Prove that $\angle DAP=\angle CAB$ in a parallelogram $ABCD$ Let $ABCD$ be a parallelogram, and let $K$ be on $BC$ and $L$ on $CD$ so that $BK\cdot BC=DL\cdot DC$. Let point $P$ be where $DK$ and $BL$ intersect. Prove that $\angle DAP=\angle CAB$ (angles $DAP$ and $CAB$ are equal).
I got that $ADL$ and $ABC$ are similar so it is enough to prove that $\angle LAP=\angle CAK$, but I think there is an better way to use the given statement.
 A: This is a solution using oblique coordinates. 

The main purpose is to show that the proof is entirely linear. This means that one can construct a synthetic geometry proof using only similarity of triangles, after drawing lines through $K,L,P$ parallel to the sides of the parallelogram.

Let's use $AB$ and $AD$ as the axes. So $A=(0,0)$, $B=(0,\ell_1)$, $D=(\ell_2,0)$ and $C=(\ell_2,\ell_1)$. Call $r:=\frac{\ell_1}{\ell_2}$.
The line $AC$ is $y=rx$. 

We want to prove that the line $y=\frac{1}{r}x$ is the locus of the points $P$. This is enough to get the conclusion.

Take a point $P:=(a,a/r)$ on this line.
Then the line from $D=(\ell_2,0)$ to $P=(a,a/r)$ is given by 
$$y=\frac{\frac{a}{r}-0}{a-\ell_2}x-\ell_2\frac{\frac{a}{r}-0}{a-\ell_2}$$
This intersects $BC=(y=\ell_1)$ at $K=\left(\left[\ell_1+\frac{\ell_2a}{r(a-\ell_2)}\right]\cdot\frac{r(a-\ell_2)}{a},\ell_1\right)$, i.e. $$BK=\left[\ell_1+\frac{\ell_2a}{r(a-\ell_2)}\right]\cdot\frac{r(a-\ell_2)}{a}$$
The line from $B=(0,\ell_1)$ to $P=(a,a/r)$ is given by 
$$y=\frac{\frac{a}{r}-\ell_1}{a-0}x+\ell_1$$
This line intersects $DC=(x=\ell_2)$ at $L=\left(\ell_2,\frac{a-r\ell_1}{ra}\ell_2+\ell_1\right)$, i.e.
$$DL=\frac{a-r\ell_1}{ra}\ell_2+\ell_1$$
Now we only need to divide
$$\frac{BK}{DL}=\frac{\left[\ell_1+\frac{\ell_2a}{r(a-\ell_2)}\right]\cdot\frac{r(a-\ell_2)}{a}}{\frac{a-r\ell_1}{ra}\ell_2+\ell_1}=r=\frac{\ell_1}{\ell_2}=\frac{DC}{BC}$$
A: With knowledge of projective geometry this is pretty easy problem. 

Let $a=AB$, $b = BC$, $BK = x$ and $DL = y$. Then we have $$y= {b\over a} x$$ 
so $L$ is lineary dependent on $K$, thus the transformation $$K\mapsto L$$ from a line $BC$ to a line $CD$ is projective. This one induces new projective transformation $$DK \mapsto BL$$ from a bundle of lines through $D$ to a bundle of lines through $B$. Since clearly $BD$ goes to it self, this transformation is perspective so the intersection point $P$ describes some line. 
Clearly when $K=\infty $ then $P=A$ and let $C$ maps to $E$ (so $DE = {b^2\over a}$). We see that $E$ must be on $CD$. All we have to see now is that the triangle $ADE$ is similar to the triangle $ABC$ which is simple since: $${DE\over DA} = {{b^2\over a }\over b} = {b\over a} = {BC\over BA}$$
