4.2 Rectangles and Squares

Let us pin down a definition you already half-hold.

Definition: Rectangle A rectangle is a quadrilateral in which (i) every angle is a right angle (90^\circ), and (ii) opposite sides have equal length.

Deduction 1: the diagonals are equal

Place the rods so their tips are the corners of a rectangle ABCD, with diagonal AC = 10\ \text{cm}. Set \triangle ADC beside \triangle DAB:

  • AB = CD (opposite sides of a rectangle),
  • \angle BAD = \angle CDA = 90^\circ,
  • AD belongs to both triangles.

The SAS condition gives \triangle ADC \cong \triangle DAB. So AC = BD (matching parts of congruent triangles). Hence a rectangle’s diagonals are equal in length, the second rod must also measure 10\ \text{cm}. Question 1 is answered.

Rectangle ABCD with diagonals AC and BD; triangles ADC and DAB are congruent, so the diagonals are equal A B C D △ADC and △DAB share the side AD — SAS gives AC = BD
Figure 4.2 · Deduction 1: △ADC and △DAB share side AD; SAS congruence forces AC = BD.

Deduction 2: the diagonals bisect each other

Let the diagonals cross at O. Examine \triangle AOB and \triangle COD:

  • \angle AOB = \angle COD (vertically opposite angles),
  • AB = CD (opposite sides),
  • and the base angles \angle 1 and \angle 2 match, because at the right angle B we have \angle 3 + \angle 1 = 90^\circ, while inside \triangle BCD we have \angle 3 + \angle 2 = 90^\circ; subtracting gives \angle 1 = \angle 2 = 90^\circ - \angle 3.

By AAS, \triangle AOB \cong \triangle COD, so OA = OC and OB = OD. Thus O is the midpoint of each diagonal, the diagonals bisect each other. Question 2 is answered: fasten the rods at their midpoints.

Rectangle ABCD with diagonals meeting at O; angles 1, 2 and 3 mark the AAS congruence of triangles AOB and COD 1 3 2 A B C D O ∠1 = ∠2 (each is 90° − ∠3), OA = OC, OB = OD
Figure 4.3 · Deduction 2: with ∠1 = ∠2 (each 90° − ∠3), triangles AOB and COD are congruent, so OA = OC and OB = OD.

Math Talk We could instead pair \triangle AOD with \triangle COB, using AO = CO, \angle AOD = \angle COB (vertically opposite), and AD = CB. Write out that congruence yourself and confirm which condition (SAS) closes the argument.

Deduction 3: the crossing angle is free

Now the twist. Suppose we draw two equal diagonals that bisect each other, but let them meet at a chosen angle, say 50^\circ. What figure appears?

Because OA = OB (each is half of equal diagonals), \triangle AOB is isosceles, so its base angles are equal, say a apiece. With apex 50^\circ: a + a + 50 = 180 \;\Rightarrow\; 2a = 130 \;\Rightarrow\; a = 65. Carry the same reasoning around the figure and every corner of the quadrilateral turns out to be 90^\circ, while \triangle AOB \cong \triangle COD and \triangle AOD \cong \triangle COB force AB = CD and AD = CB. So ABCD is a rectangle, regardless of the crossing angle.

Let us argue it in full generality. Let the angle between the diagonals be x. The four angles at O read x,\ x,\ 180^\circ - x,\ 180^\circ - x. In the isosceles triangle with apex x, each base angle is a = \frac{180 - x}{2} = 90 - \frac{x}{2}. In the adjacent isosceles triangle with apex 180 - x, each base angle is b = \frac{180 - (180 - x)}{2} = \frac{x}{2}. Each corner of the quadrilateral is built from one angle of each kind, so it measures a + b = \left(90 - \frac{x}{2}\right) + \frac{x}{2} = 90^\circ. All four corners equal 90^\circ whatever the value of x. Question 3 is answered: for a rectangle, the crossing angle simply does not matter.

Two equal rods fastened at their midpoints at an angle x; the wire around the tips always forms a rectangle x a b a = 90° − x/2 and b = x/2, so every corner is a + b = 90°
Figure 4.4 · Deduction 3: whatever the crossing angle x, every corner is a + b = (90° − x/2) + x/2 = 90°.

The Frame-Maker’s Answer To build a rectangle from two rods: make them equal in length and fasten them at their midpoints. The angle between them is yours to choose. This is not idle theory, builders and metalworkers use exactly this midpoint-crossing trick to lay out a true rectangular frame on site.

A leaner definition of a rectangle

Our definition asked for two things: all angles 90^\circ and opposite sides equal. Is the second part really necessary? Could we get away with “a quadrilateral whose every angle is 90^\circ”?

Deduction 4: four right angles force equal opposite sides

Take quadrilateral ABCD with all four angles 90^\circ and draw diagonal BD. As in Deduction 2, at the right angle B we have \angle 3 + \angle 1 = 90^\circ, and inside \triangle BCD we have \angle 3 + \angle 2 = 90^\circ, so \angle 1 = \angle 2. With BD shared, the AAS condition gives \triangle BAD \cong \triangle DCB. Hence AD = CB and DC = BA, the equal opposite sides appear for free. The definition can therefore be trimmed.

