Question

Hey friends.....

Please solve it fast....

Its urgent.....

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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

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