1. **Problem 1: Rectangle JKLM with diagonals intersecting at N.** Given: $JN = x + 3$ and $JL = 3x + 1$. We need to find which values among A to E are correct. 2. **Key properties:** - In a rectangle, diagonals are equal in length. - Point $N$ is the intersection of diagonals $JN$ and $LM$, so $N$ is the midpoint of both diagonals. - Therefore, $JN = NL$ and $JN = NM$ (since diagonals bisect each other). 3. **Step 1: Find $x$ using the diagonal length $JL$** - $JL$ is the full diagonal, so $JL = 3x + 1$. - Since $N$ is midpoint, $JN$ is half of $JL$, so: $$JN = \frac{JL}{2} = \frac{3x + 1}{2}$$ - But given $JN = x + 3$, so: $$x + 3 = \frac{3x + 1}{2}$$ 4. **Step 2: Solve for $x$** - Multiply both sides by 2: $$2(x + 3) = 3x + 1$$ $$2x + 6 = 3x + 1$$ - Rearranged: $$6 - 1 = 3x - 2x$$ $$5 = x$$ 5. **Step 3: Calculate lengths** - $JN = x + 3 = 5 + 3 = 8$ - $JL = 3x + 1 = 3(5) + 1 = 16$ - Since $N$ is midpoint, $NL = JN = 8$ - $NM = NL = 8$ - $KM$ is a side of the rectangle; since rectangle sides are not given, we cannot determine $KM$ directly. 6. **Step 4: Check options** - A. $NL = 4$? No, $NL = 8$ - B. $JN = 5$? No, $JN = 8$ - C. $NM = 8$? Yes - D. $KM = 10$? Cannot confirm without side lengths - E. $JL = 16$? Yes **Correct answers for problem 1:** C and E. --- 7. **Problem 2: Parallelogram ABCD with point E inside, triangles AED and CEB.** Given: ABCD is a parallelogram. 8. **Goal:** Determine which triangle congruence theorems can prove $\triangle AED \cong \triangle CEB$. 9. **Key properties:** - Opposite sides of parallelogram are equal and parallel. - Diagonals bisect each other. - Triangles $AED$ and $CEB$ share certain sides and angles. 10. **Step 1: Analyze triangles** - $AE$ and $CE$ are parts of diagonal $AC$. - $DE$ and $BE$ are parts of diagonal $BD$. - Since diagonals bisect each other, $E$ is midpoint of both diagonals, so: $$AE = CE$$ $$DE = BE$$ 11. **Step 2: Identify congruence criteria** - Side $AE = CE$ - Side $DE = BE$ - Angle $AED$ and angle $CEB$ are vertical angles, so: $$\angle AED = \angle CEB$$ 12. **Step 3: Apply triangle congruence theorems** - SAS (Side-Angle-Side): Two sides and the included angle are equal. - ASA (Angle-Side-Angle): Two angles and the included side are equal. - SSS (Side-Side-Side): All three sides equal (not confirmed here). - AA (Angle-Angle): Only similarity, not congruence. 13. **Step 4: Conclusion** - SAS applies because two sides and included angle are equal. - ASA applies because two angles and included side are equal. - SSS cannot be confirmed without third side. - AA proves similarity, not congruence. **Correct answers for problem 2:** B (ASA), C (SAS), and possibly D (SSS) if third side equality is known, but since not given, exclude D. --- **Final answers:** - Problem 1: C and E - Problem 2: B and C **Note:** The user asked to select three for problem 2, so including A (AA) is incorrect for congruence, so only B, C, and possibly D if more info is given. Since D is uncertain, select B, C, and exclude A and D. "slug":"rectangle diagonals","subject":"geometry","content":"1. **Problem 1: Rectangle JKLM with diagonals intersecting at N.**\n\nGiven: $JN = x + 3$ and $JL = 3x + 1$. We need to find which values among A to E are correct.\n\n2. **Key properties:**\n- In a rectangle, diagonals are equal in length.\n- Point $N$ is the intersection of diagonals $JN$ and $LM$, so $N$ is the midpoint of both diagonals.\n- Therefore, $JN = NL$ and $JN = NM$ (since diagonals bisect each other).\n\n3. **Step 1: Find $x$ using the diagonal length $JL$**\n- $JL$ is the full diagonal, so $JL = 3x + 1$.\n- Since $N$ is midpoint, $JN$ is half of $JL$, so:\n $$JN = \\frac{JL}{2} = \\frac{3x + 1}{2}$$\n- But given $JN = x + 3$, so:\n $$x + 3 = \\frac{3x + 1}{2}$$\n\n4. **Step 2: Solve for $x$**\n- Multiply both sides by 2:\n $$2(x + 3) = 3x + 1$$\n $$2x + 6 = 3x + 1$$\n- Rearranged:\n $$6 - 1 = 3x - 2x$$\n $$5 = x$$\n\n5. **Step 3: Calculate lengths**\n- $JN = x + 3 = 5 + 3 = 8$\n- $JL = 3x + 1 = 3(5) + 1 = 16$\n- Since $N$ is midpoint, $NL = JN = 8$\n- $NM = NL = 8$\n- $KM$ is a side of the rectangle; since rectangle sides are not given, we cannot determine $KM$ directly.\n\n6. **Step 4: Check options**\n- A. $NL = 4$? No, $NL = 8$\n- B. $JN = 5$? No, $JN = 8$\n- C. $NM = 8$? Yes\n- D. $KM = 10$? Cannot confirm without side lengths\n- E. $JL = 16$? Yes\n\n**Correct answers for problem 1:** C and E.\n\n---\n\n7. **Problem 2: Parallelogram ABCD with point E inside, triangles AED and CEB.**\n\nGiven: ABCD is a parallelogram.\n\n8. **Goal:** Determine which triangle congruence theorems can prove $\\triangle AED \\cong \\triangle CEB$.\n\n9. **Key properties:**\n- Opposite sides of parallelogram are equal and parallel.\n- Diagonals bisect each other.\n- Triangles $AED$ and $CEB$ share certain sides and angles.\n\n10. **Step 1: Analyze triangles**\n- $AE$ and $CE$ are parts of diagonal $AC$.\n- $DE$ and $BE$ are parts of diagonal $BD$.\n- Since diagonals bisect each other, $E$ is midpoint of both diagonals, so:\n $$AE = CE$$\n $$DE = BE$$\n\n11. **Step 2: Identify congruence criteria**\n- Side $AE = CE$\n- Side $DE = BE$\n- Angle $AED$ and angle $CEB$ are vertical angles, so:\n $$\\angle AED = \\angle CEB$$\n\n12. **Step 3: Apply triangle congruence theorems**\n- SAS (Side-Angle-Side): Two sides and the included angle are equal.\n- ASA (Angle-Side-Angle): Two angles and the included side are equal.\n- SSS (Side-Side-Side): All three sides equal (not confirmed here).\n- AA (Angle-Angle): Only similarity, not congruence.\n\n13. **Step 4: Conclusion**\n- SAS applies because two sides and included angle are equal.\n- ASA applies because two angles and included side are equal.\n- SSS cannot be confirmed without third side.\n- AA proves similarity, not congruence.\n\n**Correct answers for problem 2:** B (ASA), C (SAS).\n\n---\n\n**Final answers:**\n- Problem 1: C and E\n- Problem 2: B and C\n\n