Question

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

Answer 1

SOLUTION :

Given : ∆ABC is an equilateral ∆ in which AB = BC = AC and AD ⊥ BC

In ∆ADB and ∆ADC

∠ADB = ∠ADC       [Each 90°]

AB = AC                  [Given]

AD = AD                [Common]

∆ADB ≅ ∆ADC     [By RHS condition]

Therefore, BD = CD  

[By CPCT]

BD = DC =  BC/2  

[In equilateral triangle altitude AD bisects the opposite side BC]

BC = 2 BD ………….(1)

In, ∆ABD, by Pythagoras theorem

AB² = AD² + BD²

BC² = AD² + BD²

[Given : AB = BC ]

(2BD)² = AD² + BD²

[From eq 1]

4BD² - BD² = AD²

3BD² = AD²

Hence, AD² = 3BD²

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Answer 2

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