = 45^\circ$.

Now the fun begins. $\angle PEG=45^\circ=\angle PCG$, hence $PECG$ is cyclic. $\angle EPG$ being opposite to $\angle ECG$ is a right angle. Similarly $AFQE$ is cyclic, making $\angle FQE$ another right angle. Observe that $FG$ subtends $90^\circ$ at $B,P,Q$. As a result, $B,F,P,Q,G$ lie on same circle $!!$ Its center is $D$, midpoint of $FG$.

Now $DP=DF=$ radius of circle, $\angle DPF = \angle DFP = \angle PFA$ implying that $DP \parallel BA$. Hence $\triangle BPF$ and $\triangle BDF$ have same area. Take away their common area and we get $\triangle FRP$ and $\triangle BRD$ have same area. Similarly $DQ \parallel BC$ resulting in $\triangle GSQ$ and $\triangle BSD$ having same area. Adding the common area of $PRSQ$ to these, we see $FPQG$ and $\triangle BPQ$ have same area.
We do angle chasing one more time to find some lengths. $\angle BPC =$$ \angle PAB + \angle ABP = \angle PBQ + \angle ABP = \angle ABQ $. Therefore $\triangle ABQ \sim \triangle CPB$ by $AA$ similarity. So
$$\frac{AB}{PC}=\frac{AQ}{BC} \Rightarrow AB^2=9\times 8 \Rightarrow AB=6\sqrt{2} \Rightarrow AC=12$$
From this it is found $AP=3, PQ=5, QC=4 \, (!)$ Consequently, $$[FPQG]=[BPQ]=\frac{PQ}{AC}\times [ABC]=\frac{5}{12}\times 36=15 \quad \square$$
A: 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 $[PFGQ] = 15$.
A: While at the first sight it seems a setup with no leads, one can start seeing multiple ways to approach this - for example, use the fact that $PFGQ$ is cyclic or use construction that sets up a harmonic range on segment $AC$ or using the fact that incenter and excenter of $\triangle BGF$ are both on the perpendicular bisector of $AC$. Now all of them may not lead to a solution but I will leave them for you to pursue. Here is a solution that I propose -

We know that $\alpha + \theta = 135^\circ$. Reflect point $B$ about $AC$. Now using symmetry,  $\angle PNQ = 45^\circ$. But also note that $BN$ is angle bisector of $\angle B$ in $\triangle FBG$ and $FP$ and $GQ$ are external angle bisectors, so they must meet at a point on $BN$ and $FG$ should make angle of  $45^\circ$ at the point considering $\angle GFN$ and $\angle FGN$ add to $135^\circ$. In other words, that point must be $N$. Using similarity of  $\triangle CNG$ and $\triangle MBP$ and as $BM:CN = 1:\sqrt2$, $NG = \sqrt2 BP, CG = \sqrt2 PM$ and similarly, $NF = \sqrt2 BQ, AF = \sqrt2 MQ$. So we conclude $FG = \sqrt2 PQ$,  $S_{FNG} = 2 \cdot S_{PBQ}$ and as $\triangle PBQ \cong PNQ$, $S_{PFGQ} = S_{PBQ}$.
$BF = AB - AF = \sqrt2 CM - \sqrt2 MQ = \sqrt2 CQ$ and we also get $BG = \sqrt2 AP$
Applying Pythagoras in $\triangle FBG$,
$2 (AP^2 + CQ^2) = 2 PQ^2 \implies AP^2 + CQ^2 = PQ^2$. Now, $AC = AQ + CP - PQ = 17 - PQ$,
$AP = 8 - PQ, CQ = 9 - PQ$
So $(8-PQ)^2 + (9-PQ)^2 = PQ^2$ and we get $PQ = 5, AC = 12, BM = AM = 6$.
That leads to $~S_{PFGQ} = 15$.
