Question

3. In Fig. 6.58, ABC is a triangle in which Z ABC >90° and AD ICB produced,
AC?=ABP + BC2 + 2 BC.BD.
thank u

​

Answer 1

In Fig. 6.58, ABC is a triangle in which Z ABC >90° and AD ICB produced,

AC?=ABP + BC2 + 2 BC.BD.

Given,

ABC is a triangle,

∠ ABC > 90°

AD ⊥ CB

To prove:

AC² = AB² + BC² + 2 BC·BD

Proof:

In Δ ADB,

AB² = AD² + BD² ........(1)

In Δ ADC,

AC² = AD² + CD² ........(2)

AC² = AD² + (BD+BC)²            (CD = BD+BC)

AC² = AD² + BD² + BC² + 2BD·BC

AC² = (AD² + BD²) + BC² + 2BD·BC

From (1), we get

AC² = AB² + BC² + 2BD·BC

Hence proved