1. Problem statement. In the cyclic pentagon ABCDE the points A, C, D and E lie on a circle with centre O, AB is parallel to DC, $\angle EAC = 36^\circ$ and $\angle CAB = 78^\circ$. Find the values of $x$, $y$ and $z$ as marked at C, D and E respectively. 2. Relevant facts and formulae. Inscribed angle theorem: an inscribed angle equals half the measure of its intercepted arc, i.e. $\text{inscribed angle} = \frac{1}{2}\times\text{(intercepted arc measure)}$. In a cyclic quadrilateral opposite angles are supplementary, i.e. $\angle_1+\angle_2 = 180^\circ$ for opposite angles. Parallel lines give equal alternate interior angles, so an angle formed with AB equals the corresponding angle formed with DC when AB \parallel DC. 3. Find $x$. Because AB is parallel to DC, the angle $\angle CAB$ equals the angle $\angle ACD$ by alternate interior angles. Hence $x = \angle ACD = \angle CAB = 78^\circ$. 4. Find $y$. The inscribed angle $\angle EAC = 36^\circ$ intercepts arc EC, so the measure of arc EC is $2\times36^\circ = 72^\circ$. The arc EC which does not contain A has measure $72^\circ$, so the other arc CE (the arc not through D) has measure $360^\circ-72^\circ = 288^\circ$. The inscribed angle at D, $\angle CDE$, intercepts that arc CE, so $\angle CDE = \tfrac{1}{2}\times 288^\circ$. $$\frac{288^\circ}{\cancel{2}} = 144^\circ$$ Therefore $y = 144^\circ$. (Equivalently, since $\angle EAC=36^\circ$ and A and D are opposite in the cyclic quadrilateral ACDE, $y=180^\circ-36^\circ=144^\circ$.) 5. Find $z$. In the cyclic quadrilateral ACDE the angles at C and E are opposite, so they are supplementary. Thus $z = 180^\circ - x = 180^\circ - 78^\circ = 102^\circ$. 6. Final answers. $x = 78^\circ$ by alternate interior angles from AB \parallel DC. $y = 144^\circ$ because opposite angles in cyclic quadrilateral ACDE sum to 180^\circ (or by inscribed-angle/arc calculation as shown). $z = 102^\circ$ because $z$ is supplementary to $x$ in the cyclic quadrilateral.