Explanation:
Let ABDC be a square. Connect AD and BC.
CD=DB and CA=AB (since ABDC is a square) AD=AD, then by SSS, triangle ADC=triangle ADB. Since these triangles are congruent, then their angles are congruent. Thus, angle CDA=angle BDA and angle CAD= angle BAD. Therefore AD bisects angles CDB and BAC.
The same reasoning can be used to show that CD=DB and CA=AB (since ABDC is a square) BC=BC, then by SSS, triangle CAB=triangle BDC. Thus angle ACB= angle DCB and angle ABC= angle DBC. Therefore BC bisects angles ACD and DBA.
Since ABDC is a square, angle BDC= angle DCA. Angle EDC = 1/2angle BDC (since it is bisected by AD) and angle ECD= 1/2angle DCA. Thus angle ECD=angle EDC (since they are both half of congruent angles). Then, by properties of isosceles triangles, since base angles ECD and EDC are equal, then side EC=DE.
This same reasoning can be applied to all other angles, concluding that angle ECA=angle CAE and thus CE=AE. Similarly, angle EAB= angle EBA and AE=EB. Finally, angle EBD= angle EDB and ED=EB.
Thus, AE=BE=DE=CE. Therefore, these all act as radii with E as the center. Thus circle ABDC has been circumscribed.