+130077 Draw DB and AC
Since ABCD is a quadrilateral inscribed in a circle angles DAB and DCB are supplementary
And angle DCB is acute so angle DAB is obtuse
Using the Law of Cosines
DB^2 = 3^2 + 9^2 - 2 (3)(9) cos DCB (1)
DB^2 = 6^2 +2^2 +2 ( 6)(2) cos DCB (2)
Subtract (1) from (2)
0 = -50 + 78 cos DCB
50 = 78 cosDCB
cos DCB = 50/78 = 25 / 39
And DCB and BCP are supplementary
So cos BCP = - 25 /39
arccos (-25/39) =BCP
Using the Law of Cosines again and noting that ADC and ABC are supplementary
AC^2 = 6^2 + 9^2 -2(6)(9)cos ADC (3)
AC^2 = 2^2 + 3^2 + 2(2)(3)cos ADC (4)
Subtract (3) from (4)
0 = -104 + 120 cos ADC
cos ADC = 104/120 = 13 / 15
arccos (13/15)= ADC = PBC
angle BPC = [ 180 - angle BCP - angle PBC] = [ arccos(-1) - arccos(-25/39) -arccos (13/15)]
Then by the Law of Sines
BP / sin [ BCP] = 3 / sin [ BPC]
BP = sin (arccos (-25/39) ) * 3 / sin [ arccos(-1) - arccos (-25/39) -arccos (13/15) ] = 20 / 3
