CQ=t

AP=4t

(1) solution: Because the side length of the square ABCD is 12.

So AB=BC=CD=AD= 12.

Angle ABC= Angle ADC= Angle DCB=90 degrees.

Angle ABD= Angle CBD= Angle BDC=45 degrees.

Because PQ is parallel to BD

So CQP angle BDC angle =45 degrees.

QPC angle CBD angle =45 degrees.

So CQP angle = QPC angle =45 degrees.

So CQ=CP

Because CQ = AB+BC-AP =12+12-4t = 24-4t.

So 24-4t=t

So t=4.8

So when t=4.8, PQ is parallel to BD.

(2) Solution: Since DP is perpendicular to CE, let the vertical foot be O.

So the angle POD=90 degrees

Because pod+angle ODC+ angle DCO= 180 degrees.

So ODC angle +DCO angle =90 degrees.

Because angle DCO+ angle BCO= angle DCB=90 degrees

So Angel ·ODC = Angel ·BCO

Because AB=BC (authentication)

Angle ABD= Angle CBD=45 degrees (confirmed)

Yes = yes

So triangle ABE and triangle CBE are congruent (SAS)

So BCO angle BAP angle

So angle BAP= angle ODC

Because AB=DC (authentication)

Angle DCB= angle ABC=90 degrees

So triangle ABP and triangle DCP are congruent (ASA)

So BP=CP= 1/2BC=6.

Because CP=AB+BC-AP=24-4t.

So 24-4t=6.

So t=4.5

So when t=4.5, DP is perpendicular to CE.

(3) Because the distance AF from point A to PQ is 12, the vertical foot F is set.

So AFQ angle AFP angle =90 degrees.

Because AB=AD= 12 (authentication)

So AD=AF=AB

Because the angle ABC=90 degrees

So triangle AFP and triangle ABP are right triangles.

Because AP=AP

So right triangle ABP and right triangle AFP are congruent (HL)

So BP=PF

The same can be proved.

DQ=FQ

Because PQ=PF+FQ

BQ=AP-AB=4t- 12

DQ=DC-CQ= 12-t

CP=AB+BC-AP=24-4t

Because the angle DCB=90 degrees

So the triangle CPQ is a right triangle.

So pq 2 = CQ 2+CP 2.

So (24-4t) 2+t2 = (12-t+4t-12) 2.

So t= 12+6 times the root number 2 (0

T= 12-6 times the root number 2.

So when t= 12-6 is multiplied by the root number 2, the distance from point A to PQ is 12.