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Question

In ABC, ACB=90,CDAB. Prove that
BC2AC2=BDAD


Solution

ΔADC and ΔACB
ADC = ACB=90
CAD=CAB (common)
So, ΔADCΔACB by AA
ADAC=ACAB
AC2=AD.AB       …. (i)
ΔBCDandΔBAC
CAD=CBA (common)
BDC=BCA=90
So, ΔBCDΔBAC
By AA similarity
BCBA=BDBC
BC2=BA×BD...(ii)
From (i) and (ii)
AC2BC2=ADBC     (Proved)



 

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