From the condition vector PB-vector AD + vector-vector CM = vector PA-vector BM + vector AO find vector X

PB – OD + X – CM = RA – BM + AO.

Move the vectors to the right side of the equality, leaving one X to the left, while changing the sign in front of the vector to the opposite.

X = РA – BM + AO – PB + OD + CM.

Let’s group the vectors.

X = (РA + AO) + CM + OD – BM – PB.

The sum of the vectors RA + AO = PO then,

X = PO + CM + OD – BM – PB.

Let’s change the signs in front of the vectors ВМ and РВ, while changing the direction of the vectors.

– ВM = MВ, – PB = BP.

X = PO + CM + OD + MB + BP.

MВ + BP = MР.

X = (PO + OD) + MP + CM.

PO + OD = PD, then

X = PD + (CM + MP).

CM + MР = CP, then

X = PD + CP.

The sum of vectors РD + СР = СD.

X = CD.