Question

ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD

Answer 1

HELLO DEAR,

1 ) . IN ΔADB and  ΔCAB

∠DAB = ∠ ACB  (Each 90°)

∠ADB = ∠CBA  (Common angle)

∴ ΔADB ∼ ΔCAB

AB/CB = BD/AB

AB² = BD*CB

2 ) .  IN  ΔCBA

 ∠CAB = x

∠CBA = 180° - 90° -x

∠CBA = 90° -x 

simillarly,

IN ∠CAD,

∠CAD =  90° - ∠CBA

 =  90° - x 

∠CDA = 180° - 90° - ( 90° - x )

∠CDA = x

In ΔCBA AND ΔCAD

∠CBA = ∠CAD

∠CAB = ∠CDA 

 ∠ACB = ∠DCA

∴ ΔCBA ∼ ΔCAD

AC/DC = BC/AC

AC² = BC * DC

3 ) .  IN ΔDCA = ΔDAB

∠DCA = ∠DAB

 ∠CDA = ∠ADB

∴ ΔDCA ∼ ΔDAB

DC/DA = DA/DB

AD² = CD * BD

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

Given :

In ∆ADB ,

<A= 90° and AC perpendicular to BD

Prove : i ) AB² = BC . BD

ii ) AC² = BC . DC

iii ) AD² = BD . CD

Proof :

i ) From ∆ABD , ∆ACB

<B is common angle -----( 1 )

<DAB = <ACB = 90°-------( 2 )

From ( 1 ) and ( 2 ) ,

∆ABD ~ ∆CBA ( A.A similarity )

DA/AC = DB/AB = AB/BC

BD/AB = AB/BC

=> AB² = BD . BC ------( i )

ii ) From ∆DAB , ∆DAC

<D is common angle

<DAB = <DCA = 90°

Therefore ,

∆DAB ~ ∆DCA ( A.A similarity )

DA/DC = DB/DA

AD/CD = BD/AD

=> AD² = BD .CD -----( iii )

iii ) Since ∆ABD ~ ∆ACB and

∆DAB ~ ∆DCA

=> ∆ACB ~ ∆DCA

then ,

DC/AC = DA/AB = AC/BC

=> AC² = DC . BC --------( ii )

I hope this helps you.

: )