In △ABC, if AB = AC and D is a point on BC. Prove that AB² - AD² = BD × CD.

ABC is a triangle in which AB = AC and D is any point on BC.

In △ ABE and △ ACE,

⇒ AB = AC (Given)

⇒ AE = AE (Common)

⇒ ∠AEB = ∠AEC (both are 90°)

Using RHS congruency criterion,

△ ABE ≅ △ ACE

⇒ BE = CE (by C.P.C.T.)

In △ ABE, using Pythagorean theorem,

Hypotenuse² = Base² + Height²

⇒ AB² = AE² + BE² ……(1)

In △ ADE, using Pythagorean theorem,

⇒ AD² = AE² + DE² ……(2)

Subtracting equation (2) from (1), we get:

⇒ AB² - AD² = (AE² + BE²) - (AE² + DE²)

⇒ AB² - AD² = AE² + BE² - AE² - DE²

⇒ AB² - AD² = BE² - DE²

⇒ AB² - AD² = (BE - DE)(BE + DE)

⇒ AB² - AD² = (BE - DE)(CE + DE) [∴ BE = CE]

⇒ AB² - AD² = BD × CD

Hence, proved that AB² - AD² = BD × CD.