nickthegreek wrote:

A different approach:

Your solution is actually identical to hatchguy's. Take a look:

hatchguy wrote:

The concurrent with $AC$ bisectors condition is equivalent to $\frac{AB}{BC}=\frac{AD}{DC}$ $=> \frac{AD*BC}{AB*DC} =1$

Since $\angle DAB+ \angle DCB= 180$ and $\angle DCP+ \angle DCB= 180$ we have $\angle DAB= \angle DCP$ .

Therefore we have that $ARD$ is similar to $CPD$ and therefore $\frac{AR}{PC}=\frac{AD}{DC}$ . (1)

By Menelaus theorem for transversal $CQA$ in triangle $BRP$ we obtain:

$\frac{AB*RQ*PC}{AR*PQ*BC}=1$ . Therefore $\frac{RQ}{QP}=\frac{AR*BC}{AB*PC}=\frac{AD*BC}{AB*DC}$ (because of (1)). The conclusion is now obvious.