And because ∠C=3∠A, we can see ∠D=2∠A=∠BAD.

Therefore, BD = BA =10; BC=8

CD=2

And because AC is the bisector of ∠BAD, therefore, AD:AB=DC:BC= 1:4.

AD=2.5。

After a, AE is vertical BD in e, BF is vertical AD in f,

Available, delta △ADE is similar to delta △BDF.

Therefore, DE = 5/ 16, AE = 15 √ 7/ 16.

So AC = √ ((CD-de) 2+AE 2) = 3.