Solution :
(i) Since, AC \(\perp\) BD, therefore,
\(\triangle\) ABC ~ \(\triangle\) DBA and each triangle is similar to whole triangle ABD.
\(\implies\) \(AB\over BD\) = \(BC\over AB\)
So, \({AB}^2\) = BC.BD
(ii) Since, \(\triangle\) ABC ~ \(\triangle\) DAC, therefore,
\(\implies\) \(AC\over BC\) = \(DC\over AC\)
So, \({AC}^2\) = BC.DC
(iii) Since, \(\triangle\) ACD ~ \(\triangle\) BAD, therefore,
\(\implies\) \(AD\over CD\) = \(BD\over AD\)
So, \({AD}^2\) = BD.CD