AE=t, AP=AC-CP= 10-t, t/( 10-t)=4/5, and the solution is t=40/9.

(2)EF∥AD，AC=CD= 10，CG=CF，∠ PGE > 90。 Before the two points of p and g overlap, if △PEG is an isosceles triangle, there are EG=PG, ∠AEF =∠CFG =∞.

According to the topic, the quadrilateral AEFD is a parallelogram, AE=FD, so AE = Ag = FD = T.

EG=PG，AG+PG+CP=AC，AE+EG+CP=AC，EG= 10-2t

According to Pythagorean theorem, BC=6 is calculated. If it passes through point A, it can be regarded as AH⊥CD, AH=BC=6, CH=AB=8 (abbreviated proof), and AD=2√ 10.

EF=AD，FG=2√ 10- 10+2t，FG:AD=CF:CD，(2√ 10- 10+2t):2√ 10 =( 10-t): 10

T=(50-5√ 10)/9。

△AEG is an isosceles triangle. When p and g coincide, when △ PEG is an isosceles triangle, if EP=PG, point P must coincide with point A, but AE = t < AB = 8, AP = 10-t > 2, and point A and point P cannot coincide. Therefore, if △ PEG is an isosceles triangle, it is impossible to overlap.

△AEG∽△CAD, EG:AD=AE:AC, EG=√ 10t/5, and△ △PEG∽△AEG (abbreviated proof).

Then △PEG∽△CAD, PG:EG=AE:AC, PG=2t/5, CG=CF= 10-t, CG+PG=t,

The solution is t=25/4.

To sum up, when t=(50-5√ 10)/9 or 25/4, △PEG is an isosceles triangle.

PS: I typed by hand. I haven't typed for a long time. His grandmother is exhausted.