# In ΔABC, AB = AC. Side BC is produced to D. Prove that (AD2 —AC2) = BD . CD Construction: Draw an altitude from A on BC and named it O. Given: ABC is an isosceles triangle with AB = AC

To Prove: AD2 —AC2 = BD × CD

In right triangle ∆AOD, using Pythagoras theorem, we have

(Perpendicular)2 + (Base)2 = (Hypotenuse)2

AO2 + OD2 = AD2 …(i)

Now, in right triangle ∆AOB, using Pythagoras theorem, we have

AO2 + BO2 = AB2 …(ii)

Subtracting eq (ii) from (i), we get

AD2 – AB2 = AO2 + OD2 – AO2 – BO2

AD2 – AB2 = OD2 – BO2

AD2 – AB2 = (OD + BO)(OD – OB)

[ (a2 – b2)= (a + b)(a – b)]

AD2 – AB2 = (BD)(OD – OC) [OB = OC]

AD2 – AC2 = (BD)(CD) [AB =AC]

Hence Proved

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