In an equilateral ΔABC,AD⊥BC. prove that AD2=3BD2
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 = BC2 --- (i) Corresponding parts of similar triangles
In, ΔABD, by Pythagoras theorem
Hence proved