In the given figure, $$ \triangle ABC $$ is a right angled triangle with $$\angle ABC = 90^\circ$$. D, E, F are points on AB, AC, BC respectively such that $$AD = AE$$ and $$CE = CF$$. Then, $$\angle DEF =$$ (in degree)

Since $$AD=AE$$, triangle $$ADE$$ is isosceles with base angles $$90-\frac{A}{2}$$, and since $$CE=CF$$, triangle $$CEF$$ is isosceles with base angles $$90-\frac{C}{2}$$. Since $$D$$, $$E$$, $$F$$ lie such that $$A$$, $$E$$, $$C$$ are collinear, $$\angle DEF = 180-(90-\frac{A}{2})-(90-\frac{C}{2})=\frac{A+C}{2}=\frac{90}{2}=45$$.