1. Problem: Prove triangle HGE is congruent to triangle FGE given that segment GE is an angle bisector of both angle HEF and angle FGH. Step 1: Since GE bisects angle HEF, we have \(\angle HGE = \angle EGF\). Step 2: GE is common to both triangles HGE and FGE, so \(GE = GE\) (common side). Step 3: Since GE bisects angle FGH, \(\angle HGE = \angle FGE\). Step 4: Triangles HGE and FGE share side GE and have two pairs of equal angles, so by the Angle-Side-Angle (ASA) congruence criterion, \(\triangle HGE \cong \triangle FGE\). 2. Problem: Triangles ACD and BCD are isosceles. Given \(\angle BAC = 33^\circ\) and \(\angle BDC = 35^\circ\), find \(\angle ABD\). Step 1: Since ACD and BCD are isosceles, sides AC = DC and BC = DC respectively. Step 2: In triangle ACD, base angles are equal. Let \(\angle CAD = \angle CDA = x\). Step 3: Sum of angles in triangle ACD: \(x + x + 33^\circ = 180^\circ\) so \(2x = 147^\circ\) and \(x = 73.5^\circ\). Step 4: In triangle BCD, base angles are equal. Let \(\angle CBD = \angle BDC = y\). Step 5: Given \(\angle BDC = 35^\circ\), so \(y = 35^\circ\). Step 6: Sum of angles in triangle BCD: \(y + y + \angle BCD = 180^\circ\) so \(2(35^\circ) + \angle BCD = 180^\circ\) and \(\angle BCD = 110^\circ\). Step 7: Since points A, B, C, D are arranged with B inside, \(\angle ABD = 180^\circ - (\angle BAC + \angle BCD) = 180^\circ - (33^\circ + 110^\circ) = 37^\circ\). 3. Problem: Which conjecture is possible to prove? Step 1: Analyze each option: A. All triangles with at least one side length of 5 are congruent - False, triangles can vary in other sides and angles. B. All pentagons with at least one side length of 5 are congruent - False, pentagons can have many different shapes. C. All rectangles with at least one side length of 5 are congruent - False, rectangles can have different other side lengths. D. All squares with at least one side length of 5 are congruent - True, because all sides of a square are equal, so if one side is 5, all sides are 5, making all such squares congruent. Final answer: D is possible to prove.