(i) In ⊿ABD and In ⊿CAB
∠DAB=∠ACB=90°
∠ABD=∠CBA [Common]
∠ADB=∠CAB [remaining angle]
So,
⊿ADB≅⊿CAB [By AAA Similarity]
∴ AB/CB=BD/AB
AB2 = BC x BD
(ii)
Let ∠CAB = x
InΔCBA = 180 - 90° - x
∠CBA = 90° - x
Similarly in ΔCAD
∠CAD = 90° - ∠CAD = 90° - x
∠CDA = 90° - ∠CAB
= 90° - x
∠CDA = 180° - 90° - (90° - x)
∠CDA = x
Now in ΔCBA and ΔCAD we may observe that
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA = 90°
Therefore ΔCBA ~ ΔCAD ( by AAA rule)
Therefore AC/DC = BC/AC
AC2 = DC x BC
(iii) In DCA and ΔDAB
<DCA = DAB (both angles are equal to 90°)
<CDA =, <ADB (common)
<DAC = <DBA
ΔDCA= ΔDAB (AAA condition)
Therefore DC/DA=DA/DB
AD2 = BD x CD
(iv) From part (I) AB2=CBxBD
From part (II) AC2 = DC x BC
Hence AB2/AC2 = CB x BD/DC x BC
AB2/AC2 = BD/DC
Hence proved