Let $$\triangle ABC$$ be a triangle with $$Ab = AC$$ and $$D$$ be a point on $$BC$$ such that $$\angle BAD = 30^\circ$$. If E is a point on $$AC$$ such that $$AD = AE$$, then $$\angle CDE$$ equals

Based on the information provided, make the diagram as given above. We realise that the two angles on the unequal side in an isosceles triangle are equal. That both $$\triangle ABC$$ and $$\triangle ADE$$ are isosceles triangles.

Let $$\angle DAE$$ be $$2x$$, it follows that in $$\triangle ABC$$, $$\angle ABC + \angle ACB = 180-(30+2x)$$. Also, since $$\angle ABC = \angle ACB$$ we get $$\angle ABC = \angle ACB = 75-x$$

We also find in $$\triangle ADE$$, $$\angle ADE + \angle AED = 180-2x$$, and since $$\angle ADE = \angle AED$$ we get $$\angle AED = 90-x$$. 

Since $$\angle AED$$ and $$\angle DEC$$ are collinear, $$\angle DEC$$ becomes $$180-\angle AED = 180-(90-x) = 90+x$$

Finally, in $$\triangle CDE$$, we need to find the value of $$\angle CDE$$, which will be equal to $$180-(\angle DEC + \angle ACB) = 180- (75-x+90+x) = 180-165= 15^{\circ}$$