Prove and apply theorems about parallelograms: opposite sides and angles, diagonals that bisect each other and its converse, and rectangles as parallelograms with congruent diagonals

Learning goal: Prove and apply theorems about parallelograms, formally and informally: opposite sides are congruent; opposite angles are congruent; the diagonals of a parallelogram bisect each other, and conversely a quadrilateral whose diagonals bisect each other is a parallelogram; and rectangles are exactly the parallelograms with congruent diagonals.

Starting point. A parallelogram is a quadrilateral with both pairs of opposite sides parallel. That is the definition and the only thing we may assume about a parallelogram at the start; everything else is proved. The tools are the theorems about parallel lines (alternate interior angles are congruent), the triangle congruence criteria (SAS, ASA, SSS) and the fact that corresponding parts of congruent triangles are congruent. Label the parallelogram ABCD in order, so that AB ∥ DC and AD ∥ BC.

Opposite sides are congruent. Draw diagonal AC. Two-column proof: (1) AB ∥ DC and AD ∥ BC; definition of parallelogram. (2) ∠BAC ≅ ∠DCA; alternate interior angles, transversal AC crossing parallel lines AB and DC. (3) ∠BCA ≅ ∠DAC; alternate interior angles, transversal AC crossing parallel lines BC and AD. (4) AC = AC; a segment is congruent to itself. (5) Triangle ABC ≅ triangle CDA; ASA, using (2), (4), (3). (6) AB = CD and BC = DA; corresponding parts of congruent triangles. The diagonal cuts the parallelogram into two congruent triangles, and the two pairs of opposite sides are corresponding sides of those triangles.

Opposite angles are congruent. From the same congruent triangles, ∠B ≅ ∠D as corresponding parts. For the other pair, ∠BAD is the sum of ∠BAC and ∠CAD, and ∠BCD is the sum of ∠DCA and ∠ACB; by steps (2) and (3) the parts are congruent in pairs, so the sums are congruent: ∠A ≅ ∠C. A consequence: consecutive angles are supplementary, because ∠A and ∠B are same-side interior angles for the parallel lines AD and BC cut by AB. Applying it: if m∠A = (3x + 15)° and m∠C = (5x − 25)°, then 3x + 15 = 5x − 25, x = 20, and both angles are 75°; then ∠B = 180° − 75° = 105°.

The diagonals bisect each other. Let the diagonals AC and BD meet at E. Two-column proof: (1) AB ∥ DC; definition. (2) ∠BAE ≅ ∠DCE; alternate interior angles, transversal AC. (3) ∠ABE ≅ ∠CDE; alternate interior angles, transversal BD. (4) AB = CD; opposite sides of a parallelogram (proved above). (5) Triangle ABE ≅ triangle CDE; ASA. (6) AE = CE and BE = DE; corresponding parts. So E is the midpoint of both diagonals: each diagonal bisects the other. In coordinates, A(0, 0), B(6, 0), C(8, 3), D(2, 3) is a parallelogram (AB and DC both horizontal; AD and BC both with run 2 and rise 3); the midpoint of AC is (4, 1.5) and the midpoint of BD is (4, 1.5), the same point.

The converse: diagonals that bisect each other make a parallelogram. Suppose quadrilateral ABCD has diagonals meeting at E with AE = CE and BE = DE. (1) ∠AEB ≅ ∠CED; vertical angles. (2) Triangle AEB ≅ triangle CED; SAS, using AE = CE, the vertical angles and BE = DE. (3) ∠BAE ≅ ∠DCE; corresponding parts. (4) AB ∥ DC; if a transversal makes congruent alternate interior angles, the lines are parallel (the converse of the alternate interior angle theorem, which is proved separately). The same argument with triangles AED and CEB gives AD ∥ BC. Both pairs of opposite sides are parallel, so ABCD is a parallelogram by definition. This is why "the diagonals bisect each other" is a criterion for parallelograms, not just a property of them.

Rectangles are parallelograms with congruent diagonals. A rectangle is a quadrilateral with four right angles. It is a parallelogram: each pair of opposite sides is cut by a side that makes two right angles with them, and two lines perpendicular to the same line are parallel. Theorem, forward direction: the diagonals of a rectangle are congruent. In rectangle ABCD, triangles ABC and DCB have AB = DC (opposite sides of a parallelogram), ∠ABC ≅ ∠DCB (both right angles) and BC = CB, so they are congruent by SAS, and AC = DB. Converse: a parallelogram with congruent diagonals is a rectangle. In parallelogram ABCD with AC = BD, triangles ABC and DCB have AB = DC, BC = CB and AC = DB, so they are congruent by SSS, giving ∠ABC ≅ ∠DCB. These angles are same-side interior angles for the parallel lines AB and DC cut by BC, so they are supplementary; congruent and supplementary means each is 90°. Opposite angles of a parallelogram are congruent, so all four angles are 90° and ABCD is a rectangle. The word "exactly" in the learning goal is these two directions together. In coordinates, the rectangle P(0, 0), Q(6, 0), R(6, 4), S(0, 4) has diagonals PR = √(36 + 16) = √52 and QS = √(36 + 16) = √52.

