Question

In right angled ∆ABD, right angled at A and AC perpendicular to BD . prove that AB^2=BC×BD.

Answer 1

in triangle ABC & ACD
<ACB=<ACD (90°)
AC=AC (common)
by RHS congruence triangle ABC congruence to triangle ACD
now AB=AD,BC=CD
by Pythagoras theorem
BD^2=AB^2+AD^2
BC×BD^2=AB^2+AB^2
BC×BD=whole root AB^4
root and four cancel and gives AB^2
BC×BD=AB^2

Answer 2

$\underline{\underline{\Huge\mathfrak{Answer ;}}}$

Dear ,

1) In ,

ΔADB and  ΔCAB
∠DAB = ∠ ACB  (Each 90°)
∠ADB = ∠CBA  (Common angle)

∴ ΔADB ∼ ΔCAB
AB/CB = BD/AB
AB2 = BD*CB

2)  ∠CAB = x

• In  ΔCBA ;-

∠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

Hence , Proved !
__________________________

- Regards
@ItsDmohit