In the figure given, A, B and C are right angles. If $$\angle AEB = 40^\circ$$ and $$\angle BED = \angle BDE$$, then find $$\angle CDE$$.

Since $$\angle B = 90^\circ$$ and $$\angle BED = \angle BDE$$, triangle BED is isosceles right-angled, so both base angles equal $$45^\circ$$. Working through the right angles at A, B and C step by step, $$\angle 1 = 180^\circ - (90^\circ+40^\circ) = 50^\circ$$, giving $$\angle 2 = 50^\circ$$, then $$\angle 3 = 180^\circ-(90^\circ+50^\circ) = 40^\circ$$, giving $$\angle 4=40^\circ$$, and finally $$\angle 5 = 180^\circ-(90^\circ+40^\circ) = 50^\circ$$. So $$\angle CDE = \angle 5 + 45^\circ = 50^\circ+45^\circ = 95^\circ$$.