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

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in triangle ABd and ABC

<B = <B               (common angle)

<BAD = <ACB    (90)

so ABD ~ CBA    (AA cong. )

therefore AB/CB  = BD /AB

so AB 2 = BC * BD

similarly you can prove the other 2!

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