Eight Solutions (more detailed)

1. Draw segment $DF$ parallel to $BC$ with $F$ on $AB$. Draw $CF$ intersecting $BD$ at $G$. Triangles $BGC$ and $DGF$ are equilateral. Triangle CEB is isosceles ( $\angle BEC = 50= \angle BCE$), so $BE=BC=BG$. Then triangle $BEG$ is isosceles with a vertex angle of $20^\circ$ and base angles of $80^\circ$. Since $80^\circ+\angle EGF +60^\circ=180^\circ$, we have $\angle EGF = 40^\circ$. But in triangle $BFC$, $\angle BFC=40^\circ$, so triangle $EFG$ is isosceles and $EFGD$ is a kite.

2. Use the Law of Sines in triangle $BED$ and triangle $BCD$ and $BE=BC$ to get $$\frac{\sin(160^\circ-x)}{\sin x} =\frac{BD}{BE}=\frac{BD}{BC}= \frac{\sin(80^\circ)}{\sin 40^\circ}$$. Simplify to get $\sin(20^\circ +x) = 2\cos 40^\circ \sin x$.

3. Draw lines through $D$ and $H$ parallel to $BC$ and $DC$, respectively, intersecting at $H$. Draw $CG$ with $G$ on $BD$ and $\angle GCB = 60^\circ$. $BE=BC=CG$, $BH=CD$ and $\angle EBH =\angle GCD =20^\circ$, so $\Delta EBH \cong \Delta GCD$. Therefore $\angle BHE = \angle CDG = 40^\circ$. But $\angle BHD =80^\circ$, so $\angle EHD=40^\circ=\angle BHE$ and $E$ is the incenter of triangle $BDH$.

4. Mark $K$ on $AC$ such that $\angle KBC = 20^\circ$. Draw $KB$ and $KE$. $BE=BC=BK$ and $\angle EBK= 60^\circ$, so triangle $EBK$ is equilateral and triangle $KBC$ is isosceles with $\angle BKC=80^\circ$. $\angle EKD = 40^\circ$ since $\angle EKC = 140^\circ$. In triangle $BDK,\; \angle BDK = 40^\circ$, so that triangle $BKD$ is isosceles with $KD=KB=KE$. Then triangle $KDE$ is isosceles with a $40^\circ$ vertex angle at $K$. The base angles are $70^\circ$, so $\angle EDC = 30^\circ$.

5. Reflect $E$ through $AC$ to point $H$. Triangle $ECH$ is equilateral with line $CD$, the perpendicular bisector of $EH$. But $BE=BC$, so $BCHE$ is a kite. Draw symmetry line $BH$. Ray $BD$ bisects angle $EBH$. Consider the circumcircle of triangle $BEH$. Line $BD$ passes through the midpoint of arc $EH$ and ray $BD$ passes through the midpoint of arc $EH$. Since $D$ is on the ray and the line it must be on the circle. Then the measure of inscribed angle $EDB$ has the same measure as inscribed angle $EHB$

6. Let the bisector of $\angle ABC$ intersect $AC$ at point $T$. $BD$ bisects $\angle EBT$. $\Delta BET\cong\Delta BCT$ by SAS, so $\angle ETB= \angle CTB = 60^\circ$. Therefore $\angle ETD = 60^\circ$ which is the measure of the angle formed by the extension of $BT$ and $AT$ and $D$ is an excenter of triangle $BET$. ($D$ is equidistant from the lines $BT$, $ET$, and $BE$.) Therefore $ED$ bisects $\angle AET$ and we have $\angle BED = 50^\circ + 30^\circ + 50^\circ = 130^\circ$.

7. Let $O$ be the circumcenter of triangle $DEC$. Then central angle $EOD$ is twice the measure of inscribed angle $ECD$. So $\angle EOD=60^\circ$ and triangle $EOD$ is equilateral. Now $D$ lies on the perpendicular bisector of $EO$ and $BD$ bisects angle $BEO$. If the perpendicular bisector of $EO$ is different from $BD$, then $D$ would lie on the circumcircle of triangle $EOB$. Then opposite angles of cyclic quadrilateral $EBOD$ would be supplementary, but $\angle EBO+\angle EDO= 40^\circ+ 60^\circ\ne180^\circ$. Therefore $BD$ bisects $\angle EDO$.

8. Reflect triangle $ABC$ through $AB$ to triangle $ABC^\prime$ and also reflect it through $AC$ to triangle $ACB^\prime$. $AC^\prime B^\prime$ is equilateral and $\angle BC^\prime B^\prime=20^\circ$. Draw $C^\prime E$. Therefore $C^\prime E$ bisects $\angle AC^\prime B^\prime$. Draw $DB$. $AD=DB=DB^\prime$, so $D$ is on the perpendicular bisector of $AB^\prime$ which coincides with the angle bisector since the triangle is equilateral.