12. 如图,在菱形ABCD中,对角线AC,BD相交于点O,BP//AC,CP//BD. 求证:OP=AD.

12. 证明:$\because BP// AC$,$CP// BD$,
$\therefore$四边形$BPCO$为平行四边形.
$\because$在菱形$ABCD$中,对角线$AC⊥ BD$,
$\therefore∠ DOC=90°$,
$\therefore$四边形$BPCO$是矩形,
$\therefore OP=BC$,
$\because$在菱形$ABCD$中,$AD=BC$,$\therefore OP=AD$.

13. 如图,在菱形ABCD中,∠ABC=100°,点P在对角线AC上,作PE//BC交AB于点E,作PF//AB交BC于点F.
(1)求∠EPF的度数.
(2)连结PD,当∠DPC=60°时,判断PD与PF的数量关系并证明.

13. 解:(1)$\because PE// BC$,$PF// AB$,
$\therefore$四边形$EBFP$是平行四边形,
$\therefore∠ EPF=∠ ABC=100°$.
(2)$PD=PF$
理由如下:如图,连结$PB$,在菱形$ABCD$中,
$CB=CD$,$∠ DCA=∠ BCA$,
$\because PC=PC$,
$\therefore△ CDP≌△ CBP$(或直接由轴对称性得),
$\therefore∠ BPC=∠ DPC=60°$,
$\because∠ ABC=100°$,$BA=BC$,
$\therefore∠ BAC=∠ BCA=40°$,
$\because PF// AB$,
$\therefore∠ CPF=∠ CAB=40°$,
$\therefore∠ BPF=60°-40°=20°$,$∠ PFB=40°+40°=80°$.
$\therefore∠ PBF=∠ PFB=80°$,
$\therefore PB=PF$,$\therefore PD=PF$.