Question

In an equilateral triangle prove that 3 times the square of one side is equal to 4 times the square of it's one of the altitudes​

Answer 1

$\huge{\textbf{\underline{\underline{Answer:-}}}}$

★ Given :-

→ A ∆ABC in which AB = BC = CA and AD \perp BC .

★ To prove :-

→ 3AB² = 4AD².

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$\huge{\textbf{\underline{\underline{Explanation:-}}}}$

★ Proof :-

In ∆ADB and ∆ADC, we have

→ AB = AC. [ Given ]

→ ∠B = ∠C = 60°

→ ∠ADB = ∠ADC = 90°

∴ ∆ADB ≅ ∆ADC [ AAS - Congruence ]

∴ BD = DC = ½BC

➡ From right ∆ADB, we have

AB² = AD² + BD² [ By Pythagoras theorem ]

= AD² + ( ½ BC )² .

➡ From right ∆ADB, we have

AB² = AD² + BD² [ By Pythagoras' theorem ]

= AD² + ( ½ BC )²

= AD² + ¼ BC²

=> 4AB² = 4AD² + BC²

=> 3AB² = 4AD² . [∴BC = AB ]

Hence, 3AB² = 4AD²

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