$$\angle ABC\ $$is an isosceles triangle and $$\ \overline{AB}\ $$=$$ \overline{AC}\ $$= 2a unit $$\ \overline{BC}\ $$= a unit, Draw $$ \overline{AD}\ \bot\ \overline{BC},\ $$and find the length of $$\ \overline{AD}$$

Given that AB=BC=2a units and BC=a units

AD$$\bot$$BC $$\Rightarrow$$ 'D' is midpoint of BC

BD=DC=$$\frac{a}{2}$$

Here \triangle ABD is a right angled triangle where AB is hypotenuse

$$AB^{2}$$=$$BD^{2}+AD^{2}$$
$$\Rightarrow AD^{2}$$=$$AB^{2}-BD^{2}$$
$$\Rightarrow AD$$= $$\sqrt{2a^{2}-\frac{a}{2}}$$
                    = $$\sqrt{4a^{2}-\frac{a^2}{4}}$$
                    = $$\sqrt{\frac{15a^{2}}{4}}$$ = $$\frac{\sqrt{15}a}{2}$$ units