What's the measure of the segment $QC$ in the triangle $ABC$ below?

Question

For reference: the acute triangle $ABC$ are traced the bisectors $AQ$ and $CP$ ($Q$ in $BC$ and $P$ in $AB$). Calculate $QC$, if $AP=2, PB=3$ and $BQ=4$.

My progress:Here's the drawing and relationships I found

Bissector Th.$\triangle ACB - PC$:

$\frac{BC}{3}=\frac{AC}{2}\implies BC =\frac{3AC}{2}$

Bissector Th.$\triangle ACB-AQ$:

$\frac{BQ}{5}=\frac{CQ}{AC}$

$PD \parallel AD \implies \frac{5}{4} = \frac{2}{DQ}\\ \therefore DQ = 1,6$

and

$BD = 1,4$

I tried to trace the HQ parallel but it had no effect..there is still some similarity to finish...

Correft drawing

Answer 1

We can apply angle bisector theorem twice to the triangle $\small ABC$, with respect to two bisectors.

$\frac{BP}{PA}=\frac{BC}{AC}\implies\frac32=\frac{4+QC}{AC}$

$\frac{BQ}{QC}=\frac{AB}{AC}\implies\frac{4}{QC}=\frac5{AC}$

Now eliminate $\small AC$.