1. **Stating the problem:** We have a quadrilateral ABCD with angles $x^\circ$ at A, $y^\circ$ at D, $38^\circ$ at C, and a right angle (90$^\circ$) at B. Sides AD and DC are equal, so $AD = DC$. We need to find the relationship between the angles $x$ and $y$. 2. **Formula and rules:** The sum of interior angles in any quadrilateral is 360$^\circ$. So, $$x + 90 + 38 + y = 360$$ Also, since $AD = DC$, triangle ADC is isosceles, so angles opposite these equal sides are equal. The angles at A and D in triangle ADC are equal, so $$x = y$$ 3. **Intermediate work:** Substitute $x = y$ into the angle sum equation: $$x + 90 + 38 + x = 360$$ $$2x + 128 = 360$$ 4. **Simplify:** $$2x = 360 - 128$$ $$2x = 232$$ 5. **Solve for $x$:** $$x = \frac{232}{2}$$ $$x = 116$$ 6. **Find $y$:** Since $x = y$, $$y = 116$$ **Final answer:** $$x = 116^\circ, \quad y = 116^\circ$$