Draw a circle of a radius #AB=10# with a center #A# and choose on it a point and label it #B#.
From point #B#, using the same radius #10#, draw an arc that intersects a circle at point #F#.
Obviously, #Delta ABF# is equilateral triangle, #AB=BF=AF=10# and #/_BAF=60^o#.
Bisect #/_BAF# by a radius #AD#, so #/_BAD=/_DAF=/_30^o#.
Bisect #/_BAD# by a radius #AC#, so #/_BAC=/_CAD=/_15^o#.
We are ready now to derive the length #BC# from #BD#, which, in turn, we can derive from #BF#, that we know is equal to 10.
Let #ADnnBF=P#
Using Pythagorean Theorem, calculate #BD# from #BP# and #PD#.
#BD^2 = BP^2+PD^2#
#BP = (BF)/2 = 10/2 = 5#
#PD = AD - AP#
#AD = 10#
#AP^2 = AF^2 - PF^2 = 10^2-(10/2)^2 = 100-25=75#
#AP = sqrt(75) = 5sqrt(3)#
#PD = 10 - 5sqrt(3)#
#BD^2 = (10/2)^2+(10-5sqrt(3))^2=25+100-100sqrt(3)+75=#
#=100(2-sqrt(3))#
#BD = 10sqrt(2-sqrt(3)) ~~5.176381#
Analogously, calculate #BC# by knowing #BD#.
Let #ACnnBD=Q#
Using Pythagorean Theorem, calculate #BC# from #BQ# and #QC#.
#BQ = (BD)/2 = 5sqrt(2-sqrt(3))#
#QC = AC - AQ#
#AC=10#
#AQ^2=AB^2-BQ^2=10^2-25(2-sqrt(3))=25(2+sqrt(3))#
#AQ = 5sqrt(2+sqrt(3))#
#QC=10-5sqrt(2+sqrt(3))#
#BC^2=BQ^2+QC^2=#
#=25(2-sqrt(3))+100-100sqrt(2+sqrt(3))+25(2+sqrt(3))=#
#=100(2-sqrt(2+sqrt(3)))#
#BC=10sqrt(2-sqrt(2+sqrt(3)))~~2.610524#
