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

Circle the best answer.
Read this first
Use this
Given: DFGH is a parallelogram. Prove: DF ≅ GH. Ladder: 1. DFGH is a parallelogram -- Given; 2. DF ∥ HG and DH ∥ FG -- Definition of parallelogram; 3. ∠FDG ≅ ∠HGD -- Alternate interior angles theorem; 4. ∠HDG ≅ ∠FGD -- Alternate interior angles theorem; 5. DG ≅ DG -- Reflexive property; 6. triangle DFG ≅ triangle GHD -- ASA; 7. DF ≅ GH -- ?.
Milo proves the parallelogram opposite sides theorem on a window frame DFGH, using diagonal DG. Which reason fills the blank?
- Parallelogram opposite sides theorem
- ASA
- Given
- CPCTC
Use this
Given: DFGH is a parallelogram. Prove: GH ≅ DF. Ladder: 1. DFGH is a parallelogram -- Given; 2. DF ∥ HG and DH ∥ FG -- Definition of parallelogram; 3. ∠DFH ≅ ∠GHF -- Alternate interior angles theorem; 4. ∠GFH ≅ ∠DHF -- Alternate interior angles theorem; 5. FH ≅ FH -- Reflexive property; 6. triangle FGH ≅ triangle HDF -- ASA; 7. GH ≅ DF -- ?.
Noah proves the parallelogram opposite sides theorem on a rug DFGH, using diagonal FH. 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: arrows on DF and HG, and double arrows on DH and FG. Given: DFGH is a parallelogram. Prove: GH ≅ DF. Ladder: 1. DFGH is a parallelogram — Given; 2. DF ∥ HG and DH ∥ FG — Definition of parallelogram; 3. ∠DFH ≅ ∠GHF — Alternate interior angles theorem; 4. ∠GFH ≅ ∠DHF — Alternate interior angles theorem; 5. FH ≅ FH — Reflexive property; 6. △FGH ≅ △HDF — ASA; 7. GH ≅ DF — ?. 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. Given: DFGH is a parallelogram. Prove: GH ≅ DF. Ladder: 1. DFGH is a parallelogram — Given; 2. DF ∥ HG and DH ∥ FG — Definition of parallelogram; 3. ∠DFH ≅ ∠GHF — Alternate interior angles theorem; 4. ∠GFH ≅ ∠DHF — Alternate interior angles theorem; 5. FH ≅ FH — Reflexive property; 6. △FGH ≅ △HDF — ASA; 7. GH ≅ DF — ?. D F G H Q Not to scale. - Parallelogram opposite sides theorem
- ASA
- CPCTC
- Given
Use this
Given: DFGH is a parallelogram. Prove: ∠DFG ≅ ∠GHD. Ladder: 1. DFGH is a parallelogram -- Given; 2. DF ∥ HG and DH ∥ FG -- Definition of parallelogram; 3. ∠FDG ≅ ∠HGD -- ?; 4. ∠HDG ≅ ∠FGD -- Alternate interior angles theorem; 5. DG ≅ DG -- Reflexive property; 6. triangle DFG ≅ triangle GHD -- ASA; 7. ∠DFG ≅ ∠GHD -- CPCTC.
