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

1.	∆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» (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

∆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»

(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