Question

6. 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.
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Answer 1

Answer:

Hey there !!

➡ Given :-

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

➡ To prove :-

→ 3AB² = 4AD².

➡ Proof :-

In ∆ADB and ∆ADC, we have 

→ AB = AC. [ Given ]

→ \angle∠ B = \angle∠ C = 60° .

→ \angle∠ ADB = \angle∠ ADC = 90° .

•°• ∆ADB \cong≅ ∆ADC . [ AAS - Congruence ] 

•°• BD = DC = ½BC .

▶ 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 ✅✅.

Answer 2

plz refer to this attachment