Pia proves the parallelogram opposite angles theorem on a tile DFGH, using diagonal DG. 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: arrows on DF and HG, and double arrows on DH and FG. Given: DFGH is a parallelogram. Prove: ∠DFG ≅ ∠GHD. Ladder: 1. DFGH is a parallelogram — Given; 2. DF ∥ HG and DH ∥ FG — Definition of parallelogram; 3. ∠FDG ≅ ∠HGD — ?; 4. ∠HDG ≅ ∠FGD — Alternate interior angles theorem; 5. DG ≅ DG — Reflexive property; 6. △DFG ≅ △GHD — ASA; 7. ∠DFG ≅ ∠GHD — CPCTC. 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. Given: DFGH is a parallelogram. Prove: ∠DFG ≅ ∠GHD. Ladder: 1. DFGH is a parallelogram — Given; 2. DF ∥ HG and DH ∥ FG — Definition of parallelogram; 3. ∠FDG ≅ ∠HGD — ?; 4. ∠HDG ≅ ∠FGD — Alternate interior angles theorem; 5. DG ≅ DG — Reflexive property; 6. △DFG ≅ △GHD — ASA; 7. ∠DFG ≅ ∠GHD — CPCTC. D F G H Q Not to scale. - Corresponding angles postulate
- Converse of the alternate interior angles theorem
- CPCTC
- Alternate interior angles theorem
Use this
Given: DFGH is a parallelogram. Prove: ∠FGH ≅ ∠HDF. Ladder: 1. DFGH is a parallelogram -- Given; 2. DF ∥ HG and DH ∥ FG -- Definition of parallelogram; 3. ∠DFH ≅ ∠GHF -- Alternate interior angles theorem; 4. ∠GFH ≅ ∠DHF -- ?; 5. FH ≅ FH -- Reflexive property; 6. triangle FGH ≅ triangle HDF -- ASA; 7. ∠FGH ≅ ∠HDF -- CPCTC.
Ren proves the parallelogram opposite angles theorem on a table top DFGH, using diagonal FH. 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: arrows on DF and HG, and double arrows on DH and FG. Given: DFGH is a parallelogram. Prove: ∠FGH ≅ ∠HDF. Ladder: 1. DFGH is a parallelogram — Given; 2. DF ∥ HG and DH ∥ FG — Definition of parallelogram; 3. ∠DFH ≅ ∠GHF — Alternate interior angles theorem; 4. ∠GFH ≅ ∠DHF — ?; 5. FH ≅ FH — Reflexive property; 6. △FGH ≅ △HDF — ASA; 7. ∠FGH ≅ ∠HDF — CPCTC. 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. Given: DFGH is a parallelogram. Prove: ∠FGH ≅ ∠HDF. Ladder: 1. DFGH is a parallelogram — Given; 2. DF ∥ HG and DH ∥ FG — Definition of parallelogram; 3. ∠DFH ≅ ∠GHF — Alternate interior angles theorem; 4. ∠GFH ≅ ∠DHF — ?; 5. FH ≅ FH — Reflexive property; 6. △FGH ≅ △HDF — ASA; 7. ∠FGH ≅ ∠HDF — CPCTC. Sketch, not to scale D F G H Q - Alternate exterior angles theorem
- Alternate interior angles theorem
- CPCTC
- Converse of the alternate interior angles theorem
Use this
Given: DFGH is a parallelogram. Prove: FG ≅ HD. Ladder: 1. DFGH is a parallelogram -- Given; 2. DF ∥ HG and DH ∥ FG -- Definition of parallelogram; 3. ∠FDG ≅ ∠HGD -- Alternate interior angles theorem; 4. ∠HDG ≅ ∠FGD -- Alternate interior angles theorem; 5. DG ≅ DG -- Reflexive property; 6. triangle DFG ≅ triangle GHD -- ?; 7. FG ≅ HD -- CPCTC.
Tala proves that FG ≅ HD on a parallelogram door panel DFGH, using diagonal DG. 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: arrows on DF and HG, and double arrows on DH and FG. Given: DFGH is a parallelogram. Prove: FG ≅ HD. Ladder: 1. DFGH is a parallelogram — Given; 2. DF ∥ HG and DH ∥ FG — Definition of parallelogram; 3. ∠FDG ≅ ∠HGD — Alternate interior angles theorem; 4. ∠HDG ≅ ∠FGD — Alternate interior angles theorem; 5. DG ≅ DG — Reflexive property; 6. △DFG ≅ △GHD — ?; 7. FG ≅ HD — CPCTC. 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. Given: DFGH is a parallelogram. Prove: FG ≅ HD. Ladder: 1. DFGH is a parallelogram — Given; 2. DF ∥ HG and DH ∥ FG — Definition of parallelogram; 3. ∠FDG ≅ ∠HGD — Alternate interior angles theorem; 4. ∠HDG ≅ ∠FGD — Alternate interior angles theorem; 5. DG ≅ DG — Reflexive property; 6. △DFG ≅ △GHD — ?; 7. FG ≅ HD — CPCTC. D F G H Q Not to scale. - SAS
- CPCTC
- ASA
- Definition of parallelogram
Use this
Given: DFGH is a parallelogram. Prove: FG ≅ HD. Ladder: 1. DFGH is a parallelogram -- Given; 2. DF ∥ HG and DH ∥ FG -- Definition of parallelogram; 3. ∠DFH ≅ ∠GHF -- Alternate interior angles theorem; 4. ∠GFH ≅ ∠DHF -- Alternate interior angles theorem; 5. FH ≅ FH -- Reflexive property; 6. triangle FGH ≅ triangle HDF -- ?; 7. FG ≅ HD -- CPCTC.
