Which angle pairs prove that two lines are parallel?
Grade 10 Math Worksheet

Circle the best answer, or write your answer on the line.
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Sketch, not to scale: lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠UGD = 108° and m∠GHS = 72°. Step 2: m∠UGF = 72°, because ∠UGD and ∠UGF are a linear pair, so they add to 180°. Step 3: m∠UGF = m∠GHS, so DF ∥ QS. Reason for Step 3: ?
Jun is writing a two-column proof that DF ∥ QS. Which reason justifies Step 3?
- If lines are parallel, then corresponding angles are congruent (Corresponding Angles Theorem)
- If alternate interior angles are congruent, then the lines are parallel (converse of the Alternate Interior Angles Theorem)
- If corresponding angles are congruent, then the lines are parallel (converse of the Corresponding Angles Theorem)
- Linear pair angles add to 180° (the reason for Step 2)
Sketch, not to scale
Lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠DGH = 64° and m∠SHV = 116°. Step 2: m∠GHQ = 116°, because ∠SHV and ∠GHQ are vertical angles, so they are congruent. Step 3: m∠DGH + m∠GHQ = 180°, so DF ∥ QS. Reason for Step 3: ?
Kira is writing a two-column proof that DF ∥ QS. Which reason justifies Step 3?
- If lines are parallel, then same-side interior angles are supplementary (Same-Side Interior Angles Theorem)
- If corresponding angles are congruent, then the lines are parallel (converse of the Corresponding Angles Theorem)
- Vertical angles are congruent (the reason for Step 2)
- If same-side interior angles are supplementary, then the lines are parallel (converse of the Same-Side Interior Angles Theorem)
Use this
Proof. Step 1: Given m∠UGF = 125° and m∠GHQ = 55°. Step 2: m∠FGH = 55°, because ∠UGF and ∠FGH are a linear pair, so they add to 180°. Step 3: m∠FGH = m∠GHQ, so DF ∥ QS. Reason for Step 3: ?
Luz is writing a two-column proof that DF ∥ QS. Which reason justifies Step 3?
Sketch, not to scale: lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠UGF = 125° and m∠GHQ = 55°. Step 2: m∠FGH = 55°, because ∠UGF and ∠FGH are a linear pair, so they add to 180°. Step 3: m∠FGH = m∠GHQ, so DF ∥ QS. Reason for Step 3: ? Sketch, not to scale: lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠UGF = 125° and m∠GHQ = 55°. Step 2: m∠FGH = 55°, because ∠UGF and ∠FGH are a linear pair, so they add to 180°. Step 3: m∠FGH = m∠GHQ, so DF ∥ QS. Reason for Step 3: ? D F Q S U V G H 125° 55° - If lines are parallel, then alternate interior angles are congruent (Alternate Interior Angles Theorem)
- If same-side interior angles are supplementary, then the lines are parallel (converse of the Same-Side Interior Angles Theorem)
- Linear pair angles add to 180° (the reason for Step 2)
- If alternate interior angles are congruent, then the lines are parallel (converse of the Alternate Interior Angles Theorem)
Sketch, not to scale
Lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠UGF = 100° and m∠SHV = 80°. Step 2: m∠UGD = 80°, because ∠UGF and ∠UGD are a linear pair, so they add to 180°. Step 3: m∠UGD = m∠SHV, so DF ∥ QS. Reason for Step 3: ?
Milo is writing a two-column proof that DF ∥ QS. Which reason justifies Step 3?
- If lines are parallel, then alternate exterior angles are congruent (Alternate Exterior Angles Theorem)
- If alternate exterior angles are congruent, then the lines are parallel (converse of the Alternate Exterior Angles Theorem)
- Linear pair angles add to 180° (the reason for Step 2)
- If alternate interior angles are congruent, then the lines are parallel (converse of the Alternate Interior Angles Theorem)
Noah is proving DF ∥ QS from an equation. Step 3 is missing. Which statement belongs there?
Sketch, not to scale: lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠UGF = (3x + 16)°, m∠GHS = (5x − 12)° and x = 14. Step 2: m∠UGF = 58° and m∠GHS = 58°, by substitution. Step 3: ?. Step 4: DF ∥ QS. Sketch, not to scale: lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠UGF = (3x + 16)°, m∠GHS = (5x − 12)° and x = 14. Step 2: m∠UGF = 58° and m∠GHS = 58°, by substitution. Step 3: ?. Step 4: DF ∥ QS. (3x + 16)° (5x - 12)° D F G Q S H U V Sketch, not to scale. Proof: Step 1: Given m∠UGF = (3x + 16)°, m∠GHS = (5x - 12)° and x = 14. Step 2: m∠UGF = 58° and m∠GHS = 58°, by substitution. Step 3: ?. Step 4: DF ∥ QS. - x = 14, from the given
- ∠UGF and ∠GHS are corresponding angles, and they are congruent
- ∠UGF and ∠GHS are alternate interior angles, and they are congruent
- ∠UGF and the angle next to it on line UV are a linear pair
Pia is proving DF ∥ QS from an equation. Step 3 is missing. Which statement belongs there?
