Q. 253.8( 6 Votes )

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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


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


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