Two Definitions, One Shape A rectangle may be defined in either equivalent way:

  • a quadrilateral whose four angles are each 90^\circ; or
  • a quadrilateral whose diagonals are equal and bisect each other.

Both pick out exactly the same collection of figures.

Are the opposite sides parallel?

They are. Treat side AB as a transversal crossing AD and BC; then \angle A + \angle B = 90^\circ + 90^\circ = 180^\circ. When co-interior angles on one side of a transversal sum to 180^\circ, the lines are parallel, so AD \parallel BC. The identical argument yields AB \parallel DC.

Properties of a Rectangle 1. All four angles are 90^\circ. 2. Opposite sides are equal. 3. Opposite sides are parallel. 4. The diagonals are equal in length and bisect each other.

The square: a rectangle with a bonus

A square is a rectangle whose four sides happen to be equal.

Definition: Square A square is a quadrilateral in which all angles are 90^\circ and all sides are equal in length.

So every square is a rectangle, yet not every rectangle is a square, much as every sparrow is a bird, but not every bird is a sparrow. In a Venn diagram, the squares form a region tucked inside the rectangles.

Deduction 5: a square’s diagonals cross at 90^\circ

Return to the frame-maker, now after a square of diagonal 10\ \text{cm}. We already know the diagonals must be equal and must bisect each other. What further condition delivers equal sides? Compare \triangle BOA and \triangle BOC: OA = OC (halves of the diagonal), OB shared, and BA = BC (equal sides of the square). By SSS, \triangle BOA \cong \triangle BOC, so \angle BOA = \angle BOC. These two angles sit along a straight line, so \angle BOA + \angle BOC = 180^\circ, forcing \angle BOA = \angle BOC = 90^\circ. A square’s diagonals bisect each other at right angles.

Deduction: the diagonals halve the corner angles

In a square ABCD, look at \triangle ADC. Here AD = DC and \angle ADC = 90^\circ, so the base angles match: \angle 1 = \angle 3, and \angle 1 + \angle 3 + 90 = 180 \;\Rightarrow\; \angle 1 = \angle 3 = 45^\circ. The same happens at every corner, so each diagonal slices its corner angles into two 45^\circ pieces.

Square ABCD whose diagonals are equal, cross at right angles at O, and split every corner into two 45-degree angles 45° 45° A B C D O
Figure 4.5 · The square: equal diagonals cross at 90° at O and slice every corner into 45° + 45°.

Properties of a Square 1. All four sides are equal. 2. Opposite sides are parallel. 3. All four angles are 90^\circ. 4. The diagonals are equal and bisect each other at 90^\circ. 5. The diagonals bisect the corner angles (into 45^\circ each).

From Observing to Proving Measuring a handful of rectangles and noticing that their diagonals bisect each other yields only a conjecture, a statement you trust but have not shown to hold every time. Might the 500th rectangle misbehave? The only path to certainty is a proof, exactly as in Deduction 2. Experiment to spot the pattern; then deduce to lock it down.

Figure it Out: Rectangles & Squares

Practice

  1. Find the remaining angles inside these rectangles.
    1. Rectangle ABCD with diagonals meeting at the centre, where one diagonal makes a 35^\circ angle at a vertex (\angle DBA = 35^\circ).
    2. Rectangle PQRS with diagonals meeting at O and \angle QOR = 130^\circ.
  2. Draw a quadrilateral whose diagonals are equal (10\ \text{cm}) and bisect each other, meeting at an angle of (i) 35^\circ (ii) 50^\circ (iii) 90^\circ (iv) 130^\circ. Which one is a square?
  3. A circle has centre O. Segments PL and AM are two perpendicular diameters. What figure is APML?
  4. You have two rods of equal length and a wire (no paper). How can you set out an exact 90^\circ angle?
  5. Is “opposite sides parallel and equal” a full definition of a rectangle? That is, must every quadrilateral with opposite sides parallel and equal be a rectangle?
  1. (i) \angle ABD = 35^\circ,\ \angle CAB = 35^\circ,\ \angle CAD = 55^\circ,\ \angle ADB = 55^\circ,\ \angle BDC = 35^\circ,\ \angle ACB = 55^\circ (the diagonal splits each right angle into 35^\circ and 55^\circ, and the half-diagonal triangles are isosceles). (ii) \angle QOR = 130^\circ, so \angle POS = 130^\circ and \angle QOP = \angle ROS = 50^\circ. The triangle on the 130^\circ angle is isosceles, giving \angle OQR = \angle ORQ = 25^\circ; the triangle on the 50^\circ angle gives \angle OQP = \angle OPQ = \angle ORS = \angle OSR = 65^\circ.
  2. Each is drawn the same way: a 10\ \text{cm} segment, its midpoint O, then 5\ \text{cm} arcs from O at the stated angle. The case (iii) 90^\circ gives a square (equal diagonals bisecting at right angles).
  3. APML is a square. The diameters are equal (PL = AM), they bisect each other at O, and they are perpendicular, precisely the conditions for a square.
  4. Lay the equal rods AB and CD so their midpoints coincide at O, then run a wire from A through C to B. Since diagonals AB and CD are equal and bisect each other, ACBD is a rectangle, so \angle C = 90^\circ.
  5. No. A quadrilateral with opposite sides parallel and equal is a parallelogram, which need not have right angles, so it need not be a rectangle.