Question

∆ABC is a right triangle right-angled at A and AC ⊥ BD. Show that

(i) \(AB^2=BC.BD\)

(ii) \(AC^2=BC.DC\)

(iii) \(AD^2=BD.CD\)

(iv) \(\frac{AB^2}{AC^2}\) = \(\frac{BD}{CD}\)

Answer 1

(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