Quadrilaterals: parallelogram, rectangle, rhombus, square and trapezoid

When builders lay the foundation of a house, they check that it is really rectangular in a simple way: they measure the two diagonals. If the opposite sides are equal and the diagonals are equal too, the corners are right angles — no protractor needed. Why does this work? By the end of this lesson you will be able to prove it. A foundation, a window, a sheet of paper, a phone screen and a chessboard are all quadrilaterals; in this lesson we arrange them into one “family” and learn the properties of each member.

All the proofs rest on the lesson “Congruent triangles” and on the angles at parallel lines (alternate, corresponding and co-interior angles): we cut a quadrilateral into two triangles with a diagonal and prove that they are congruent.

The angle sum of a polygon

From one vertex of a convex n-gon draw all the diagonals. The vertex itself and its two neighbours give no diagonal, and the other n − 3 diagonals cut the polygon into n − 2 triangles. The angles of each triangle add up to 180°, and together they exactly fill the angles of the polygon. For example, one diagonal cuts a quadrilateral into two triangles, so its angles add up to 2 · 180° = 360°.

The parallelogram: properties and tests

Proof: diagonal AC cuts the parallelogram into triangles ABC and CDA. Because AB ∥ CD, ∠BAC = ∠DCA, and because BC ∥ AD, ∠BCA = ∠DAC (alternate angles); AC is a common side. By ASA, △ABC ≅ △CDA, so AB = CD, BC = DA and ∠B = ∠D; diagonal BD gives ∠A = ∠C in the same way. Triangles AOB and COD are congruent by ASA as well (AB = CD and two pairs of alternate angles), so AO = OC and BO = OD. Consecutive angles (the two angles at one side) add up to 180° because they are co-interior angles at parallel lines.

Each test is proved with congruent triangles again. For test 3: AO = OC, BO = OD and ∠AOB = ∠COD (vertical angles), so △AOB ≅ △COD by SAS. Then ∠OAB = ∠OCD; these are alternate angles for lines AB, CD and the transversal AC, so AB ∥ CD, and also AB = CD — test 1 holds.

Rectangle, rhombus and square

These three figures are special parallelograms, so everything true for a parallelogram is true for them too; on top of that, each has properties of its own.

Proof: let AC = BD in parallelogram ABCD. Triangles ABD and DCA have AB = DC (opposite sides), the common side AD and BD = CA, so they are congruent by SSS and ∠A = ∠D. But ∠A + ∠D = 180° (consecutive angles), so ∠A = ∠D = 90°. This is exactly the builders’ trick from the beginning of the lesson: equal opposite sides give a parallelogram (test 2), and equal diagonals give right angles. The other direction uses the same triangles: in a rectangle, triangles ABD and DCA are congruent by two legs, so BD = CA.

The reason: in a rhombus AB = AD and O is the midpoint of BD, so AO is the median of the isosceles triangle ABD — and therefore also its altitude and bisector: AC ⊥ BD and ∠BAO = ∠DAO. Conversely, if AC ⊥ BD in a parallelogram, line AC is the perpendicular bisector of BD, so AB = AD and all sides are equal. A square is both a rectangle and a rhombus, so its diagonals are equal, perpendicular and bisect each other, and they cut its angles into 45° parts.

Thales’s theorem and the midline of a triangle

Idea of the proof: through B₁ and B₂ draw segments B₁C₁ and B₂C₂ parallel to the first side of the angle (C₁ lies on A₂B₂ and C₂ on A₃B₃). The quadrilaterals formed are parallelograms, so B₁C₁ = A₁A₂ = A₂A₃ = B₂C₂. In triangles B₁C₁B₂ and B₂C₂B₃ the angles next to these equal sides are equal as corresponding angles at parallel lines, so the triangles are congruent by ASA and B₁B₂ = B₂B₃.

Thales’s theorem lets you divide a segment into any number of equal parts with compass and ruler. Dividing segment AB into 5 equal parts:

1. 1
   Helper ray

   From A draw a ray at any angle to AB.

2. 2
   Equal segments

   Mark off 5 equal segments one after another on the ray with the compass: AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.

3. 3
   Join

   Join A₅ to B.

4. 4
   Parallels

   Through A₁, A₂, A₃, A₄ draw lines parallel to A₅B (by copying an angle). By Thales’s theorem they divide AB into 5 equal parts.

Proof: extend MN beyond N by NK = MN. Triangles MBN and KCN have BN = NC, MN = NK and ∠MNB = ∠KNC (vertical angles), so they are congruent by SAS. Hence KC = MB = AM and ∠MBN = ∠KCN, which means KC ∥ AB. In quadrilateral AMKC the sides AM and KC are equal and parallel, so it is a parallelogram. Therefore MK ∥ AC and MK = AC, and MN is half of MK. Conversely, by Thales’s theorem, a line through the midpoint of one side parallel to a second side bisects the third side.

The trapezoid and its midline

In a trapezoid the two angles at one leg add up to 180°: they are co-interior angles for the parallel bases and the leg. From now on, in trapezoid ABCD the bases are AD and BC and the legs are AB and CD.

The reason: drop the heights BH and CK from B and C to AD. Right triangles ABH and DCK are congruent by the hypotenuse and a leg (AB = DC, BH = CK), so ∠A = ∠D and AH = KD = (AD − BC) ÷ 2. For the diagonals, look at triangles ABD and DCA: AB = DC, ∠A = ∠D and AD is common, so they are congruent by SAS and BD = CA.

Proof: let diagonal AC cross MN at P. By Thales’s theorem, the line through M parallel to the bases bisects AC and CD, so MP is the midline of triangle ABC and PN is the midline of triangle ACD: MP = b ÷ 2 and PN = a ÷ 2. Hence MN = (a + b) ÷ 2.