Q. 253.8( 6 Votes )

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]

So, ADB≅⊿CAB [By AAA Similarity]

AB/CB=BD/AB

AB2=BCxBD

(ii)

Let <CAB= x

InΔCBA=180-90°-x

<CBA=90°-x

<CDA=90°-<CAB

=90°-x

<CDA=180°-90°-(90°-x)

<CDA=x

Now in ΔCBA and ΔCAD we may observe that

<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°)

<DAC=<DBA

ΔDCA= ΔDAB (AAA condition)

Therefore DC/DA=DA/DB

(iv) From part (I) AB2=CBxBD

From part (II) AC2=DCxBC

Hence AB2/AC2=CBxBD/DCxBC

AB2/AC2=BD/DC

Hence proved

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