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प्रश्न
In an equilateral ΔABC, AD ⊥ BC, prove that AD2 = 3BD2.
उत्तर
We have, ΔABC is an equilateral Δ and AD ⊥ BC
In ΔADB and ΔADC
∠ADB = ∠ADC [Each 90°]
AB = AC [Given]
AD = AD [Common]
Then, ΔADB ≅ ΔADC [By RHS condition]
∴ BD = CD =BC/2 .......(i) [corresponding parts of similar Δ are proportional]
In, ΔABD, by Pythagoras theorem
AB2 = AD2 + BD2
⇒ BC2 = AD2 + BD2 [AB = BC given]
⇒ [2BD]2 = AD2 + BD2 [From (i)]
⇒ 4BD2 − BD2 = AD2
⇒ 3BD2 = AD2
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