Zee proves that FG ≅ HD on a parallelogram quilt patch DFGH, using diagonal FH. 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: arrows on DF and HG, and double arrows on DH and FG. Given: DFGH is a parallelogram. Prove: FG ≅ HD. Ladder: 1. DFGH is a parallelogram — Given; 2. DF ∥ HG and DH ∥ FG — Definition of parallelogram; 3. ∠DFH ≅ ∠GHF — Alternate interior angles theorem; 4. ∠GFH ≅ ∠DHF — Alternate interior angles theorem; 5. FH ≅ FH — Reflexive property; 6. △FGH ≅ △HDF — ?; 7. FG ≅ HD — CPCTC. 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. Given: DFGH is a parallelogram. Prove: FG ≅ HD. Ladder: 1. DFGH is a parallelogram — Given; 2. DF ∥ HG and DH ∥ FG — Definition of parallelogram; 3. ∠DFH ≅ ∠GHF — Alternate interior angles theorem; 4. ∠GFH ≅ ∠DHF — Alternate interior angles theorem; 5. FH ≅ FH — Reflexive property; 6. △FGH ≅ △HDF — ?; 7. FG ≅ HD — CPCTC. D F G H Q Not to scale. - SSS
- SAS
- ASA
- Reflexive property
Use this
Given: DFGH is a rectangle. Prove: DG ≅ FH. Ladder: 1. DFGH is a rectangle -- Given; 2. DFGH is a parallelogram, and ∠DHG and ∠FGH are right angles -- Definition of rectangle; 3. ? -- Right angles congruence theorem; 4. DH ≅ FG -- Parallelogram opposite sides theorem; 5. HG ≅ HG -- Reflexive property; 6. triangle DHG ≅ triangle FGH -- SAS; 7. DG ≅ FH -- CPCTC.
Bo proves that the braces of a rectangle door DFGH are equal, using the top edge HG. 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: right-angle boxes at all four corners. Given: DFGH is a rectangle. Prove: DG ≅ FH. Ladder: 1. DFGH is a rectangle — Given; 2. DFGH is a parallelogram, and ∠DHG and ∠FGH are right angles — Definition of rectangle; 3. ? — Right angles congruence theorem; 4. DH ≅ FG — Parallelogram opposite sides theorem; 5. HG ≅ HG — Reflexive property; 6. △DHG ≅ △FGH — SAS; 7. DG ≅ FH — CPCTC. 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: right-angle boxes at all four corners. Given: DFGH is a rectangle. Prove: DG ≅ FH. Ladder: 1. DFGH is a rectangle — Given; 2. DFGH is a parallelogram, and ∠DHG and ∠FGH are right angles — Definition of rectangle; 3. ? — Right angles congruence theorem; 4. DH ≅ FG — Parallelogram opposite sides theorem; 5. HG ≅ HG — Reflexive property; 6. △DHG ≅ △FGH — SAS; 7. DG ≅ FH — CPCTC. Sketch, not to scale D F G H Q - ∠DHG ≅ ∠HGD
- DH ≅ FG
- DG ≅ FH
- ∠DHG ≅ ∠FGH
Esme proves that the diagonals of a rectangle book cover DFGH are equal, using the left edge DH. 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: right-angle boxes at all four corners. Given: DFGH is a rectangle. Prove: FH ≅ GD. Ladder: 1. DFGH is a rectangle — Given; 2. DFGH is a parallelogram, and ∠FDH and ∠GHD are right angles — Definition of rectangle; 3. ∠FDH ≅ ∠GHD — Right angles congruence theorem; 4. DF ≅ HG — Parallelogram opposite sides theorem; 5. DH ≅ DH — Reflexive property; 6. ? — SAS; 7. FH ≅ GD — CPCTC. 