∴ DE/DA=DP/DB

And DE=t, DP= 10-t,

∴ t/6= 10-t/ 10

∴t= 15/4

∴, PE‖AB when t= 15/4

(2)∵EF is parallel and equal to CD,

Quadrilateral CDEF is a parallelogram.

∴∠DEQ=∠C,∠DQE=∠BDC.

∫BC = BD = 10，

∴∠DEQ=∠C=∠DQE=∠BDC.

∴△DEQ∽△BCD.

∴DE/BC=EQ/CD

t/ 10=EQ/4

EQ=2t/5

B is BM⊥CD, CD is M, P is PN⊥EF, EF is N,

BM= root sign (102-22) = square root of 4 6.

∫ED = DQ = BP = t，

∴PQ= 10-2t.

And △PNQ∽△BMD,

PQ/BD=PN/BM

10-2t/10 = the square root of pn/46.

Square root of PN = 4 (1-t/5)

∴S△PEQ= emotional intelligence? PN =1/2 * 2t/5 * square root of 46 (1-t/5) = square root of-46/25 * t2+square root of 4 6t/5.

3) s △ BCD =1/2 * CD * BM = the square root of 8 6.

If S△PEQ=2/25*S△BCD

The radical 6t/5 with the square root of -4 6/25 * t 2+4 = the square root of 8 6.

The solution is t 1= 1, and t2 = 4.

4) at △PDE and △FBP,

DE = BP = t，PD=BF= 10-t，∠PDE=∠FBP，

∴△PDE≌△FBP.

∴S pentagonal PFCDE=S△PDE+S quadrilateral PFCD=S△FBP+S quadrilateral pfcd = s △ BCD = 8.

∴ During the exercise, the area of pentagonal PFCDE remains unchanged.