Let’s denote the midpoints of AB and DC as E and F respectively. Since E is the midpoint of AB, we have AE = EB. Similarly, since F is the midpoint of DC, we have DF = FC.

Since ABCD is a parallelogram, we know that AD

BC and AB

CD. Therefore, we can conclude that triangle AED is similar to triangle BFC by the AA similarity criterion.

Since AE = EB and DF = FC, we can conclude that the corresponding sides of triangles AED and BFC are proportional. Therefore, we have:

AD/BC = AE/DF

Since AD = BC (opposite sides of a parallelogram are equal in length), we can simplify the equation to:

1 = AE/DF

Multiplying both sides of the equation by DF, we get:

DF = AE

Since AF and EC intersect the diagonal BD at points P and Q respectively, we can conclude that triangle APB is similar to triangle DQC by the AA similarity criterion.

Since triangles APB and DQC are similar, we have:

BP/PD = AQ/QC

Since DF = AE, we can conclude that AQ = BP and QC = PD. Therefore, we have:

BP/PD = AQ/QC = 1

So, BP:PD = 1:1.