Prove that $\angle DAP=\angle CAB$ in a parallelogram $ABCD$

Question

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.

Answer 1

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

$$\frac{BK}{DL}=\frac{\left[\ell_1+\frac{\ell_2a}{r(a-\ell_2)}\right]\cdot\frac{r(a-\ell_2)}{a}}{\frac{a-r\ell_1}{ra}\ell_2+\ell_1}=r=\frac{\ell_1}{\ell_2}=\frac{DC}{BC}$$