Given, ABC is an equilateral triangle of side 2a
Draw, AD $\perp$ BC
In ΔADB and ΔADC
AB = AC
AD = AD
$\angle ADB = \angle ADC$ [Both are 90°]
Therefore, ΔADB ≅ ΔADC by RHS congruence
Hence, BD = DC [by CPCT]
In right angled ΔADB
$AB^2 = AD^2 + BD^2$
$(2a)^2 = AD^2 + a^2 $
$AD^2 = 4a^2 – a^2$
$AD^2 = 3a^2$
AD = $\sqrt{3a}$
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