help

I have an answer for this, but it takes a lot of steps -- I welcome a quicker solution.

1) Find the length of the diameter AC: 

          Since AC is a diameter, I can use triangle ABC to find this length.

          Both AB and BC = 5 and triangle ABC is an isosceles right triangle, AC  =  5·sqrt(2).

2) Find the length PC:

          Triangle ABC is a right triangle (angle ABC is inscribed in a semicircle), so use the Pythagorean Theorem.

          PC² + PA²  =  AC²     --->     PC² + 4²  =  [5·sqrt(2)]²     --->     PC²  +  16  =  50     --->     PC²  =  34     --->     PC  =  sqrt(34).

3) Find the size of angle(PAC):

          Use the Law of Cosines for triangle PAC).

           PC²  =  PA² + AC²  -  2·PA·AC·cos(PAC)     --->     [sqrt(34)]²  =  (4)² + [5·sqrt(2)]² - 2·(4)·5(sqrt(2)·cos(PAC)

                   --->     34  =  16 + 50 - 40·sqrt(2)·cos(PAC)     --->     -32  =  -40·sqrt(2)·cos(PAC)     --->     PAC  =  55.5501^o

4) angle(PAD)  =  45^o + 55.5501^o  =  100.5501^o

5) Use the Law of Cosines on Triangle(APD):

          AP = 5; AD = 5; angle(PAD) = 100.501^o     --->     DP²  =  AP² + AD² - 2·AD·AP·cos(PAD)

                  --->     DP²  =  (4)² + (5)² - 2·(4)·(5)·cos(100.5501)      --->     DP²  =  48.3238      --->      DP  =  6.95

Two comments:

1)  I hope that I haven't made a mistake!

2)  I hope that someone can find an easier way!

AB = 5

PA = 4

BD = ?                         BD = sqrt [(AB)² + (AD)²]      BD = 7.072      r = 3.536  

Let the circle's origin be an "O", and a midpoint of PA an "M"                                                                            ∠POM = q                   sin(q) = (PM)/r     q = 34.44° 

∠POA = 2*q                             ∠POA= 68.88° 

∠AOB = 90°

∠BOP = ∠AOB - ∠POA            ∠BOP = 21.12° 

∠(BOP)/2 = w                      w = 10.56°  

PB = ?                             sin(w) =[(PB)/2] / r       (PB)/2 = 0.648      PB = 1.296  

Finally...                      PD = sqrt [(BD)² - (PB)²]                PD = 6.952