AD is an altitude of an …

Clearly, {tex}BD=\frac{BC}{2}=\frac{a}{2}{/tex} [In an equilateral triangle , altitude bisects the base]
In {tex}\triangle ABC{/tex}, {tex}\angle ADB=90^o{/tex}
Using pythgoras theorem,
AB² = AD² + BD² 
{tex}\Rightarrow a^2=AD^2+(\frac{a}{2})^2{/tex} 
{tex}\Rightarrow a^2=AD^2+(\frac{a^2}{4}){/tex}
{tex}\Rightarrow a^2 - (\frac{a^2}{4})=AD^2{/tex}
{tex}\Rightarrow (\frac{4a^2 - a^2}{4})=AD^2{/tex} 
{tex}\Rightarrow AD=\frac{\sqrt3 a }{2}{/tex} 
{tex}\therefore \triangle ABC{/tex} and {tex}\triangle ADE{/tex} are equilateral triangles and so equiangular. 
{tex}\therefore \triangle ABC \sim \triangle ADE{/tex} 
{tex}\frac{ar(\triangle ADE)}{ar(\triangle ABC)}=\frac{AD^2}{AB^2}{/tex}
{tex}\frac{ar(\triangle ADE)}{ar(\triangle ABC)}=\frac{\frac{\sqrt{3}a}{2}}{a^2}{/tex}
{tex}\frac{ar(\triangle ADE)}{ar(\triangle ABC)}=\frac{3}{4}{/tex}