Question 884309
<pre>
{{{drawing(400,400,-4,4,-4,4,

circle(0,0,3.287286115),

line(3.287286115,0,-2.188355303,2.453028958),

line(-2.188355303,2.453028958,-3.269074003,-.3455505189),

line(-3.269074003,-.3455505189,-.5152426472,-3.246655974),
green(line(3.287286115,0,-3.269074003,-.3455505189)),
line(-.5152426472,-3.246655974,3.287286115,0),
locate(0.6,1.5,6),locate(1.6,-1.5,5), locate(-2.3,2.8,D),
locate(3.4,.2,A),locate(-.6,-3.26,B), locate(-3.5,-.3,C),
locate(-1.9,-1.5,4),locate(-2.65,1.1,3) 



)}}}{{{

matrix(21,1,

matrix(1,3,LAW,OF,COSINES),
AC^2=CD^2+DA^2-2CD*DA*cos(D),
AC^2=3^2+6^2-2*3*6*cos(D),
AC^2=9+36-36cos(D),
AC^2=45-36cos(D),
"",
AC^2=BC^2+AB^2-2BC*AB*cos(B),
AC^2=4^2+5^2-2*4*5*cos(B),
AC^2=16+25-40cos(B),
AC^2=41-40cos(B),
"",
45-36cos(D)=41-40cos(B),
40cos(B)-36cos(D)=-4,
D="180°"-B,
40cos(B)-36cos("180°"-B)=-4,
40cos(B)-36(-cos(B))=-4,
40cos(B)+36(cos(B))=-4,
76cos(B)=-4,
cos(B)=-4/76,
B=arccos(-4/76) ,
B="93.01696131°")

}}}

Next we can use Ptolemy's theorem: 
The product of the two diagonals of a cyclic 
quadrilateral is equal to the sum of the 
products of opposite sides.

{{{drawing(400,400,-4,4,-4,4,

circle(0,0,3.287286115),

line(3.287286115,0,-2.188355303,2.453028958),

line(-2.188355303,2.453028958,-3.269074003,-.3455505189),

line(-3.269074003,-.3455505189,-.5152426472,-3.246655974),
green(line(3.287286115,0,-3.269074003,-.3455505189)),
red(line(-2.188355303,2.453028958,-.5152426472,-3.246655974)),
line(-.5152426472,-3.246655974,3.287286115,0),
locate(0.6,1.5,6),locate(1.6,-1.5,5), locate(-2.3,2.8,D),
locate(3.4,.2,A),locate(-.6,-3.26,B), locate(-3.5,-.3,C),
locate(-1.9,-1.5,4),locate(-2.65,1.1,3) 



)}}}{{{matrix(16,1,
AC*BD = AB*CD+AD*BC, 
AC*BD = 5*3+6*4, 
AC*BD = 15+24, 
AC*BD = 39,
BD=39/AC,
"",
matrix(1,4,We,calculate,diagonal,AC),
AC^2=45-36cos(D),
AC^2=45-36(-4/76),
AC^2=891/19,
AC=sqrt(891/19),
"",
BD=39/AC,
BD=39/sqrt(891/19),
BD=39sqrt(19/891),
BD=5.695115752)}}}

Edwin</pre>