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.