Question

Chapter Pythagoras theorem and intercept theorem

Answer 1

EXPLANATION.

In a ΔABC.

⇒ AB = AC.

D is a point on BC.

As we know that,

We can write equation as,

⇒ AB = AC [given].

⇒ AE = AE [common].

⇒ BE = EC.

⇒ ∠AEB = ∠AEC [ by 90°].

⇒ ΔABE ≅ ΔACE [By RHS congruent conditions].

As we know that,

Formula of Pythagoras Theorem.

⇒ H² = P² + B².

Hypotenuse > Perpendicular > Base.

Using this concept in the equation, we get.

In ΔABE.

⇒ AB² = AE² + BE². - - - - - (1).

In ΔADE.

⇒ AD² = AE² + DE². - - - - - (2).

Subtract equation (1) and (2), 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 = CD x BD.

Hence Proved.