According to the properties of axisymmetric graphics,

AC bisects the straight line PQ, that is, QF=PF.

AC is perpendicular to PQ,

That is, ∠ QFC = ∠ PFA = 90,

Therefore, RT△QFC≌RT△PFA.

So, AF=CF

Therefore, the line segment AC and the line segment PQ are vertically bisected,

Therefore, the quadrilateral APCQ is a diamond.

(2)

Circle o is tangent to straight line AQ,

According to the meaning of the question, the straight line AQ and the straight line AP are symmetrical about the straight line AC passing through the center o,

So the circle o must be tangent to the straight line AP, let the tangent point be h and connect OH,

Then OH perpendicular to AB,

In rectangular ABCD, CD is perpendicular to AB,

So, oh ∑CD,

Therefore, AO/OH=AC/BC.

AC * AC = AB * AB+CD * CD = 15 * 15，AC= 15

Let the radius of circle o be r, then AF=AC-2*r, OH=r,

AO=AF+r=AC-r= 15-r

Therefore, (15-r)/r =15/12 = 5/3.

r=45/8

AF=AC-2*r= 15/4

Since OQ is perpendicular to AC, ∠PFA is a right angle, * * * uses ∠CAB,

RT△AFP∽RT△ABC

AP/AF=AC/AB

AP = AF * AC/AB =( 15/4)* 15/ 12 = 75/ 16