Recall that tangent segments to a circle from an external point are equal in length,
#=> AD=AE=5, BD=BF=3#,
let #CE=x, => CF=CE=x#
By the law of cosines,
#BC^2=AB^2+AC^2-2*AB*AC*cosangleA#
#=> (3+x)^2=8^2+(5+x)^2-2*8*(5+x)*cos60#
#=> 9+6x+x^2=64+25+10x+x^2-2*8*(5+x)*1/2#
#=> 6x-10x+8x=64+25-40-9#
#=> 4x=40, => x=10# cm
Hence, #BC=BF+FC=3+x=3+10=13# cm
Solution 2 :
Let #O and r# be the center and the radius of the circle, respectively, as shown in the figure.
As #AD=AE, OD=OE=r, and OA# is the common hypotenuse,
#=> DeltaAOD and DeltaAOE# are congruent,
#=> AO# bisects #angleDAE, => angleOAD=60/2=30^@#
#=> r=AD*tan30=5*tan30=5/sqrt3#
Area of #ABC=|ABC|=1/2*AB*AE*sin60#
#=1/2*8*(5+x)*sqrt3/2=10sqrt3+2sqrt3x-----Eq(1)#
#|ABC|=1/2*r*(AB+AC+BC)#
#=1/2*5/sqrt3*(8+5+x+3+x)#
#=5/(2sqrt3)*(16+2x)-----Eq(2)#
#Eq(1)=Eq(2)#,
#=> 10sqrt3+2sqrt3x=5/(2sqrt3)(16+2x)#
#=> 60+12x=80+10x#,
#=> 2x=20, => x=10# cm
#=> BC=3+x=3+10=13# cm
