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 – AB² = (BD)(CD)

⇒ AD² – AC² = (BD)(CD) [∵AB =AC]

Hence Proved