ABD is a right angled triangle at A and AC_|_BD show that

(I) AB²=BC.BD 

(II) AC²=BC.DC

(III) AD²=BD.CD

Asked by mulkallaswapna | 15th Jan, 2020, 07:32: AM

Expert Answer:

1.
ΔADB and  ΔCAB
∠DAB = ∠ ACB  (Each 90°)
∠ADB = ∠CBA  (Common angle)
∴ ΔADB ∼ ΔCAB
AB/CB = BD/AB
AB² = BD*CB

2.
ΔCBA
∠CAB = x
∠CBA = 180° - 90° - x
∠CBA = 90° - x 
simillarly,
∠CAD = 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.
ΔDCA and ΔDAB
∠DCA = ∠DAB
∠CDA = ∠ADB
∴ ΔDCA ∼ ΔDAB
DC/DA = DA/DB
AD² = CD * BD

Answered by Arun | 19th Jan, 2020, 05:06: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.