(a) \( \Large \triangle ABC - \triangle ADB - \triangle CDB \)



From \( \Large \triangle ABC\ and\ \triangle ADB \)

\( \Large \angle ABC = \angle ADB = 90 ^{\circ} \)

AB = AB, common

From A-S-A, \( \Large \triangle ABC - \triangle ADB \)

Again, from \( \Large \triangle ABC\ and\ \triangle CDB \)

\( \Large \angle C = common \)

\( \Large BC = common \)

\( \Large \angle ABC = \angle CBD \)

From A-S-A, \( \Large \triangle ABC - \triangle ADB \)

From (i) and (ii)

\( \Large \triangle ABC = \triangle ADB - \triangle CDB \)

(b) From \( \Large \triangle ADB - \triangle BDC \)

\( \Large \frac{BD}{AD} = \frac{DC}{BD} \)

\( \Large \therefore BD^{2} = AD \times DC \)

And from \( \Large \triangle ADB - \triangle ABC \)

\( \Large \frac{AB}{AD} = \frac{AC}{AB} \)

(c) \( \Large \frac{\triangle ADB}{\triangle CDB} = \frac{\frac{1}{2}AD \times BD}{\frac{1}{2}CD \times BD} \)

= \( \Large \frac{AD}{CD} = \frac{AB^{2}}{BC^{2}} \ne \frac{AB}{BC} \)