Showing $(sp-bc)(sq-bc)=bc(s-b)(s-c)$, for $s$ the semiperimeter of a triangle, with $p$ and $q$ determined by a line tangent to the incircle I've the following construction, as shown in the figure. The line $QP$ is tangent to the incircle of $\triangle ABC$. The triangle has side lengths given by $a,b,c$. I am trying to prove the result that 
$$(sp-bc)(sq-bc)=bc(s-b)(s-c)$$
where $s$ is the semiperimeter of the triangle, $p = AP$, and $q = AQ$.
 
I've tried applying the Law of Cosines separately for $\triangle APQ$ and $\triangle ABC$, as they share the angle $\angle CAB$, but I've had no luck. 
 A: Let the usual $x=s-a=AD=AE$.  We know $PQ=PE+QD=2x-p-q$, so the cosine rule gives
$$
\frac{(b+c)^2-a^2}{2bc}=1+\cos A=\frac{(p+q)^2-(2x-p-q)^2}{2pq}=\frac{2x(p+q-x)}{pq}.
$$
Clearing denominators, we have
$$
pq((b+c)^2-a^2)=2bc\cdot 2x(p+q-x)
$$
i.e.
$$
pqs-bc(p+q)=-bcx.
$$
Multiply by $s$ and add $b^2c^2$:
$$
pqs^2-bc(sp+sq)+b^2c^2=bc(bc-xs).
$$
The LHS is $(sp-bc)(sq-bc)$.  In the RHS, recall $x=s-a=b+c-s$, so
$$
bc(bc-xs)=bc[bc-(b+c-s)s]=bc(s-b)(s-c)
$$
as desired.
A: Consider the following:

Due to pairs of congruent triangles, we can write
$$|\triangle ABC| = |\triangle APQ| + 2\,|\triangle BIP| + 2\,|\triangle CIQ| \tag{1}$$
where $I$ is the incenter of $\triangle ABC$. Thus,
$$\begin{align}
|\triangle ABC| - |\triangle APQ| &= r \left(\;(c - p) + (b - q)\;\right)
&\left(\;|\triangle BIP|=\frac12r(c-p), \text{etc}\;\right) \tag{2}\\[6pt]
s\left(\;|\triangle ABC| - |\triangle APQ|\;\right) &= |\triangle ABC|\, (b+c-p-q) &\left(\;rs=|\triangle ABC|\;\right)\tag{3}\\[6pt]
s\left(\;\frac12b c \sin A - \frac12 p q \sin A\;\right) &= \frac12 b c \sin A\, (b+c-p-q) \tag{4}\\[6pt]
s\left(\;b c - p q \;\right) &= b c\, (b+c-p-q) \tag{5}\\[6pt]
\end{align}$$
The reader readily verifies that $(5)$ is, for $s \neq 0$, equivalent to the desired relation. $\square$
