## Elementary Geometry for College Students (6th Edition)

1- $\triangle AEB$ is an equilateral so each angle is equal to $60^{\circ}$ And all side of equilateral are congruent and they are equal to the sides of the square ABCD. 2- in a triangle BEC is an isosceles triangle ( EB and CB are congruent ) so $\angle CEB = \angle ECB= 2x+30= 90, x= 75$ And the same procedure for triangle DEC. 3- in a triangle DEC, where the goal is to find $\angle DEC$ , we find that $\angle EDC=\angle ECD= 90-75=15$ 4- last step that since $\triangle DEC = 15+ 15 + \angle DEC = 150$ We conclude that $\angle DEC= 150^{\circ}$.
m$\angle$A+m$\angle$D=180 m$\angle$D=180-58 m$\angle$D=122$^{\circ}$ m$\angle$B+m$\angle$C=180 m$\angle$B+125=180 m$\angle$B=180-125 m$\angle$B=55$^{\circ}$