In an equilateral Δ ABC, AD⊥BC, prove that AD² = 3BD².

Given: AD⊥BC

To prove: AD² = 3BD²

Theorem Used:

Right Hand Side Congruence Condition (RHS):

Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and side of the other triangle.

Explanation:

We have,

Δ ABC is an equilateral triangle and AD⊥BC.

In Δ ADB and Δ ADC

∠ADB = ∠ADC = 90°

AB=AC (Given)

AD=AD (Common)

Δ ADB ≅ Δ ADC (By RHS condition)

∴ BD = CD = BC/2 ……. (i)

In Δ ABD

_BC²=AD²+BD²

_BC²=AD²+BD² [Given AB=BC]

_(2BD)²= AD²+BD² [From (i)]

_4BD²-BD²=AD²

_AD²=3BD²