Answer:
[tex]AD=\sqrt{3} \ a[/tex]
Step-by-step explanation:
AD is perpendicular on BC.
BD=CD=a (as D is the mid point of BC)
Now in the triangle ABD
[tex]AB^{2} =BD^{2} +AD^{2} \ \ \ \ (using \ Pythagorean \ Theorem)\\AD^2=AB^2-BD^2\\AD^2=(2a)^{2} -a^2\\AD^2=3a^2\\AD=a\sqrt{3}[/tex]
