prove that DP^2 = BP×CP - BD×CD
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AP^2 = AD^2 + PD^2
we also know , AP^2 = PC × PB
so , PC × PB = AD^2 + PD^2
PC × PB - PD^2 = AD^2------(1)
in ∆ AOD
AD^2 + OD^2 = AO^2
AD^2 = (AO + OD)(AO - OD)
As AO = BO = OC = Radius of circle
AD^2 = (BO + OD) ( OC - OD)
AD^2 = BD × CD------(2)
using 1 & 2
AD^2 = PC×PB - BD×CD
we also know , AP^2 = PC × PB
so , PC × PB = AD^2 + PD^2
PC × PB - PD^2 = AD^2------(1)
in ∆ AOD
AD^2 + OD^2 = AO^2
AD^2 = (AO + OD)(AO - OD)
As AO = BO = OC = Radius of circle
AD^2 = (BO + OD) ( OC - OD)
AD^2 = BD × CD------(2)
using 1 & 2
AD^2 = PC×PB - BD×CD
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