Question

prove that DP^2 = BP×CP - BD×CD

Answer 1

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