is a right triangle right-angled at A and . Show that

(i) (ii)

(iii) (iv)

_{(i) In ⊿ABD and In ⊿CAB}

_{∠DAB=∠ACB=90°}

_{∠ABD=∠CBA [Common]}

_{∠ADB=∠CAB [remaining angle]}

_{So, ⊿ADB≅⊿CAB [By AAA Similarity]}

_{∴ AB/CB=BD/AB}

_AB²=BCxBD

_(ii)

Let <CAB= x

InΔCBA=180-90°-x

<CBA=90°-x

Similarly in ΔCAD

<CAD=90°-<CAD=90°-x

<CDA=90°-<CAB

=90°-x

<CDA=180°-90°-(90°-x)

<CDA=x

Now in ΔCBA and ΔCAD we may observe that

<CBA=<CAD

<CAB=<CDA

<ACB=<DCA=90°

Therefore ΔCBA~ΔCAD ( by AAA rule)

Therefore AC/DC=BC/AC

AC²=DCxBC

(iii) In DCA and ΔDAB

<DCA=<DAB (both angles are equal to 90°)

<CDA=. <ADB (common)

<DAC=<DBA

ΔDCA= ΔDAB (AAA condition)

Therefore DC/DA=DA/DB

AD²=BDxCD

(iv) From part (I) AB²=CBxBD

From part (II) AC²=DCxBC

Hence AB²/AC²=CBxBD/DCxBC

AB²/AC²=BD/DC

Hence proved