Quadrilateral proofs: fill in the missing step or reason
Grade 10 Math Worksheet - Challenge Me

Circle the best answer.
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Given: DF ∥ HG and DF ≅ HG. Prove: DFGH is a parallelogram. Ladder: 1. DF ∥ HG -- Given; 2. DF ≅ HG -- Given; 3. ∠FDG ≅ ∠HGD -- Alternate interior angles theorem; 4. DG ≅ DG -- Reflexive property; 5. triangle DFG ≅ triangle GHD -- SAS; 6. ∠FGD ≅ ∠HDG -- CPCTC; 7. DH ∥ FG -- ?; 8. DFGH is a parallelogram -- Definition of parallelogram.
Milo builds a table top DFGH with DF and HG parallel and the same length. Which reason fills the blank?
- Converse of the corresponding angles postulate
- Definition of parallelogram
- Alternate interior angles theorem
- Converse of the alternate interior angles theorem
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Marked: DF and HG with arrows and one tick each. Given: DF ∥ HG and DF ≅ HG. Prove: DFGH is a parallelogram. Ladder: 1. DF ∥ HG -- Given; 2. DF ≅ HG -- Given; 3. ∠DFH ≅ ∠GHF -- Alternate interior angles theorem; 4. FH ≅ FH -- Reflexive property; 5. triangle DFH ≅ triangle GHF -- SAS; 6. ∠DHF ≅ ∠GFH -- CPCTC; 7. DH ∥ FG -- ?; 8. DFGH is a parallelogram -- Definition of parallelogram.
Noah builds a window frame DFGH with DF and HG parallel and the same length. Which reason fills the blank?
Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: DF and HG with arrows and one tick each. Given: DF ∥ HG and DF ≅ HG. Prove: DFGH is a parallelogram. Ladder: 1. DF ∥ HG — Given; 2. DF ≅ HG — Given; 3. ∠DFH ≅ ∠GHF — Alternate interior angles theorem; 4. FH ≅ FH — Reflexive property; 5. △DFH ≅ △GHF — SAS; 6. ∠DHF ≅ ∠GFH — CPCTC; 7. DH ∥ FG — ?; 8. DFGH is a parallelogram — Definition of parallelogram. Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: DF and HG with arrows and one tick each. Given: DF ∥ HG and DF ≅ HG. Prove: DFGH is a parallelogram. Ladder: 1. DF ∥ HG — Given; 2. DF ≅ HG — Given; 3. ∠DFH ≅ ∠GHF — Alternate interior angles theorem; 4. FH ≅ FH — Reflexive property; 5. △DFH ≅ △GHF — SAS; 6. ∠DHF ≅ ∠GFH — CPCTC; 7. DH ∥ FG — ?; 8. DFGH is a parallelogram — Definition of parallelogram. Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. D F G H Q - Converse of the corresponding angles postulate
- Converse of the alternate interior angles theorem
- Alternate interior angles theorem
- Definition of parallelogram
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Given: DH ∥ FG and DH ≅ FG. Prove: DFGH is a parallelogram. Ladder: 1. DH ∥ FG -- Given; 2. DH ≅ FG -- Given; 3. ∠HDG ≅ ∠FGD -- Alternate interior angles theorem; 4. DG ≅ DG -- Reflexive property; 5. triangle HDG ≅ triangle FGD -- ?; 6. ∠HGD ≅ ∠FDG -- CPCTC; 7. DF ∥ HG -- Converse of the alternate interior angles theorem; 8. DFGH is a parallelogram -- Definition of parallelogram.
Pia builds a door frame DFGH with DH and FG parallel and the same length. Which reason fills the blank?
Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: DH and FG with arrows and one tick each. Given: DH ∥ FG and DH ≅ FG. Prove: DFGH is a parallelogram. Ladder: 1. DH ∥ FG — Given; 2. DH ≅ FG — Given; 3. ∠HDG ≅ ∠FGD — Alternate interior angles theorem; 4. DG ≅ DG — Reflexive property; 5. △HDG ≅ △FGD — ?; 6. ∠HGD ≅ ∠FDG — CPCTC; 7. DF ∥ HG — Converse of the alternate interior angles theorem; 8. DFGH is a parallelogram — Definition of parallelogram. Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: DH and FG with arrows and one tick each. Given: DH ∥ FG and DH ≅ FG. Prove: DFGH is a parallelogram. Ladder: 1. DH ∥ FG — Given; 2. DH ≅ FG — Given; 3. ∠HDG ≅ ∠FGD — Alternate interior angles theorem; 4. DG ≅ DG — Reflexive property; 5. △HDG ≅ △FGD — ?; 6. ∠HGD ≅ ∠FDG — CPCTC; 7. DF ∥ HG — Converse of the alternate interior angles theorem; 8. DFGH is a parallelogram — Definition of parallelogram. D F G H Q Not to scale. - SAS
- CPCTC
- ASA
- SSS
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Given: DH ∥ FG and DH ≅ FG. Prove: DFGH is a parallelogram. Ladder: 1. DH ∥ FG -- Given; 2. DH ≅ FG -- Given; 3. ∠DHF ≅ ∠GFH -- Alternate interior angles theorem; 4. FH ≅ FH -- Reflexive property; 5. triangle DHF ≅ triangle GFH -- ?; 6. ∠DFH ≅ ∠GHF -- CPCTC; 7. DF ∥ HG -- Converse of the alternate interior angles theorem; 8. DFGH is a parallelogram -- Definition of parallelogram.
Ren builds a box lid DFGH with DH and FG parallel and the same length. Which reason fills the blank?
Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: DH and FG with arrows and one tick each. Given: DH ∥ FG and DH ≅ FG. Prove: DFGH is a parallelogram. Ladder: 1. DH ∥ FG — Given; 2. DH ≅ FG — Given; 3. ∠DHF ≅ ∠GFH — Alternate interior angles theorem; 4. FH ≅ FH — Reflexive property; 5. △DHF ≅ △GFH — ?; 6. ∠DFH ≅ ∠GHF — CPCTC; 7. DF ∥ HG — Converse of the alternate interior angles theorem; 8. DFGH is a parallelogram — Definition of parallelogram. Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: DH and FG with arrows and one tick each. Given: DH ∥ FG and DH ≅ FG. Prove: DFGH is a parallelogram. Ladder: 1. DH ∥ FG — Given; 2. DH ≅ FG — Given; 3. ∠DHF ≅ ∠GFH — Alternate interior angles theorem; 4. FH ≅ FH — Reflexive property; 5. △DHF ≅ △GFH — ?; 6. ∠DFH ≅ ∠GHF — CPCTC; 7. DF ∥ HG — Converse of the alternate interior angles theorem; 8. DFGH is a parallelogram — Definition of parallelogram. D F G H Q Not to scale. - ASA
- SAS
- CPCTC
- SSS
Tala checks that stick DG of a rhombus kite DFGH splits the corner at D evenly. Which statement fills the blank?
Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: one tick on each side. Given: DFGH is a rhombus. Prove: DG bisects ∠HDF. Ladder: 1. DFGH is a rhombus — Given; 2. DFGH is a parallelogram, DF ≅ DH and FG ≅ HG — Definition of rhombus; 3. DG ≅ DG — Reflexive property; 4. ? — SSS; 5. ∠FDG ≅ ∠HDG — CPCTC; 6. DG bisects ∠HDF — Definition of angle bisector. Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: one tick on each side. Given: DFGH is a rhombus. Prove: DG bisects ∠HDF. Ladder: 1. DFGH is a rhombus — Given; 2. DFGH is a parallelogram, DF ≅ DH and FG ≅ HG — Definition of rhombus; 3. DG ≅ DG — Reflexive property; 4. ? — SSS; 5. ∠FDG ≅ ∠HDG — CPCTC; 6. DG bisects ∠HDF — Definition of angle bisector. D F G H Q Not to scale - △DFG ≅ △DHG
- △DFG ≅ △HDG
- △DFG ≅ △QFG
- DG bisects ∠HDF
Use this
Given: DFGH is a rhombus. Prove: DG bisects ∠FGH. Ladder: 1. DFGH is a rhombus -- Given; 2. DFGH is a parallelogram, DF ≅ DH and FG ≅ HG -- Definition of rhombus; 3. DG ≅ DG -- Reflexive property; 4. triangle DFG ≅ triangle DHG -- SSS; 5. ? -- CPCTC; 6. DG bisects ∠FGH -- Definition of angle bisector.
