Find the area of ​the shaded region: in the figure

For reference:

In the figure, calculate the area of ​​the shaded region if $ML=LN, PM=a, NQ=b, AC=c.$ (Answer: $\frac{(a+b)c}{4}$)

My progress: Let $ML=LN=BL=x.$

$S_{ALC} = S = S_{APQC}-S_{ALP}-S_{CLQ}\implies=\frac{(CQ+AP)(a+b+2x)}{2}-\frac{AP.(a+x)}{2}-\frac{CQ(b+x)}{2}$

$\therefore S =\frac{AP}{2}(b+x)+\frac{CQ}{2}(a+x)$

$S_{ABCL} = \frac{c\cdot x}{2}$

First notice that $AB$ and $CB$ are angle bisectors of $\angle CAP$ and $\angle ACQ$ respectively. Next, drop a perp from $M$ to $AC$. Say it meets $AC$ at $H$.

As $\triangle MAH \cong \triangle MAP, MH = a$. Similarly drop a perp $NG$ to $AC$ and show $NG = b$.

As $L$ is midpoint of $MN$, $ \displaystyle LE = \frac{a+b}{2}$.

That gives area of the shaded region as $~~\dfrac{(a+b) \cdot c}{4}$

Why $AB$ and $CB$ are angle bisectors of $\angle CAP$ and $\angle ACQ$.

As $L$ is midpoint of hypotenuse of right triangle $ \triangle MNB$, it is its circumcenter. So $BL = ML = LN$. So $\angle BMN = \angle ABE = 90^\circ - \angle A = \angle C$ and then $\angle BNM = \angle A$. That leads to $\angle PAB = \angle A$ and $\angle QCB = \angle C$