In the adjoining figure, $$AB=AC$$, $$\angle C=80^\circ$$ and $$BC=BD=DE$$. Then the measure in degrees of $$\angle ADE$$ is

Since $$AB=AC$$ and $$\angle C=80^\circ$$, we get $$\angle B=80^\circ$$. 

In $$\triangle BCD$$, $$BC=BD$$, so $$\angle BDC=\angle BCD=80^\circ$$ 

Applying angle sum property to the triangle we get $$\angle CBD= 180^\circ -80^\circ-80^\circ= 20^\circ$$. 

Therefore, $$\angle DBE=80^\circ-20^\circ=60^\circ$$, and because $$BD=DE$$, triangle $$BDE$$ is equilateral. 

Thus $$\angle ADE=180^\circ-80^\circ-60^\circ=40^\circ$$.