For reference: In the ABCD parallelogram, M is the point mean of CD , BM and AC intersects at P.How far is P from AD; if B is 12 from AD. (Answer:8) My progress:
Draw $CK \\BF (K \perp AD, K \in AD)\\ \triangle ACK \sim APE \implies\\ \frac{EP}{12} = \frac{AP}{AC} = \frac{AE}{AI}$
Extend $BM$ until $J, J \in AD \implies:\\\triangle BJF \sim IJK \implies:\\ \frac{PE}{IK} = \frac{JP}{IJ}=\frac{EJ}{JK}\\ \triangle BPC \sim JPA\implies \\ \frac{12-EP}{EP} = \frac{BP}{JP}=\frac{CP}{AP}$
...???
Hint MathLover: $CP:PA = 1:2$ $$AP = 2CP\\\triangle APE \sim ACK\\ \frac{PE}{AP} = \frac{CK}{AC}\implies \frac{PE}{2CP} = \frac{12}{CP+AP}\implies \frac{PE}{2CP} = \frac{12}{3CP}\implies PE =8$$