A) \( \Large \triangle \) ABC, ADB and CDB are similar. |
B) \( \Large AD \times DC \) |
C) \( \Large \triangle ADB : \triangle CDB = AB : BC \) |
D) \( \Large AB^{2} = AD \times AC \) |
C) \( \Large \triangle ADB : \triangle CDB = AB : BC \) |
(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) \( \Large \frac{\triangle ADB}{\triangle CDB} = \frac{\frac{1}{2}AD \times BD}{\frac{1}{2}CD \times BD} \)
= \( \Large \frac{AD}{CD} = \frac{AB^{2}}{BC^{2}} \ne \frac{AB}{BC} \)