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Proof: The Vertical Angles Theorem (Geometry)

Free printable Geometry geometry worksheet: complete a two-column proof that vertical angles are congruent, using the Linear Pair Postulate.

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Geometry · Math worksheet
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Geometry Proof: The Vertical Angles Theorem

Two lines intersect, forming two pairs of vertical angles. Complete each two-column proof that the vertical angles are congruent by filling in the missing statement or reason.

  1. 1.
    Two lines intersect at O, with B, O, D collinear and C, O, A collinear. Prove: ∠BOC\angle BOC ≅\cong ∠DOA\angle DOA. Complete the missing statement in step 3 of the two-column proof.
    BCDAO
    StatementReason
    1.B, O, D are collinear, and C, O, A are collinear.Given
    2.∠BOC\angle BOC and ∠COD\angle COD form a linear pair.Definition of a linear pair (BOD is a straight line)
    3. Definition of a linear pair (COA is a straight line)
    4.m∠BOCm\angle BOC + m∠CODm\angle COD = 180°, and m∠CODm\angle COD + m∠DOAm\angle DOA = 180°.Linear Pair Postulate
    5.m∠BOCm\angle BOC + m∠CODm\angle COD = m∠CODm\angle COD + m∠DOAm\angle DOA.Transitive Property of Equality
    6.m∠BOCm\angle BOC = m∠DOAm\angle DOA.Subtraction Property of Equality
    7.∠BOC\angle BOC ≅\cong ∠DOA\angle DOA.Definition of congruent angles
  2. 2.
    Two lines intersect at P, with Q, P, S collinear and R, P, T collinear. Prove: ∠QPR\angle QPR ≅\cong ∠SPT\angle SPT. Complete the missing statement in step 4 of the two-column proof.
    QRSTP
    StatementReason
    1.Q, P, S are collinear, and R, P, T are collinear.Given
    2.∠QPR\angle QPR and ∠RPS\angle RPS form a linear pair.Definition of a linear pair (QPS is a straight line)
    3.∠RPS\angle RPS and ∠SPT\angle SPT form a linear pair.Definition of a linear pair (RPT is a straight line)
    4. Linear Pair Postulate
    5.m∠QPRm\angle QPR + m∠RPSm\angle RPS = m∠RPSm\angle RPS + m∠SPTm\angle SPT.Transitive Property of Equality
    6.m∠QPRm\angle QPR = m∠SPTm\angle SPT.Subtraction Property of Equality
    7.∠QPR\angle QPR ≅\cong ∠SPT\angle SPT.Definition of congruent angles
  3. 3.
    Two lines intersect at O, with A, O, C collinear and B, O, D collinear. Prove: ∠AOB\angle AOB ≅\cong ∠COD\angle COD. Complete the missing reason in step 2 of the two-column proof.
    ABCDO
    StatementReason
    1.A, O, C are collinear, and B, O, D are collinear.Given
    2.∠AOB\angle AOB and ∠BOC\angle BOC form a linear pair. 
    3.∠BOC\angle BOC and ∠COD\angle COD form a linear pair.Definition of a linear pair (BOD is a straight line)
    4.m∠AOBm\angle AOB + m∠BOCm\angle BOC = 180°, and m∠BOCm\angle BOC + m∠CODm\angle COD = 180°.Linear Pair Postulate
    5.m∠AOBm\angle AOB + m∠BOCm\angle BOC = m∠BOCm\angle BOC + m∠CODm\angle COD.Transitive Property of Equality
    6.m∠AOBm\angle AOB = m∠CODm\angle COD.Subtraction Property of Equality
    7.∠AOB\angle AOB ≅\cong ∠COD\angle COD.Definition of congruent angles
  4. 4.
    Two lines intersect at K, with E, K, G collinear and F, K, H collinear. Prove: ∠EKF\angle EKF ≅\cong ∠GKH\angle GKH. Complete the missing reason in step 3 of the two-column proof.
