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 same-side interior angles are supplementary, then the lines are parallel (converse of the Same-Side Interior Angles Theorem)
- Vertical angles are congruent (the reason for Step 2)
- If corresponding angles are congruent, then the lines are parallel (converse of the Corresponding 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 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)
- If lines are parallel, then alternate interior angles are congruent (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?
- 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)
- 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)
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. - ∠UGF and ∠GHS are corresponding angles, and they are congruent
- x = 14, from the given
- ∠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
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. - x = 20, from the given
- ∠UGD and the angle next to it on line DF are a linear pair
- ∠UGD and ∠SHV are same-side exterior angles, and they are congruent
- ∠UGD and ∠SHV are alternate exterior angles, and they are congruent
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. Bo checks two toy-train rails. ∠UGF and ∠GHS are corresponding angles, m∠UGF = (3x + 16)° and m∠GHS = (5x − 12)°. What value of x makes DF ∥ QS?
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?
Use the picture in question 7.
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 7.
Kira checks the two rails of a ladder. ∠DGH and ∠QHV are corresponding angles, m∠DGH = (5x − 6)° and m∠QHV = (3x + 20)°. What value of x makes DF ∥ QS?
Use the picture in question 7.