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 $ED$ is not parallel to $AC$, 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 $\angle BAC = \angle ACD = x$. - $\angle EAC = 36^\circ$ and $\angle CAB = 78^\circ$ are parts of $\angle BAC$, so $x = 78^\circ$. - In triangle $ACD$ on the circle, $\angle ACD = x = 78^\circ$. - $y$ is $\angle ADC$, which subtends arc $AC$. - By the circle theorem, angles subtending the same chord are equal, so $y = \angle EAC = 36^\circ$. - $z$ is $\angle AED$, which subtends arc $CD$. - Since $AB \parallel DC$, $\angle ABC = \angle BCD = y = 36^\circ$. 3. **Why $ED$ is not parallel to $AC$:** - If $ED \parallel AC$, corresponding angles would be equal. - But $\angle EAC = 36^\circ$ and $\angle AED = z$ are not equal (since $z$ is different from $36^\circ$), so $ED$ is not parallel to $AC$. 4. **Find $\angle EOC$:** - $\angle EOC$ is the central angle subtending arc $EC$. - The inscribed angle $\angle EAC = 36^\circ$ subtends the same arc $EC$. - Central angle is twice the inscribed angle, so $\angle EOC = 2 \times 36^\circ = 72^\circ$. 5. **Find $\angle ABC$ given $AB=AC$:** - Triangle $ABC$ is isosceles with $AB=AC$. - $\angle BAC = 78^\circ$. - Sum of angles in triangle $ABC$ is $180^\circ$. - Let $\angle ABC = \angle ACB = \theta$. - Then $78^\circ + 2\theta = 180^\circ \Rightarrow 2\theta = 102^\circ \Rightarrow \theta = 51^\circ$. **Final answers:** - $x = 78^\circ$ (alternate interior angles, parallel lines) - $y = 36^\circ$ (angles subtending same chord) - $z \neq 36^\circ$ so $ED \nparallel AC$ - $\angle EOC = 72^\circ$ (central angle twice inscribed angle) - $\angle ABC = 51^\circ$ (isosceles triangle angles sum)