As shown in the figure, let AB = A (vector), AD = B, AE = C, then AM = A/2, An = B/3,? AR=2c/3

On the other hand, Ag = A+B+C, AP = UAG = UA+UB+UC:

AP=sAQ+( 1-s)AR，AQ = tAM+( 1-t)AN = ta/2+( 1-t)b/3。

AP = s[ta/2+( 1-t)b/3]+( 1-s)2c/3

=(ST/2)a+[s( 1-t)/3]b+[2( 1-s)/3]c

∴u = ST/2 = s( 1-t)/3 = 2( 1-s)/3。 Eliminate s, t, get: u = 2/ 13.

AP=(2/ 13)AG，？ PG =( 1-2/ 13)AG = 1 1/ 13AG，？ ∴AP∶PG=2∶ 1 1.