Prove and apply theorems about lines and angles: vertical angles, angle pairs formed by a transversal crossing parallel lines, and the perpendicular bisector as the set of points equidistant from the endpoints

Learning goal: Prove and apply theorems about lines and angles, both formally, with a reason for every step, and informally, as a general argument told without citing every step, including that vertical angles are congruent; that when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; and that the points on the perpendicular bisector of a line segment are exactly the points equidistant from the segment's endpoints.

What a proof is, and the two forms. A theorem is a statement that follows from definitions, earlier theorems and a small number of accepted facts. A proof is the chain of statements connecting the givens to the conclusion, each with a reason. A two-column proof lists numbered statements on the left and their reasons on the right; a paragraph proof gives the same chain in sentences. Both are formal proofs when every step has a reason; the paragraph form is not the "informal" one. An informal proof is still a proof: a general argument that covers every case but is told more loosely, for instance by describing what a rigid motion must carry where, without numbering the steps or naming every reason. What is not a proof at all is a check on one picture or one measurement, which can convince without proving because it covers only the figure drawn. The accepted facts used in this lesson are: the angles on one side of a line at a point sum to 180° (a linear pair is supplementary); through a point not on a line there is exactly one line parallel to the given line; translations and half-turns (rotations through 180°) carry every line to a parallel line or to itself, although rotations through other angles do not; and rigid motions preserve angle measure.

Vertical angles are congruent. When two lines cross at a point they form four angles; angles opposite each other, sharing only the vertex, are vertical angles. Theorem: vertical angles are congruent. Two-column proof, with ∠1 and ∠3 vertical and ∠2 between them: (1) ∠1 and ∠2 form a linear pair; reason, they are adjacent angles whose outer sides form a line. (2) m∠1 + m∠2 = 180°; reason, a linear pair is supplementary. (3) ∠3 and ∠2 form a linear pair; reason, same as (1). (4) m∠3 + m∠2 = 180°; reason, a linear pair is supplementary. (5) m∠1 + m∠2 = m∠3 + m∠2; reason, both equal 180°. (6) m∠1 = m∠3; reason, subtract m∠2 from both sides. (7) ∠1 ≅ ∠3; reason, angles with equal measure are congruent. Applying it to two crossing lines other than the ones in the diagram: if m∠1 = (3x + 10)° and m∠3 = (5x − 30)°, then 3x + 10 = 5x − 30, so x = 20 and both angles measure 70°; the two angles between them measure 110° each.

Transversals and parallel lines. A transversal is a line that crosses two other lines at two points. Number the angles 1 to 4 around the first crossing (upper left, upper right, lower right, lower left) and 5 to 8 the same way around the second. Corresponding angles occupy the same position at each crossing: ∠1 and ∠5, ∠2 and ∠6, ∠3 and ∠7, ∠4 and ∠8. Alternate interior angles are between the two lines on opposite sides of the transversal: ∠4 and ∠6, ∠3 and ∠5. Theorem: if the two lines are parallel, corresponding angles are congruent. Paragraph proof: let the transversal meet line ℓ at P and line m at Q, with ℓ parallel to m. Translate the plane by the segment from P to Q. The translation carries P to Q and carries the transversal onto itself, because Q lies on the transversal in the direction of the translation. It carries ℓ onto a line through Q parallel to ℓ; by the parallel postulate there is only one such line, and m is that line, so ℓ lands on m. The translation therefore carries ∠2 at P onto ∠6 at Q, and because a translation preserves angle measure, ∠2 ≅ ∠6. The same argument gives each pair of corresponding angles.

Alternate interior angles. Theorem: if a transversal crosses parallel lines, alternate interior angles are congruent. Two-column proof for ∠4 and ∠6: (1) ℓ ∥ m; given. (2) ∠4 ≅ ∠8; reason, corresponding angles of parallel lines are congruent. (3) ∠8 ≅ ∠6; reason, vertical angles are congruent. (4) ∠4 ≅ ∠6; reason, congruence of angles is transitive. A second proof uses a rotation: rotate the plane 180° about the midpoint M of PQ. The half turn swaps P and Q, carries the transversal onto itself, and carries ℓ to a line through Q parallel to ℓ, which must be m. So the half turn carries the interior angle at P on one side of the transversal onto the interior angle at Q on the other side, and rotations preserve angle measure. Either proof is complete; a good student can give both. A useful consequence: same-side interior angles, ∠4 and ∠5, are supplementary, because m∠4 = m∠6 and m∠6 + m∠5 = 180° as a linear pair.

Applying the transversal theorems. With ℓ ∥ m, if m∠2 = (2x + 15)° and m∠6 = (4x − 25)°, the corresponding-angles theorem gives 2x + 15 = 4x − 25, so x = 20 and both angles are 55°. If instead m∠4 = (2x − 15)° and m∠5 = (3x + 20)°, same-side interior angles are supplementary, so 5x + 5 = 180, x = 35, and m∠4 = 55° and m∠5 = 125°, matching the diagram. The theorems only apply when the lines are known to be parallel; an equation set up from a diagram with no parallel marking proves nothing.

