1. The problem states that triangle ABD is isosceles with AB = AD. 2. Given angles are \(\angle D = 74^\circ\) and \(\angle C = 28^\circ\). \(\angle B\) is marked as \(x\), and angles at \(A\) are split into \(y\) and \(z\). 3. Since AB = AD, triangle ABD is isosceles, so \(\angle B = \angle D = 74^\circ\). 4. The triangle ABD has angles \(\angle A = y + z\), \(\angle B = 74^\circ\), and \(\angle D = 74^\circ\). 5. The sum of angles in triangle ABD is \(180^\circ\), so: $$y + z + 74 + 74 = 180$$ 6. Simplify the sum of known angles: $$y + z + 148 = 180$$ $$y + z = 32$$ 7. Angle \(C = 28^\circ\) is connected to the split of \(A\) into \(y\) and \(z\), likely indicating that \(y = 28^\circ\). 8. Substituting \(y = 28^\circ\), solve for \(z\): $$28 + z = 32$$ $$z = 32 - 28 = 4$$ Final answer: \(\boxed{4^\circ}\) is the size of angle \(z\).