In Figure, ABD is a triangle right angled at A and AC \(\perp\) BD. Show that
(i) \(AB^2 = BC × BD\)
(ii) \(AC^2 = BC × DC\)
(iii) \(AD^2 = BD × CD\)

(i) In \(\triangle\) ADB and \(\triangle\) CAB,
\(\angle\) DAB = \(\angle\) ACB (Each 90°)
\(\angle\) ABD = \(\angle\) CBA (Common angles)
\(\triangle\) ADB ~ \(\triangle\) CAB [AA similarity criterion]
AB/CB = BD/AB
\(AB^2 = CB × BD\)


(ii) Let \(\angle\) CAB = x
In \(\triangle\) CBA,
\(\angle\) CBA = 180° – 90° – x
\(\angle\) CBA = 90° – x

Similarly, in \(\triangle\) CAD
\(\angle\) CAD = 90° – \(\angle\) CBA
= 90° – x
\(\angle\) CDA = 180° – 90° – (90° – x)
\(\angle\) CDA = x

In \(\triangle\) CBA and \(\triangle\) CAD, we have
\(\angle\) CBA = \(\angle\) CAD
\(\angle\) CAB = \(\angle\) CDA
\(\angle\) ACB = \(\angle\) DCA (Each 90°)
\(\triangle\) CBA ~ \(\triangle\) CAD [AAA similarity criterion]
AC/DC = BC/AC
\(AC^2 = DC × BC\)


(iii) In \(\triangle\) DCA and \(\triangle\) DAB,
\(\angle\) DCA = \(\angle\) DAB (Each 90°)
\(\angle\) CDA = \(\angle\) ADB (common angles)
\(\triangle\) DCA ~ \(\triangle\) DAB [AA similarity criterion]
DC/DA = DA/DA
\(AD^2 = BD × CD\)