1. **Problem statement:** Given pentagon ABCDE with circle center O passing through A, C, D, E, angles $\angle EAC=36^\circ$, $\angle CAB=78^\circ$, and $AB \parallel DC$. Find $x,y,z$ with reasons, explain why $ED \not\parallel AC$, find $\angle EOC$, and find $\angle ABC$ given $AB=AC$. 2. **Find $x,y,z$:** - Since $AB \parallel DC$, alternate interior angles are equal, so $x=\angle CAB=78^\circ$. - $\angle EAC=36^\circ$ is given. - $y$ and $z$ are angles on the circle at points $D$ and $E$ respectively. 3. **Use cyclic quadrilateral properties:** - Points $A,C,D,E$ lie on the circle, so opposite angles sum to $180^\circ$. - $\angle EAC$ and $\angle EDC$ are opposite angles in cyclic quadrilateral $ACDE$, so $\angle EDC = 180^\circ - 36^\circ = 144^\circ$. - Since $y=\angle EDC$, $y=144^\circ$. 4. **Find $z=\angle AED$:** - In triangle $AED$, angles sum to $180^\circ$. - $\angle EAC=36^\circ$ subtends arc $EC$, so $\angle EDC=144^\circ$ subtends the same arc. - $z=\angle AED$ subtends arc $AC$. - Arc $AC$ corresponds to $2 \times \angle EAC = 2 \times 36^\circ = 72^\circ$. - So $z=72^\circ$. 5. **Summary for part (a):** - $x=78^\circ$ (alternate interior angles from $AB \parallel DC$). - $y=144^\circ$ (opposite angle in cyclic quadrilateral). - $z=72^\circ$ (angle subtending arc $AC$). 6. **Part (b): Explain why $ED \not\parallel AC$:** - If $ED \parallel AC$, corresponding angles would be equal. - But $\angle AED = 72^\circ$ and $\angle EAC = 36^\circ$ are not equal. - Therefore, $ED$ is not parallel to $AC$. 7. **Part (c): Find $\angle EOC$:** - $O$ is center of circle. - Central angle $\angle EOC$ subtends arc $EC$. - Inscribed angle $\angle EAC=36^\circ$ subtends same arc. - Central angle is twice inscribed angle: $\angle EOC = 2 \times 36^\circ = 72^\circ$. 8. **Part (d): Find $\angle ABC$ given $AB=AC$:** - Triangle $ABC$ is isosceles with $AB=AC$. - Base angles $\angle ABC$ and $\angle ACB$ are equal. - $\angle CAB=78^\circ$ given. - Sum of angles in triangle $ABC$ is $180^\circ$. - So $\angle ABC + \angle ACB + 78^\circ = 180^\circ$. - Since $\angle ABC = \angle ACB$, $2 \times \angle ABC = 180^\circ - 78^\circ = 102^\circ$. - Therefore, $\angle ABC = 51^\circ$. **Final answers:** - $x=78^\circ$, $y=144^\circ$, $z=72^\circ$. - $ED$ is not parallel to $AC$ because corresponding angles differ. - $\angle EOC=72^\circ$. - $\angle ABC=51^\circ$.