∵ABCD is a parallelogram.

∴AD∥PE∥BC and AD = BC.

∫PQ/PR = 3/4

∴eq/qc=pq/qr=pq/(pr-pq)=3/(4-3)=3/ 1

Let DE/EQ/QC=x/3/ 1 and x > 0, then

PE/AD=EQ/QD=EQ/(DE+EQ)=3/(3+x)

= PE/BC = DE/DC = DE/(DE+EQ+QC)= x/(x+3+ 1)= x/(x+4)

That is, 3/(3+x)=x/(x+4)

3x+ 12=3x+x？

x？ = 12

∵ x > 0

∴x=2√3

∴ad/br=dp/pb=de/ec=de/(eq+qc)=x/(3+ 1)=2√3/4=√3/2