Distance from meeting the bisectors to the side of the quadrilateral In a quadrilateral $ABCD,\;AC$ and $BD$ are bisectors of $\angle BAD$ and $\angle ADC$. If $AC$ intersects $BD$ at $P,\;AB=6,\;CD=3$ and $\angle APD= 135º$, calculate the distance from $P$ to $AD$.
I even designed it but I did not find the solution;



Let $BE = b, EA = a, BA=c$ and the distance from $P$ to the three edges of $\triangle ABE$ be $r$. Then
$$r = \frac{b+a-c}{2}\tag{1}.$$
Since $BC$ is the angle bisector of $\angle EBA$,
$$\frac{a-6}{b} = \frac{6}c,\text{ or } (a-6)c = 6b\tag{2}.$$
Similarly
$$\frac{b-3}{b} = \frac3c,\text{ or }(b-3)c = 3a\tag{3}.$$
Take $2(3)-(2)$ we have
$$c(2b-a) = 6(a-b)\tag{4}.$$
On the other hand,
$$\color{red}{\frac{r}{a-6}} + \color{blue}{\frac{r}{b}} = \color{red}{\frac{BP}{BC}} + \color{blue}{\frac{PC}{BC}} = 1$$
so $$r(a+b-6) = (a-6)b.\tag{5}$$
And similarly
$$r(a+b-3) = (b-3)a.\tag{6}$$
Taking $(6)-(5)$ we obtain
$$ r = 2b - a.\tag{7}$$
Replace $r$ from $(7)$ into $(1)$ we get
$$a+b-c = 2(2b-a),\text{ or } c = 3(a-b).\tag{8}$$
Now replace $(7)$ and $(8)$ into $(4)$ we get $r=2$.
