Let ABCD be a parallelogram. Let M be the midpoint of line AB and N be the midpoint of line AD. Diagonal BD intersects line CM and line CN at P and

With GeoGebra  .....

[ and the "Snipping Tool"  ]

<CBD = <MBQ

<DQC  =<MQB

thus ΔDQC  ~ ΔBQM

And since DC = 2MB.....by similarity  DQ  = 2BQ

And

<CBD = <PDN

<BPC  =<NPD

thus ΔBPC  ~ ΔDPN

And since BC = 2DN.....by similarity  BP  = 2PD

So

BD  = DQ  + BQ                BD = BP  + PD

BD = 2BQ  + BQ               BD = 2PD + PD

BD  = 3BQ                        BD = 3PD

therefore

BQ  = PD

And

Since DQ  = 2BQ  then  DQ  = 2PD ..... and......

PD + PQ = 2PD   subtract PD from each side

PQ  = PD ...... so.....BQ = PD  = PQ

BD = PD + PQ + BQ

BD  = PQ  + PQ  + PQ

BD  = 3PQ

So

PQ / BD  = 1/3

Here's a pic :