Help with Geometry!

This problem uses isoceles triangles and sum of angles and vertical angles

___________________________________________________________________________

Lets see:

BAC is 28 degrees, and since ABCD is a rectangle, DAB is 90 degrees.

So that means DAC is 90 - 28 = 62 degrees.

That also means that DCA is 28 degrees because of vertical angle theorem (keep that in mind)

The sum of the angles of triangle DAQ is 180 degrees.

So since DQA is 90 degrees as given, and DAC is 62 degrees. 

Angle ADQ is 180 - 90 - 62 = 28 degrees.

Angle ADC is a right angle, so 90 - 28 = 62 degrees. QDC is 62 degrees.

And since DCA is 28 degrees,

Angle QDP is 62 - 28 = 34 degrees

 BAC  = 28......then because AC is a transversal cutting two parallel sgments AB and DC, then angle ACD also = 28 

And since DP = PC.....then angle PDC = 28....so angle DPC   =  (180 - 2(28))  = 124

And by the exterior angle theorem  

Angle DPC   = angle QDP + angle  DQP

124  = angle QDP + 90

34  =  angle QDP