Prove that three times the square of any side of an equilateral triangle
is equal to four times the square of the attitude.
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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 ✅✅.
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