Calculating the angles of a triangle

Question

Let $\angle{BDE}=\angle{DBE}=\angle{EBF}=20^\circ$ and $DE=BE=BF$. How to find $\angle{FDE}$ and $\angle{DFB}$?

GeoGebra gives $\angle{FDE}=10^\circ$ and $\angle{DFB}=110^\circ$.

http://picpaste.com/pics/triangle-SqYQIDFG.1511902941.jpg

Answer 1

Let $X$ be the reflection of $E$ with respect to the line $BD$. Then $XB=EB=FB$ and $$\angle FBX = \angle FBX + \angle EBD + \angle DBX = 20^\circ + 20^\circ + 20^\circ = 60^\circ.$$ It follows that the triangle $FBX$ is equilateral. In particular $XF = XB$. Clearly $X$ lies on the perpendicular bisector of $BD$ so $XB=XD$. So $XF=XB=XD$. In other words, $D,F,B$ lie on a circle centered at $X$. Thus $$\angle BDF = \frac 12 \angle BXF = \frac 12 \cdot 60^\circ = 30^\circ.$$ Therefore $\angle EDF = 30^\circ - 20^\circ = 10^\circ$.