Question

In an equilateral ΔABC, AD ⊥ BC, prove that AD² = 3BD².

Answer 1

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