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

Circle the best answer.
Milo draws a leaning window frame DFGH, a parallelogram. Which reason fills the blank?
- Given
- Parallelogram opposite sides theorem
- Definition of parallelogram
- Definition of midpoint
Noah sketches a slanted door panel DFGH, a parallelogram. 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: DF ≅ HG. Ladder: 1. DFGH is a parallelogram — Given; 2. DF ≅ HG — ?. 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: DF ≅ HG. Ladder: 1. DFGH is a parallelogram — Given; 2. DF ≅ HG — ?. D F G H Q Not to scale - Given
- Parallelogram opposite sides theorem
- Definition of parallelogram
- Reflexive property
Ren builds a parallelogram frame DFGH with two sticks crossing at Q. 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: DQ ≅ QG. Ladder: 1. DFGH is a parallelogram — Given; 2. DQ ≅ QG — ?. 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: DQ ≅ QG. Ladder: 1. DFGH is a parallelogram — Given; 2. DQ ≅ QG — ?. D F G H Q Not to scale - Parallelogram diagonals theorem
- Rectangle diagonals theorem
- Definition of midpoint
- Given
Tala flies a rhombus-shaped kite DFGH whose sticks cross at Q. 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: one tick on each side. Given: DFGH is a rhombus. Prove: ∠DQF is a right angle. Ladder: 1. DFGH is a rhombus — Given; 2. DG ⊥ FH — ?; 3. ∠DQF is a right angle — Definition of perpendicular lines. 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: ∠DQF is a right angle. Ladder: 1. DFGH is a rhombus — Given; 2. DG ⊥ FH — ?; 3. ∠DQF is a right angle — Definition of perpendicular lines. D F G H Q Not to scale - Given
- Rectangle diagonals theorem
- Parallelogram diagonals theorem
- Rhombus diagonals theorem
Zee braces a rectangle door DFGH with two boards from corner to corner. 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: right-angle boxes at all four corners. Given: DFGH is a rectangle. Prove: DG = FH. Ladder: 1. DFGH is a rectangle — Given; 2. DG ≅ FH — ?; 3. DG = FH — Definition of congruent segments. 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. DG ≅ FH — ?; 3. DG = FH — Definition of congruent segments. D F G H Q Not to scale - Parallelogram diagonals theorem
- Reflexive property
- Rectangle diagonals theorem
- Rhombus diagonals theorem
Ari lays a parallelogram rug DFGH on a floor. 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: m∠HDF + m∠DFG = 180. Ladder: 1. DFGH is a parallelogram — Given; 2. ∠HDF and ∠DFG are supplementary — ?; 3. m∠HDF + m∠DFG = 180 — Definition of supplementary angles. 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: m∠HDF + m∠DFG = 180. Ladder: 1. DFGH is a parallelogram — Given; 2. ∠HDF and ∠DFG are supplementary — ?; 3. m∠HDF + m∠DFG = 180 — Definition of supplementary angles. D F G H Q Not to scale - Given
- Parallelogram opposite angles theorem
- Definition of rectangle
- Parallelogram consecutive angles theorem
Esme ties string across a parallelogram box lid DFGH, corner to corner. 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: Q is the midpoint of DG. Ladder: 1. DFGH is a parallelogram — Given; 2. ? — Parallelogram diagonals theorem; 3. Q is the midpoint of DG — Definition of midpoint. 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: Q is the midpoint of DG. Ladder: 1. DFGH is a parallelogram — Given; 2. ? — Parallelogram diagonals theorem; 3. Q is the midpoint of DG — Definition of midpoint. D F G H Q Not to scale - DQ ≅ QG and FQ ≅ QH
- DG ≅ FH
- DQ ≅ FQ and QG ≅ QH
- Q is the midpoint of DG
Jun traces the cover of a book, rectangle DFGH. 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: m∠HDF = 90. Ladder: 1. DFGH is a rectangle — Given; 2. ? — Definition of rectangle; 3. m∠HDF = 90 — Definition of right angle. 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: m∠HDF = 90. Ladder: 1. DFGH is a rectangle — Given; 2. ? — Definition of rectangle; 3. m∠HDF = 90 — Definition of right angle. D F G H Q Not to scale - DF ≅ DH
- ∠HDF is a right angle
- m∠HDF = 90
- DG ⊥ FH
Kira draws the top of a parallelogram table DFGH. 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: m∠DFG = m∠GHD. Ladder: 1. DFGH is a parallelogram — Given; 2. ? — Parallelogram opposite angles theorem; 3. m∠DFG = m∠GHD — Definition of congruent angles. 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: m∠DFG = m∠GHD. Ladder: 1. DFGH is a parallelogram — Given; 2. ? — Parallelogram opposite angles theorem; 3. m∠DFG = m∠GHD — Definition of congruent angles. D F G H Q Not to scale - m∠DFG = m∠GHD
- DF ≅ FG
- ∠DFG ≅ ∠GHD
- ∠DFG ≅ ∠FGH
Luz folds a parallelogram flap DFGH on an envelope. 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: FG = HD. Ladder: 1. DFGH is a parallelogram — Given; 2. ? — Parallelogram opposite sides theorem; 3. FG = HD — Definition of congruent segments. 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. ? — Parallelogram opposite sides theorem; 3. FG = HD — Definition of congruent segments. D F G H Q Not to scale - FG = HD
- FG ∥ GH
- FG ≅ HD
- FG ≅ GH