Question

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Answer 1

Given: Equilateral triangle ABC in which AD is a altitude drawn to the side BC. 

To Prove : 3AB² = 4AD² 

Proof:    AB = AC [∆ is equilateral] 

∠B = ∠C = 60° [common] 

and ∠ADB = ∠ADC = 90° 

Therefore, by using AAS congruent condition

Now, in ∆ ABD right triangle we have

Answer 2

AD⊥BC 

∴ In  ΔABD,AB² = AD² +BD²

⇒AB² = AD² +BC²/4 

or 4AB² = 4AD² +BC²

⇒3AB² = 4AD² 

Comments

Why there is by four in second step

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Detailed solution is given now.