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: right-angle boxes at all four corners. Given: DFGH is a rectangle. Prove: FH ≅ GD. Ladder: 1. DFGH is a rectangle — Given; 2. DFGH is a parallelogram, and ∠FDH and ∠GHD are right angles — Definition of rectangle; 3. ∠FDH ≅ ∠GHD — Right angles congruence theorem; 4. DF ≅ HG — Parallelogram opposite sides theorem; 5. DH ≅ DH — Reflexive property; 6. ? — SAS; 7. FH ≅ GD — CPCTC. D F G H Q Not to scale - FH ≅ GD
- △FDH ≅ △FGH
- △FDH ≅ △DHG
- △FDH ≅ △GHD
Jun proves that the diagonals of a rectangle box lid DFGH are equal, using the right edge FG. 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: right-angle boxes at all four corners. Given: DFGH is a rectangle. Prove: DG ≅ HF. Ladder: 1. DFGH is a rectangle — Given; 2. DFGH is a parallelogram, and ∠DFG and ∠HGF are right angles — Definition of rectangle; 3. ∠DFG ≅ ∠HGF — Right angles congruence theorem; 4. DF ≅ HG — Parallelogram opposite sides theorem; 5. FG ≅ FG — Reflexive property; 6. ? — SAS; 7. DG ≅ HF — CPCTC. 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: right-angle boxes at all four corners. Given: DFGH is a rectangle. Prove: DG ≅ HF. Ladder: 1. DFGH is a rectangle — Given; 2. DFGH is a parallelogram, and ∠DFG and ∠HGF are right angles — Definition of rectangle; 3. ∠DFG ≅ ∠HGF — Right angles congruence theorem; 4. DF ≅ HG — Parallelogram opposite sides theorem; 5. FG ≅ FG — Reflexive property; 6. ? — SAS; 7. DG ≅ HF — CPCTC. D F G H Q Not to scale - △DFG ≅ △GFH
- DG ≅ HF
- △DFG ≅ △FGH
- △DFG ≅ △HGF
Use this
Given: DFGH is a parallelogram. Prove: HQ ≅ FQ. Ladder: 1. DFGH is a parallelogram -- Given; 2. DH ∥ FG -- Definition of parallelogram; 3. DH ≅ FG -- Parallelogram opposite sides theorem; 4. ? -- Alternate interior angles theorem; 5. ∠DHQ ≅ ∠GFQ -- Alternate interior angles theorem; 6. triangle DHQ ≅ triangle GFQ -- ASA; 7. HQ ≅ FQ -- CPCTC.
Luz proves that the folds of a parallelogram paper DFGH cut each other in half at Q. 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: arrows on DF and HG, and double arrows on DH and FG. Given: DFGH is a parallelogram. Prove: HQ ≅ FQ. Ladder: 1. DFGH is a parallelogram — Given; 2. DH ∥ FG — Definition of parallelogram; 3. DH ≅ FG — Parallelogram opposite sides theorem; 4. ? — Alternate interior angles theorem; 5. ∠DHQ ≅ ∠GFQ — Alternate interior angles theorem; 6. △DHQ ≅ △GFQ — ASA; 7. HQ ≅ FQ — CPCTC. 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. Given: DFGH is a parallelogram. Prove: HQ ≅ FQ. Ladder: 1. DFGH is a parallelogram — Given; 2. DH ∥ FG — Definition of parallelogram; 3. DH ≅ FG — Parallelogram opposite sides theorem; 4. ? — Alternate interior angles theorem; 5. ∠DHQ ≅ ∠GFQ — Alternate interior angles theorem; 6. △DHQ ≅ △GFQ — ASA; 7. HQ ≅ FQ — CPCTC. D F G H Q Not to scale. - ∠HDQ ≅ ∠FGQ
- HQ ≅ FQ
- ∠HDQ ≅ ∠QDF
- ∠HDQ ≅ ∠DFQ

