Find the area of ​$​PTHQ$ region in the figure below

Question

For reference:

On the side $AC$ of a triangle $ABC$, the points $P$ and $Q$ are marked such that $AQ = PC$ $(QC > CP)$. Then $PT$ and $QR$ were traced in parallel to $AB$ ($T$ and $R$ in $BC$). If the area of ​​the region $AHR$ is $"S"$ calculate the area of ​$​PTHQ$ region. $(QR \cap AT=H)$ (Answer: $S$)

My progress:

$[QHTP] = [ACT]-[AHQ]-[CPT]$

$\displaystyle \frac{ATP}{AHQ}=\frac{AT.AP}{AH.HQ}$

$\displaystyle \frac{S}{[AHQ]}=\frac{HR}{HQ}$

$\displaystyle \frac{[CPT]}{[APT]}=\frac{CP}{AP}$

$\displaystyle \triangle ABC \sim \triangle CPT: \frac{CT}{CB}=\frac{CP} {AC}=\frac{PT}{AB}$

$\displaystyle \triangle CRQ \sim \triangle CPT: \frac{CT}{CR}=\frac{CP}{CQ}=\frac{PT}{PQ}$

$\displaystyle \frac{[ART]}{[AHR]}=\frac{AH}{HT}$

...?

Answer 1

$$ CQ:CP=QR:PT\implies AP:AQ=QR:PT \implies AP\cdot PT=AQ\cdot QR. $$ But the last equality implies that triangles $APT$, $AQR$ have the same area.

Answer 2

Let area $\triangle ABC=\Delta$, area $AHQ=Z$, and area $PTHQ=X$

Labelling the sides of the triangle $ABC$ in the usual way, let $AQ=\lambda b=PC$, in which case, $CT=\lambda a=BR$ due to similar triangles.

Then we have the following:

area $\triangle PTC=\lambda^2\Delta$

area $\triangle QRC=(1-\lambda)^2\Delta$

Therefore area $ABRQ=\Delta-(1-\lambda)^2\Delta=(2\lambda-\lambda^2)\Delta$

Now area $\triangle ABR=\lambda\Delta$

So $$S+Z=(2\lambda-\lambda^2)\Delta-\lambda\Delta=(\lambda-\lambda^2)\Delta$$

However, area $\triangle ATC=\lambda\Delta$

So we have $$\lambda\Delta=Z+X+\lambda^2\Delta$$ $$\implies Z+X=(\lambda-\lambda^2)\Delta=S+Z\implies X=S$$