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AD is perpendicular to the side BC of an equilateral △ABC. Prove that 4AD2 = 3AB2.

Pythagoras Theorem

ICSE

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Answer

Given, AD ⊥ BC and AB = BC = CA (Equilateral triangle).

The perpendicular to base in equilateral triangle bisects the base.

∴ BD = BC2\dfrac{BC}{2}.

From figure,

AD is perpendicular to the side BC of an equilateral △ABC. Prove that 4AD^2 = 3AB^2. Pythagoras Theorem, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

In right triangle ABD,

By pythagoras theorem,

⇒ AB2 = AD2 + BD2

⇒ AB2 = AD2 + (BC2)2\Big(\dfrac{BC}{2}\Big)^2

⇒ AB2 = AD2 + BC24\dfrac{BC^2}{4}

⇒ AB2 = 4AD2+BC24\dfrac{4\text{AD}^2 + \text{BC}^2}{4}

⇒ 4AB2 = 4AD2 + BC2

⇒ 4AB2 = 4AD2 + AB2 (∵ BC = AB)

⇒ 4AD2 = 3AB2.

Hence, proved that 4AD2 = 3AB2.

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