Sketch, not to scale: lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠DGH = (4x + 2)°, m∠GHS = (2x + 40)° and x = 19. Step 2: m∠DGH = 78° and m∠GHS = 78°, by substitution. Step 3: ?. Step 4: DF ∥ QS. Sketch, not to scale: lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠DGH = (4x + 2)°, m∠GHS = (2x + 40)° and x = 19. Step 2: m∠DGH = 78° and m∠GHS = 78°, by substitution. Step 3: ?. Step 4: DF ∥ QS. (4x + 2)° (2x + 40)° D F G Q S H U V Sketch, not to scale. Proof: Step 1: Given m∠DGH = (4x + 2)°, m∠GHS = (2x + 40)° and x = 19. Step 2: m∠DGH = 78° and m∠GHS = 78°, by substitution. Step 3: ?. Step 4: DF ∥ QS. - ∠DGH and ∠GHS are alternate interior angles, and they are congruent
- x = 19, from the given
- ∠DGH and ∠GHS are corresponding angles, and they are congruent
- ∠DGH and the angle next to it on line UV are a linear pair
Ren is proving DF ∥ QS from an equation. Step 3 is missing. Which statement belongs there?
Sketch, not to scale: lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠UGD = (6x − 15)°, m∠SHV = (4x + 25)° and x = 20. Step 2: m∠UGD = 105° and m∠SHV = 105°, by substitution. Step 3: ?. Step 4: DF ∥ QS. Sketch, not to scale: lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠UGD = (6x − 15)°, m∠SHV = (4x + 25)° and x = 20. Step 2: m∠UGD = 105° and m∠SHV = 105°, by substitution. Step 3: ?. Step 4: DF ∥ QS. (6x - 15)° (4x + 25)° D F G Q S H U V Sketch, not to scale. Proof: Step 1: Given m∠UGD = (6x - 15)°, m∠SHV = (4x + 25)° and x = 20. Step 2: m∠UGD = 105° and m∠SHV = 105°, by substitution. Step 3: ?. Step 4: DF ∥ QS. - ∠UGD and ∠SHV are same-side exterior angles, and they are congruent
- x = 20, from the given
- ∠UGD and ∠SHV are alternate exterior angles, and they are congruent
- ∠UGD and the angle next to it on line DF are a linear pair
Tala is proving DF ∥ QS from an equation. Step 3 is missing. Which statement belongs there?
Sketch, not to scale: lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠DGH = (5x − 6)°, m∠QHV = (3x + 20)° and x = 13. Step 2: m∠DGH = 59° and m∠QHV = 59°, by substitution. Step 3: ?. Step 4: DF ∥ QS. Sketch, not to scale: lines DF and QS (no arrows given), crossed by line UV at G on DF and at H on QS; U is above, V below; D and Q are on the left. Proof. Step 1: Given m∠DGH = (5x − 6)°, m∠QHV = (3x + 20)° and x = 13. Step 2: m∠DGH = 59° and m∠QHV = 59°, by substitution. Step 3: ?. Step 4: DF ∥ QS. Proof (5x - 6)° (3x + 20)° D G F Q H S U V Step 1: Given m∠DGH = (5x - 6)°, m∠QHV = (3x + 20)° and x = 13. Step 2: m∠DGH = 59° and m∠QHV = 59°, by substitution. Step 3: ?. Step 4: DF ∥ QS. Not to scale. - x = 13, from the given
- ∠DGH and ∠QHV are corresponding angles, and they are congruent
- ∠DGH and ∠QHV are alternate interior angles, and they are congruent
- ∠DGH and the angle next to it on line UV are a linear pair
Sketch, not to scale: lines DF and QS, with no arrows (we do not know yet if they are parallel). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Sketch, not to scale: lines DF and QS, with no arrows (we do not know yet if they are parallel). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Sketch D G F Q H S U V Lines DF and QS, with no arrows (we do not know yet if they are parallel). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Not to scale. Esme checks two guitar strings. ∠DGH and ∠GHS are alternate interior angles, m∠DGH = (4x + 2)° and m∠GHS = (2x + 40)°. What value of x makes DF ∥ QS?
Jun checks two streets on a map. ∠UGD and ∠SHV are alternate exterior angles, m∠UGD = (6x − 15)° and m∠SHV = (4x + 25)°. What value of x makes DF ∥ QS?
Use the picture in question 9.