Q. 253.8( 6 Votes )

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

(i) (ii)

(iii) (iv)

Answer :

_{(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}

_{AB}^{2}=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

AC^{2}=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

AD^{2}=BDxCD

(iv) From part (I) AB^{2}=CBxBD

From part (II) AC^{2}=DCxBC

Hence AB^{2}/AC^{2}=CBxBD/DCxBC

AB^{2}/AC^{2}=BD/DC

Hence proved

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