Solution: As shown in Figure 3,

Because AC and HG intersect.

So Angel ·APG = Angel ·cph.

It's AF//BC again.

So angular clearance = angle HCP.

So triangle APG is similar to triangle CPH.

So AG:CH=GP:HP=AP:CP.

Similarly, it can be proved that triangle AEP is similar to triangle CFP.

AE:CF=EP:FP=AP:CP。

So AG:CH=GP:HP=AE:CF=EP:FP=AP:CP.

That is, the quadrilateral AEPG is similar to the quadrilateral CFPH.

As shown in figure 2, let the intersection of AC\HG be p. Because AC intersects with HG.

So Angel ·APG = Angel ·cph.

It's because of AD//BC again.

So angular clearance = angle HCP.

So triangle APG is similar to triangle CPH.

So AG:CH=GP:HP=AP:CP.

Similarly, it can be proved that triangle AEP is similar to triangle CFP.

AE:CF=EP:FP=AP:CP。

So AG:CH=GP:HP=AE:CF=EP:FP=AP:CP.

That is, the quadrilateral AEPG is similar to the quadrilateral CFPH.