1. **Problem 1: Given polygons ABCH and EDCH are congruent, with C on BD, m(\angle AHC) = 110^\circ, BC = 5 cm. Find the requested values.\n\n2. Since ABCH \cong EDCH, corresponding sides and angles are equal.\n\n3. (1) AB = ED (corresponding sides). Since BC = 5 cm and BC corresponds to DC, and polygons share side DC, AB = ED = 5 cm.\n\n4. (2) AH corresponds to EH, so AH = EH. We need more info to find length, but since polygons are congruent, AH = EH.\n\n5. (3) m(\angle E) = m(\angle A) (corresponding angles in congruent polygons).\n\n6. (4) m(\angle B) = m(\angle D) (corresponding angles).\n\n7. (5) m(\angle HCB) = 70^\circ because \angle AHC = 110^\circ and angles on a straight line sum to 180^\circ, so 180^\circ - 110^\circ = 70^\circ.\n\n8. (6) m(\angle EHC) = m(\angle AHC) = 110^\circ (corresponding angles).\n\n9. (7) BD = BC + CD = 5 + 5 = 10 cm (assuming CD = BC from congruence).\n\n10. (8) m(\angle AHE) = 110^\circ (since AH corresponds to EH and angles at H correspond).\n\n---\n\n11. **Problem 2: Polygon EFGH is image of ABCD by rotation. Find x and y.\n\n12. Rotation preserves side lengths and angles. Given GF = 6 cm, so AB = 6 cm.\n\n13. Angles x and y correspond to angles in ABCD. Use given angle measures to solve for x and y.\n\n---\n\n14. **Problem 3: Triangle BCD is image of BAD by reflection in DB. Find x and y.\n\n15. Reflection preserves distances and angles relative to the mirror line DB.\n\n16. Given \angle DAB = 30^\circ, \angle DBA = 2y^\circ, and \angle BCD = 100^\circ. Use triangle angle sum and reflection properties to find x and y.\n\n---\n\n**Final answers:**\n① AB = 5 cm\n② AH = EH (length equal, exact value depends on figure)\n③ m(\angle E) = m(\angle A)\n④ m(\angle B) = m(\angle D)\n⑤ m(\angle HCB) = 70^\circ\n⑥ m(\angle EHC) = 110^\circ\n⑦ BD = 10 cm\n⑧ m(\angle AHE) = 110^\circ\n\nx and y values depend on given angles and side lengths in problems 2 and 3, requiring solving equations from rotation and reflection properties.