    EFGHK
    StatementReason
    1.E, K, G are collinear, and F, K, H are collinear.Given
    2.∠EKF\angle EKF and ∠FKG\angle FKG form a linear pair.Definition of a linear pair (EKG is a straight line)
    3.∠FKG\angle FKG and ∠GKH\angle GKH form a linear pair. 
    4.m∠EKFm\angle EKF + m∠FKGm\angle FKG = 180°, and m∠FKGm\angle FKG + m∠GKHm\angle GKH = 180°.Linear Pair Postulate
    5.m∠EKFm\angle EKF + m∠FKGm\angle FKG = m∠FKGm\angle FKG + m∠GKHm\angle GKH.Transitive Property of Equality
    6.m∠EKFm\angle EKF = m∠GKHm\angle GKH.Subtraction Property of Equality
    7.∠EKF\angle EKF ≅\cong ∠GKH\angle GKH.Definition of congruent angles
  5. 5.
    Two lines intersect at N, with L, N, V collinear and U, N, J collinear. Prove: ∠LNU\angle LNU ≅\cong ∠VNJ\angle VNJ. Complete the missing reason in step 2 of the two-column proof.
    LUVJN
    StatementReason
    1.L, N, V are collinear, and U, N, J are collinear.Given
    2.∠LNU\angle LNU and ∠UNV\angle UNV form a linear pair. 
    3.∠UNV\angle UNV and ∠VNJ\angle VNJ form a linear pair.Definition of a linear pair (UNJ is a straight line)
    4.m∠LNUm\angle LNU + m∠UNVm\angle UNV = 180°, and m∠UNVm\angle UNV + m∠VNJm\angle VNJ = 180°.Linear Pair Postulate
    5.m∠LNUm\angle LNU + m∠UNVm\angle UNV = m∠UNVm\angle UNV + m∠VNJm\angle VNJ.Transitive Property of Equality
    6.m∠LNUm\angle LNU = m∠VNJm\angle VNJ.Subtraction Property of Equality
    7.∠LNU\angle LNU ≅\cong ∠VNJ\angle VNJ.Definition of congruent angles
  6. 6.
    Two lines intersect at N, with J, N, U collinear and L, N, V collinear. Prove: ∠JNL\angle JNL ≅\cong ∠UNV\angle UNV. Complete the missing statement in step 3 of the two-column proof.
    JLUVN
    StatementReason
    1.J, N, U are collinear, and L, N, V are collinear.Given
    2.∠JNL\angle JNL and ∠LNU\angle LNU form a linear pair.Definition of a linear pair (JNU is a straight line)
    3. Definition of a linear pair (LNV is a straight line)
    4.m∠JNLm\angle JNL + m∠LNUm\angle LNU = 180°, and m∠LNUm\angle LNU + m∠UNVm\angle UNV = 180°.Linear Pair Postulate
    5.m∠JNLm\angle JNL + m∠LNUm\angle LNU = m∠LNUm\angle LNU + m∠UNVm\angle UNV.Transitive Property of Equality
    6.m∠JNLm\angle JNL = m∠UNVm\angle UNV.Subtraction Property of Equality
    7.∠JNL\angle JNL ≅\cong ∠UNV\angle UNV.Definition of congruent angles
  7. 7.
    Two lines intersect at M, with X, M, Z collinear and Y, M, W collinear. Prove: ∠XMY\angle XMY ≅\cong ∠ZMW\angle ZMW. Complete the missing reason in step 5 of the two-column proof.
    XYZWM
    StatementReason
    1.X, M, Z are collinear, and Y, M, W are collinear.Given
    2.∠XMY\angle XMY and ∠YMZ\angle YMZ form a linear pair.Definition of a linear pair (XMZ is a straight line)
    3.∠YMZ\angle YMZ and ∠ZMW\angle ZMW form a linear pair.Definition of a linear pair (YMW is a straight line)
    4.m∠XMYm\angle XMY + m∠YMZm\angle YMZ = 180°, and m∠YMZm\angle YMZ + m∠ZMWm\angle ZMW = 180°.Linear Pair Postulate
    5.m∠XMYm\angle XMY + m∠YMZm\angle YMZ = m∠YMZm\angle YMZ + m∠ZMWm\angle ZMW. 
