Find the ratio of the lengths PX and ZY.

Question

Side AB of a parallelogram ABCD is drawn out to point P so BP 1/2 AB and side CD is extended to point Q so CQ = CD. The line PQ meets the diagonal AC at Z and the sides AD and BC at X and Y.

(b) Find the ratio of the lengths PX and ZY.

Now, usually I would post my attempts at this question as per the recommendations on MSE, but this time I have no idea where to start with (b). I have already tried drawing a diagram and assigning variables to ZY etc. but I was unable to continue. Because I think this still might be the way to get the answer, a step by step solution would be appreciated.

Answer 1

First step is of course to draw it

$DQY$ is similar to $APY$ and $\frac{2}{3}=\frac{DQ}{AP}=\frac{DY}{AY}$

$APY$ is similar to $BPX$ and $\frac{3}{1}=\frac{AP}{BP}=\frac{AY}{BX}$

$ABCD$ is a parallelogram, $AY+DY=BX+CX$

$AYZ$ is similar to $CXZ$ and $\frac{3}{4}=\frac{AY}{CX}=\frac{YZ}{XZ}$

Finally $APY$ is similar to $BPX$ again and $\frac{3}{1}=\frac{PY}{PX}$

I use variables $m$ and $n$ to make it easier. $\frac{PX}{YZ}=\frac{7}{6}$

Answer 2

$$PB:BA=1:2,CD:DQ=2:2\implies $AP:CQ=3:4$$

$$\implies AY:YC=3:4 \implies AV:WC=3:4\implies AV:VB=3:4$$

Therefore $AV:VB:BP=6:8:7$ and $PX:YZ=BP:AV=7:6$