Question

ABD is a triangle right angled at A and AC is perpendicular to BD.
SHOW That 1. AB²=BC×BD
2.AC²=BC×DC
3.AD²=BD×CD​

Answer 1

Answer:

Step-by-step explanation:

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