Solution:
(i) In \[\vartriangle ACD\text{ }and\text{ }\vartriangle BCA\]
\[\angle DAC\text{ }=\angle ABC\] [Given]
\[\angle ACD\text{ }=\angle BCA\][Common angles]
Hence, \[\vartriangle ACD\text{ }\sim\text{ }\vartriangle BCA\text{ }by\text{ }AA\]criterion for similarity.
(ii) Since, \[\vartriangle ACD\text{ }\sim\text{ }\vartriangle BCA\]
We have,
\[AC/BC\text{ }=\text{ }CD/CA\]\[=\text{ }AD/AB\]
So,
\[4/BC\text{ }=\text{ }CD/4\text{ }=\text{ }5/8\]
\[4/BC\text{ }=\text{ }5/8\]
So, \[BC\text{ }=\text{ }32/5\text{ }=\text{ }6.4\text{ }cm\]
And,
\[CD/4\text{ }=\text{ }5/8\]
Thus, \[CD\text{ }=\text{ }20/8\text{ }=\text{ }2.5\text{ }cm\]