In the following figure, 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
how to do this using pythagoras theorem ?
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