(02.04 HC) C is the circumcenter of isosceles triangle ABD with vertex angle ∠ABD. Does the following proof correctly justify that triangles ABE and DBE are congruent? It is given that triangle ABD is an isosceles triangle, so segments AB and DB are congruent by the definition of isosceles triangle. It is given that C is the circumcenter of triangle ABD, making segment BE a median. By the definition of perpendicular, angles AEB and DEB are 90°, so triangles ABE and DEB are right triangles. Triangles ABE and DEB share side BE making it congruent to itself by the reflexive property. Triangles ABE and DBE are congruent by HL.

(01.06 MC) Solve for x.

(02.01 MC) Triangle PAT has been reflected over the x-axis. Which of the following best describes the relationship between the x-axis and the line connecting P to P′?

(02.03 LC) If ΔABC ≅ ΔDEF, then what corresponding parts are congruent?

(02.06 MC) Look at the quadrilateral shown below: Melissa writes the following proof for the theorem: If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram: Melissa’s proof For triangles AOB and COD, angle 1 is equal to angle 2, as they are vertical angles. AO = OC and BO = OD because it is given that diagonals bisect each other. The ________ are congruent by SAS postulate. Similarly, triangles AOD and COB are congruent. By CPCTC, AB is equal to DC. By CPCTC, AD is equal to BC. As the opposite sides are congruent, the quadrilateral ABCD is a parallelogram. Which is the missing phrase in Melissa’s proof?

(01.07 MC) Daniel is constructing a fence that consists of parallel sides and . Complete the proof to explain how he can show that m∠GKB = 120° by filling in the missing justifications.   Statement Justification ∥ m∠ELJ = 120° Given m∠ELJ + m∠ELK = 180° Linear Pair Postulate m∠BKL + m∠GKB = 180° Linear Pair Postulate m∠ELJ + m∠ELK = m∠BKL + m∠GKB Transitive Property ∠ELK ≅ ∠BKL 1. m∠ELK = m∠BKL 2. m∠ELJ + m∠ELK = m∠ELK + m∠GKB Substitution Property m∠ELJ = m∠GKB Subtraction Property m∠GKB = m∠ELJ Symmetric Property m∠GKB = 120° Substitution

(01.03 MC) Jaira is completing construction of a regular hexagon inscribed in a circle, as shown below: What should be the next step in her construction?

(01.02 MC) Ben uses a compass and straightedge to bisect , as shown: Which statement best explains why Ben must open the compass to a width greater than half of ?

(02.03 MC) Angle N = 40 degrees, side NP = 8, angle Q = 40 degrees, and side QS = 8. What additional information would you need to prove that ΔNOP ≅ ΔQRS by ASA?

(02.06 MC) Look at the quadrilateral shown below: Terra writes the following proof for the theorem: If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram: Terra’s proof AO = OC because it is given that diagonals bisect each other. BO = OD because it is given that diagonals bisect each other. For triangles AOB and COD, angle 1 is equal to angle 2, as they are ________. Therefore, the triangles AOB and COD are congruent by SAS postulate. Similarly, triangles AOD and COB are congruent. By CPCTC, angle ABD is equal to angle BDC and angle ADB is equal to angle DBC. As the alternate interior angles are congruent, the opposite sides of quadrilateral ABCD are parallel. Therefore, ABCD is a parallelogram. Which is the missing phrase in Terra’s proof?