Zee checks that the line DG on a rhombus tile DFGH splits the corner at G evenly. Which statement fills the blank?
Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: one tick on each side. Given: DFGH is a rhombus. Prove: DG bisects ∠FGH. Ladder: 1. DFGH is a rhombus — Given; 2. DFGH is a parallelogram, DF ≅ DH and FG ≅ HG — Definition of rhombus; 3. DG ≅ DG — Reflexive property; 4. △DFG ≅ △DHG — SSS; 5. ? — CPCTC; 6. DG bisects ∠FGH — Definition of angle bisector. Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: one tick on each side. Given: DFGH is a rhombus. Prove: DG bisects ∠FGH. Ladder: 1. DFGH is a rhombus — Given; 2. DFGH is a parallelogram, DF ≅ DH and FG ≅ HG — Definition of rhombus; 3. DG ≅ DG — Reflexive property; 4. △DFG ≅ △DHG — SSS; 5. ? — CPCTC; 6. DG bisects ∠FGH — Definition of angle bisector. D F G H Q Not to scale. - ∠FGD ≅ ∠HGD
- ∠FGD ≅ ∠FGH
- ∠FGD ≅ ∠DFG
- DG bisects ∠FGH
Ari checks that the fold FH of a rhombus envelope flap DFGH splits the corner at F evenly. Which statement fills the blank?
Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: one tick on each side. Given: DFGH is a rhombus. Prove: FH bisects ∠DFG. Ladder: 1. DFGH is a rhombus — Given; 2. DFGH is a parallelogram, FD ≅ FG and DH ≅ GH — Definition of rhombus; 3. FH ≅ FH — Reflexive property; 4. ? — SSS; 5. ∠DFH ≅ ∠GFH — CPCTC; 6. FH bisects ∠DFG — Definition of angle bisector. Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: one tick on each side. Given: DFGH is a rhombus. Prove: FH bisects ∠DFG. Ladder: 1. DFGH is a rhombus — Given; 2. DFGH is a parallelogram, FD ≅ FG and DH ≅ GH — Definition of rhombus; 3. FH ≅ FH — Reflexive property; 4. ? — SSS; 5. ∠DFH ≅ ∠GFH — CPCTC; 6. FH bisects ∠DFG — Definition of angle bisector. D F G H Q Not to scale - FH bisects ∠DFG
- △FDH ≅ △QFG
- △FDH ≅ △FGH
- △FDH ≅ △GFH
Use this
Ladder: 1. DFGH is a rhombus -- Given; 2. DFGH is a parallelogram, and GF ≅ GH -- Definition of rhombus; 3. FQ ≅ QH -- Parallelogram diagonals theorem; 4. GQ ≅ GQ -- Reflexive property; 5. triangle FQG ≅ triangle HQG -- SSS; 6. ∠FQG ≅ ∠HQG -- CPCTC; 7. ? -- ?.
Jun proves the rhombus diagonals theorem on a rhombus tile DFGH, using the triangles at G. Which statement and reason fill the blanks?
Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: one tick on each side. Given: DFGH is a rhombus. Prove: FH ⊥ DG. Ladder: 1. DFGH is a rhombus — Given; 2. DFGH is a parallelogram, and GF ≅ GH — Definition of rhombus; 3. FQ ≅ QH — Parallelogram diagonals theorem; 4. GQ ≅ GQ — Reflexive property; 5. △FQG ≅ △HQG — SSS; 6. ∠FQG ≅ ∠HQG — CPCTC; 7. ? — ?. Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: one tick on each side. Given: DFGH is a rhombus. Prove: FH ⊥ DG. Ladder: 1. DFGH is a rhombus — Given; 2. DFGH is a parallelogram, and GF ≅ GH — Definition of rhombus; 3. FQ ≅ QH — Parallelogram diagonals theorem; 4. GQ ≅ GQ — Reflexive property; 5. △FQG ≅ △HQG — SSS; 6. ∠FQG ≅ ∠HQG — CPCTC; 7. ? — ?. D F G H Q Not to scale. - FH ≅ DG — Rectangle diagonals theorem
- FH ∥ DG — Linear pair perpendicular theorem
- FH ⊥ DG — Linear pair perpendicular theorem
- FH ⊥ DG — Rhombus diagonals theorem
Kira measures a 72° corner at D on a parallelogram rug DFGH. Which statement and reason fill the blanks?
Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: arrows on DF and HG, and double arrows on DH and FG; ∠HDF labeled 72°. Given: DFGH is a parallelogram, and m∠HDF = 72. Prove: m∠DFG = 108. Ladder: 1. DFGH is a parallelogram — Given; 2. m∠HDF = 72 — Given; 3. DH ∥ FG — Definition of parallelogram; 4. ? — ?; 5. m∠HDF + m∠DFG = 180 — Definition of supplementary angles; 6. 72 + m∠DFG = 180 — Substitution property; 7. m∠DFG = 108 — Subtraction property of equality. Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: arrows on DF and HG, and double arrows on DH and FG; ∠HDF labeled 72°. Given: DFGH is a parallelogram, and m∠HDF = 72. Prove: m∠DFG = 108. Ladder: 1. DFGH is a parallelogram — Given; 2. m∠HDF = 72 — Given; 3. DH ∥ FG — Definition of parallelogram; 4. ? — ?; 5. m∠HDF + m∠DFG = 180 — Definition of supplementary angles; 6. 72 + m∠DFG = 180 — Substitution property; 7. m∠DFG = 108 — Subtraction property of equality. D F G H Q 72° Not to scale - ∠HDF and ∠DFG are supplementary — Same-side interior angles theorem
- ∠HDF and ∠DFG are complementary — Same-side interior angles theorem
- ∠HDF and ∠DFG are supplementary — Alternate interior angles theorem
- ∠HDF ≅ ∠DFG — Parallelogram opposite angles theorem
Luz measures a 64° corner at F on a parallelogram table top DFGH. Which statement and reason fill the blanks?
Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: arrows on DF and HG, and double arrows on DH and FG; ∠DFG labeled 64°. Given: DFGH is a parallelogram, and m∠DFG = 64. Prove: m∠FGH = 116. Ladder: 1. DFGH is a parallelogram — Given; 2. m∠DFG = 64 — Given; 3. DF ∥ HG — Definition of parallelogram; 4. ? — ?; 5. m∠DFG + m∠FGH = 180 — Definition of supplementary angles; 6. 64 + m∠FGH = 180 — Substitution property; 7. m∠FGH = 116 — Subtraction property of equality. Sketch, not to scale: quadrilateral DFGH (D bottom left, F bottom right, G top right, H top left), with diagonals DG and FH crossing at Q. Marked: arrows on DF and HG, and double arrows on DH and FG; ∠DFG labeled 64°. Given: DFGH is a parallelogram, and m∠DFG = 64. Prove: m∠FGH = 116. Ladder: 1. DFGH is a parallelogram — Given; 2. m∠DFG = 64 — Given; 3. DF ∥ HG — Definition of parallelogram; 4. ? — ?; 5. m∠DFG + m∠FGH = 180 — Definition of supplementary angles; 6. 64 + m∠FGH = 180 — Substitution property; 7. m∠FGH = 116 — Subtraction property of equality. D F G H Q 64° Not to scale - ∠DFG and ∠FGH are supplementary — Same-side interior angles theorem
- ∠DFG and ∠FGH are supplementary — Alternate interior angles theorem
- ∠DFG ≅ ∠FGH — Parallelogram opposite angles theorem
- ∠DFG and ∠FGH are complementary — Same-side interior angles theorem