Applying the diagonal theorems. In parallelogram ABCD with diagonals meeting at E, if AE = 2x + 3 and EC = 4x − 7, then 2x + 3 = 4x − 7, x = 5, AE = EC = 13 and AC = 26. If a parallelogram has diagonals of lengths 26 and 26, it is a rectangle; if the diagonals are 26 and 20, it is not, and no angle needs to be measured to know.

The misconception. Students often use a property of parallelograms to prove a figure is a parallelogram, without a converse. "Opposite sides are congruent" is a property; the statement "a quadrilateral with congruent opposite sides is a parallelogram" is a converse that needs its own proof (by SSS on a diagonal, then alternate interior angles). The standard names one converse explicitly, the diagonal one, and it is proved above; any other converse you use must be proved or cited as a proved theorem, not assumed.

Worked, opposite angles with algebra: Parallelogram ABCD has m∠A = (3x + 15)° and m∠C = (5x − 25)°. Opposite angles of a parallelogram are congruent, so 3x + 15 = 5x − 25, giving x = 20 and m∠A = m∠C = 75°. Consecutive angles are supplementary (same-side interior angles of parallel sides), so m∠B = m∠D = 180° − 75° = 105°. Check: 75 + 105 + 75 + 105 = 360°, the angle sum of any quadrilateral.

Worked, diagonals with algebra: In parallelogram PQRS the diagonals meet at T with PT = 2x + 3 and TR = 4x − 7, and QT = 9. The diagonals bisect each other, so PT = TR: 2x + 3 = 4x − 7, x = 5, PT = TR = 13, PR = 26. Also TS = QT = 9, so QS = 18. The diagonals have different lengths, 26 and 18, so PQRS is not a rectangle.

Worked, coordinate check of both diagonal theorems: A(0, 0), B(6, 0), C(8, 3), D(2, 3). AB and DC are horizontal of length 6; AD and BC both have run 2 and rise 3, length √13, so both pairs of opposite sides are parallel and ABCD is a parallelogram. Midpoint of AC: (4, 1.5). Midpoint of BD: ((6 + 2) ÷ 2, (0 + 3) ÷ 2) = (4, 1.5). Same point, so the diagonals bisect each other. Their lengths: AC = √(64 + 9) = √73 ≈ 8.54 and BD = √(16 + 9) = 5, unequal, so ABCD is not a rectangle, which agrees with the angle at A: between (6, 0) and (2, 3) its cosine is 12 ÷ (6√13) ≈ 0.555, about 56.3°, not 90°.

Worked, paragraph proof that a rectangle's diagonals are congruent, with a check: In rectangle PQRS with P(0, 0), Q(6, 0), R(6, 4), S(0, 4), consider triangles PQR and SRQ. PQ = SR because opposite sides of a parallelogram are congruent (both 6 here); ∠PQR ≅ ∠SRQ because both are right angles; QR = RQ. By SAS the triangles are congruent, so PR = SQ. Numerically PR = √(6² + 4²) = √52 and SQ = √((0 − 6)² + (4 − 0)²) = √52, both about 7.21.

Recording task: Draw parallelogram ABCD with A(0, 0), B(6, 0), C(8, 3), D(2, 3) and rectangle PQRS with P(0, 0), Q(6, 0), R(6, 4), S(0, 4) on grid paper, with both diagonals in each. Record a table with rows "opposite sides", "opposite angles", "midpoints of the diagonals" and "lengths of the diagonals", and a column for each figure: for ABCD the entries are 6 and 6, √13 and √13; about 56.3° and 56.3°, 123.7° and 123.7°; (4, 1.5) and (4, 1.5); √73 and 5. For PQRS: 6 and 6, 4 and 4; four right angles; (3, 2) and (3, 2); √52 and √52. Below the table, write which theorem each row confirms and which figure the rectangle-diagonal theorem distinguishes.

Independent performance: Write two formal proofs, one in two-column form and one in paragraph form, chosen from: opposite sides of a parallelogram are congruent; the diagonals of a parallelogram bisect each other; a quadrilateral whose diagonals bisect each other is a parallelogram; a parallelogram with congruent diagonals is a rectangle. Then design a quadrilateral with grid-point vertices whose diagonals bisect each other, and another whose diagonals are congruent but do not bisect each other, and classify each. Success criteria: every statement in the two-column proof has a reason drawn from a definition, a given, or a theorem proved earlier in this unit; the paragraph proof names the congruent triangles and the criterion (SAS, ASA or SSS) that establishes their congruence; the two designed quadrilaterals are checked by computing the diagonal midpoints and lengths; each classification cites the theorem or converse that justifies it; and a closing paragraph explains why a property of parallelograms cannot be used as a criterion for parallelograms unless its converse has been proved.