    6.m∠XMYm\angle XMY = m∠ZMWm\angle ZMW.Subtraction Property of Equality
    7.∠XMY\angle XMY ≅\cong ∠ZMW\angle ZMW.Definition of congruent angles
  8. 8.
    Two lines intersect at N, with J, N, U collinear and L, N, V collinear. Prove: ∠JNL\angle JNL ≅\cong ∠UNV\angle UNV. Complete the missing reason in step 5 of the two-column proof.
    JLUVN
    StatementReason
    1.J, N, U are collinear, and L, N, V are collinear.Given
    2.∠JNL\angle JNL and ∠LNU\angle LNU form a linear pair.Definition of a linear pair (JNU is a straight line)
    3.∠LNU\angle LNU and ∠UNV\angle UNV form a linear pair.Definition of a linear pair (LNV is a straight line)
    4.m∠JNLm\angle JNL + m∠LNUm\angle LNU = 180°, and m∠LNUm\angle LNU + m∠UNVm\angle UNV = 180°.Linear Pair Postulate
    5.m∠JNLm\angle JNL + m∠LNUm\angle LNU = m∠LNUm\angle LNU + m∠UNVm\angle UNV. 
    6.m∠JNLm\angle JNL = m∠UNVm\angle UNV.Subtraction Property of Equality
    7.∠JNL\angle JNL ≅\cong ∠UNV\angle UNV.Definition of congruent angles
  9. 9.
    Two lines intersect at K, with E, K, G collinear and F, K, H collinear. Prove: ∠EKF\angle EKF ≅\cong ∠GKH\angle GKH. Complete the missing statement in step 3 of the two-column proof.
    EFGHK
    StatementReason
    1.E, K, G are collinear, and F, K, H are collinear.Given
    2.∠EKF\angle EKF and ∠FKG\angle FKG form a linear pair.Definition of a linear pair (EKG is a straight line)
    3. Definition of a linear pair (FKH is a straight line)
    4.m∠EKFm\angle EKF + m∠FKGm\angle FKG = 180°, and m∠FKGm\angle FKG + m∠GKHm\angle GKH = 180°.Linear Pair Postulate
    5.m∠EKFm\angle EKF + m∠FKGm\angle FKG = m∠FKGm\angle FKG + m∠GKHm\angle GKH.Transitive Property of Equality
    6.m∠EKFm\angle EKF = m∠GKHm\angle GKH.Subtraction Property of Equality
    7.∠EKF\angle EKF ≅\cong ∠GKH\angle GKH.Definition of congruent angles
  10. 10.
    Two lines intersect at K, with E, K, G collinear and F, K, H collinear. Prove: ∠EKF\angle EKF ≅\cong ∠GKH\angle GKH. Complete the missing statement in step 2 of the two-column proof.
    EFGHK
    StatementReason
    1.E, K, G are collinear, and F, K, H are collinear.Given
    2. Definition of a linear pair (EKG is a straight line)
    3.∠FKG\angle FKG and ∠GKH\angle GKH form a linear pair.Definition of a linear pair (FKH is a straight line)
    4.m∠EKFm\angle EKF + m∠FKGm\angle FKG = 180°, and m∠FKGm\angle FKG + m∠GKHm\angle GKH = 180°.Linear Pair Postulate
    5.m∠EKFm\angle EKF + m∠FKGm\angle FKG = m∠FKGm\angle FKG + m∠GKHm\angle GKH.Transitive Property of Equality
    6.m∠EKFm\angle EKF = m∠GKHm\angle GKH.Subtraction Property of Equality
    7.∠EKF\angle EKF ≅\cong ∠GKH\angle GKH.Definition of congruent angles
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