如你图,过O作OE⊥AB,OF⊥CD垂足为E,F,连OD,AO,大圆半径为R,小圆半径为r
所以CF=DF,AE=BE,
所以AC^2=(CF-AF)^2=CF^2-2CF*AF+AF^2,
AD^2=(AF+DF)^2=AF^2+2AF*DF+DF^2
AB^2=(2AE)^2=4AE^2
因为AB⊥CD
所以四边形AEOF是矩形
所以OF=AE
所以AB^2=4AE^2=4OF^2
所以AB^2+AC^2+AD^2
=4OF^2+(CF^2-2CF*AF+AF^2)+(AF^2+2AF*DF+DF^2)
=4OF^2+CF^2-2CF*AF+AF^2+AF^2+2AF*DF+DF^2
=4OF^2+CF^2+AF^2+AF^2+DF^2
=4OF^2+2DF^2+2AF^2
=（2OF^2+2DF^2)+(OF^2+2AF^2)
=2(OF^2+DF^2)+2(OF^2+AF^2)
在直角三角形ODF中,由勾股定理,得OF^2+DF^2=OD^2=R^2
在直角三角形AOF中,由勾股定理,得OF^2+AF^2=OA^2=r^2
所以AB²+AC²+AD²
=2(R^2+r^2),为定值