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 – AB2 = (BD)(CD)
⇒ AD2 – AC2 = (BD)(CD) [∵AB =AC]
Hence Proved