The perpendicular bisector theorem, both directions. The perpendicular bisector of segment AB is the line through the midpoint M of AB perpendicular to AB. Theorem: a point P is on the perpendicular bisector of AB if and only if PA = PB. Forward direction, paragraph proof: let P be on the perpendicular bisector. If P = M then PA = PB because M is the midpoint. Otherwise, in triangles AMP and BMP, AM = BM (M is the midpoint), ∠AMP ≅ ∠BMP (both right angles, since the bisector is perpendicular to AB), and MP = MP. By SAS the triangles are congruent, and corresponding sides give PA = PB. Backward direction: suppose PA = PB and let M be the midpoint of AB. If P is on line AB, then P must be M, because the only point of the line equidistant from A and B is the midpoint, and M is on the bisector. Otherwise, triangles AMP and BMP have AM = BM, PA = PB and MP = MP, so they are congruent by SSS, giving ∠AMP ≅ ∠BMP. These two angles form a linear pair, so each is 90°, which means PM is perpendicular to AB at its midpoint: P is on the perpendicular bisector. The word "exactly" in the theorem is the two directions together: every point on the bisector is equidistant, and every equidistant point is on the bisector.

A worked application. A(0, 3) and B(6, 3) have midpoint (3, 3) and a horizontal segment, so the perpendicular bisector is the vertical line x = 3. Is P(3, 7) equidistant from A and B? PA = √(9 + 16) = 5 and PB = √(9 + 16) = 5, equal, and indeed P has x-coordinate 3. Is Q(4, 7) equidistant? QA = √(16 + 16) = √32 and QB = √(4 + 16) = √20, not equal, and indeed Q is not on x = 3. The theorem converts a distance question into a location question and back.

The misconception. Students often use a theorem's converse as if it were the theorem. "Corresponding angles are congruent, so the lines are parallel" is the converse of what was proved here, and it needs its own proof. Likewise "PA = PB, so P is on the bisector" is the backward direction and must be proved separately, as it was above. A theorem and its converse can both be true, as they are here, but each is proved on its own.

Worked, vertical angles with algebra: Two lines (not the diagram's) cross, forming ∠1 = (7x − 12)° and its vertical angle ∠3 = (4x + 24)°. Vertical angles are congruent, so 7x − 12 = 4x + 24, giving 3x = 36 and x = 12. Then ∠1 = 84 − 12 = 72° and ∠3 = 48 + 24 = 72°. Each of the other two angles is 180° − 72° = 108°, and the four angles sum to 72 + 108 + 72 + 108 = 360°, as the angles around a point must.

Worked, two-column proof that same-side interior angles are supplementary: Given ℓ ∥ m with transversal t, ∠4 and ∠5 same-side interior. (1) ℓ ∥ m; given. (2) ∠4 ≅ ∠6, so m∠4 = m∠6; alternate interior angles of parallel lines are congruent, and congruent angles have equal measure. (3) ∠6 and ∠5 form a linear pair; they are adjacent with outer sides on line m. (4) m∠6 + m∠5 = 180°; a linear pair is supplementary. (5) m∠4 + m∠5 = 180°; substitute m∠4 for m∠6 from (2). Therefore ∠4 and ∠5 are supplementary. Applied: m∠4 = (2x − 15)° and m∠5 = (3x + 20)° give 5x + 5 = 180, x = 35, so m∠4 = 55° and m∠5 = 125°, which add to 180°.

Worked, paragraph proof for corresponding angles by translation: Lines ℓ and m are parallel and transversal t meets them at P and Q. Translate every point by the segment from P to Q. The translation carries t onto itself, since sliding a line along itself does not change it, and carries ℓ onto the line through Q parallel to ℓ, which is m by the uniqueness of parallels. The angle at P between t and ℓ, on the upper right, is carried onto the angle at Q between t and m on the upper right. Translations preserve angle measure, so the corresponding angles are congruent. With m∠2 = (2x + 15)° and m∠6 = (4x − 25)°, the theorem gives x = 20 and both angles 55°; the other angles at each crossing are 125°.

Worked, using the perpendicular bisector theorem to find a point: A phone tower is to be placed on the road y = 2x, equally far from towns at A(1, 8) and B(7, 2) (a hypothetical layout). The set of points equidistant from A and B is the perpendicular bisector of AB: midpoint (4, 5), slope of AB = (2 − 8) ÷ (7 − 1) = −1, so the bisector has slope 1 and equation y = x + 1. Intersect with y = 2x: 2x = x + 1, x = 1, y = 2, so the tower goes at (1, 2). Check: distance to A is √(0 + 36) = 6 and to B is √(36 + 0) = 6. Equal, as the theorem promised.

Recording task: Draw two parallel lines cut by a transversal at 55° and number the eight angles as in the diagram. Record a table with columns "pair", "name of the relationship" (vertical, corresponding, alternate interior, same-side interior), "theorem or fact used" and "measures" for the pairs ∠1 and ∠3, ∠2 and ∠6, ∠4 and ∠6, ∠4 and ∠5; the measures are 125° and 125°, 55° and 55°, 55° and 55°, 55° and 125°. Then draw segment AB with A(0, 3) and B(6, 3), its perpendicular bisector x = 3, and the points P(3, 7) and Q(4, 7), recording the four distances PA = 5, PB = 5, QA = √32, QB = √20 and stating what the theorem says about P and about Q.

Independent performance: Write two complete proofs, one in two-column form and one in paragraph form, choosing from: vertical angles are congruent; alternate interior angles of parallel lines are congruent (by the corresponding-angle theorem or by a half turn); a point equidistant from the endpoints of a segment lies on its perpendicular bisector. Then apply two of the theorems to problems with algebraic angle measures or coordinates that you write yourself. Success criteria: every statement in the two-column proof has a reason that is a definition, a given, an accepted fact or an earlier theorem; the paragraph proof names the rigid motion or the congruent triangles it uses and says why the motion carries one line onto the other; each application states the theorem used before setting up the equation, and the solution is checked by substituting back; and a closing paragraph explains the difference between a theorem and its converse, with the perpendicular bisector theorem as the example.