(1)
A1B1=(1/2)*（1^2）=1/2
A2B2=(1/2)*(2^2)=2
A3B3=(1/2)*(3^2)=9/2
根據(jù)中位線的性質(zhì)得：
CB2=(A1B1+A3B3)/2
=(1/2+9/2)/2
=5/2
CA2=CB2-A2B2
=5/2-2
=1/2
(2)假設(shè)A1的橫坐標(biāo)為x,則A2、A3的橫坐標(biāo)分別為x+1,x+2;
A1B1=(1/2)*(x^2)-x+1
A2B2=(1/2)*[(x+1)^2]-(x+1)+1
A3B3=(1/2)*[(x+2)^2]-(x+2)+1
根據(jù)中位線的性質(zhì)得：
CB2=(A1B1+A3B3)/2
={(1/2)*(x^2)-x+1+(1/2)*[(x+2)^2]-(x+2)+1}/2
=(x^2+2)/2
=(1/2)*x^2+1
CA2=CB2-A2B2
=(1/2)*x^2+1-{(1/2)*[(x+1)^2]-(x+1)+1}
=(1/2)*x^2+1-(1/2)*(x^2+1)
=1/2