14. 如图,在矩形ABCD中,AB=6. 点P、点Q同时从点A出发,沿AB方向匀速运动,点P的速度为1,点Q的速度为3,点Q到达点B时停留在点B,待点P继续运动到点B时结束. 设运动时间为t,已知当t=1时,线段DC上有一点M，使四边形PQMD是菱形. 若运动过程中,线段DC上另有一点N,使四边形PQND是菱形,则此时t=.

14. $\dfrac{11}{4}$

15. 如图,在菱形ABCD中,∠B=60°,点E在边BC上,点F在边CD上.
(1)如图1所示,若E是BC的中点,∠AEF=60°.求证:BE=DF.
(2)如图2所示,若∠EAF=60°,求证:△AEF是等边三角形.

15. 证明:(1)连结$AC$,在菱形$ABCD$中,$AB=BC=DC$,
$\because∠ B=60°$,$\therefore△ ABC$是等边三角形.
$\because E$是$BC$的中点,
$\therefore AE⊥ BC$,$\therefore∠ AEB=90°$,
$\because∠ AEF=60°$,
$\therefore∠ FEC=180°-∠ AEB-∠ AEF=30°$,
$\because∠ C=180°-∠ B=120°$,
$\therefore∠ CFE=30°$,$\therefore CF=CE$,
$\therefore CD - CF=BC - EC$,$\therefore DF=BE$.
(2)连结$AC$,在菱形$ABCD$中,$AB=BC=DC$,
$\because∠ B=60°$,$\therefore△ ABC$是等边三角形.
$\therefore AB=AC$,$\because∠ BAC=60°$,
$\because∠ EAF=60°$,$\therefore∠ BAE=∠ CAF$,
$\because$在菱形$ABCD$中,$∠ C=180°-∠ B=120°$,
$\therefore∠ ACF=\dfrac{1}{2}∠ BCD=60°$,$\therefore∠ ACF=∠ B$,
$\therefore△ BAE≌△ CAF(\mathrm{ASA})$,$\therefore AE=AF$,
$\therefore△ AEF$是等边三角形.