Answer:
[tex]m\angle AED=70^{\circ}[/tex]
[tex]m\angle ACE=x=10^{\circ}[/tex]
Step-by-step explanation:
Consider quadrilateral ABDE. This quadrilateral is inscribed in the circle, so the sum of the measures of two opposite angles of quadrilateral is 180°. So,
[tex]m\angle AED+m\angle ABD=180^{\circ}\\ \\m\angle AED+110^{\circ}=180^{\circ}\\ \\m\angle AED=180^{\circ}-110^{\circ}=70^{\circ}[/tex]
Consider triangle ABD. The sum of the measures of all interior angles in triangle ABD is 180°, so
[tex]m\angle BAD+m\angle ABD+m\ange BDA=180^{\circ}\\ \\m\angle BAD+110^{\circ}+40^{\circ}=180^{\circ}\\ \\m\angle BAD=180^{\circ}-110^{\circ}-40^{\circ}=30^{\circ}[/tex]
Consider triangle ADE. This triangle is isosceles triangle, because AD = DE. Angles adjacent to the base AE are congruent, so
[tex]m\angle DAE=m\angle AED=70^{\circ}[/tex]
Consider triangle ACE. The sum of the measures of all interior angles in triangle ACE is 180°, so
[tex]m\angle CAE+m\angle ACE+m\ange CEA=180^{\circ}\\ \\m\angle CAD+m\angle DAE+m\angle ACE+m\ange CEA=180^{\circ}\\ \\m\angle ACE+30^{\circ}+70^{\circ}+70^{\circ}=180^{\circ}\\ \\m\angle ACE=180^{\circ}-30^{\circ}-70^{\circ}-70^{\circ}=10^{\circ}[/tex]
