过A作AM//CD交BE延长线于M.
则∠PCE=∠MAE,∠CEP=∠AEM;
又点E为AC中点知CE=AE,
∴△CEP≌△AEM,
∴CP=AM,EP=EM;
∵AM//CD
∴∠BDP=∠BAM,
又∠DBP=∠ABM
∴△BDP∽△BAM
∴BP/BM=DP/AM=BD/AB=(AB-AD)/AB=1-AD/AB=2/3;
∴BP=2/3BM=2/3(BP+EP+EM)=2/3BP+4/3EP;DP=2/3AM=2/3CP;
有EP=1/4BP;
∴BP:BE=BP/(BP+EP)=4/5=4:5,
CP:CD =CP/(DP+CP)=CP/(2/3CP+CP)=3/5=3:5.
答:(略).