Quadrilateral

A quadrilateral is a closed figure with four straight sides. You met its family in Grade 8:

Quadrilateral	What makes it one
trapezium	one pair of opposite sides parallel
parallelogram	both pairs of opposite sides parallel
rectangle	a parallelogram with every angle 90∘°
rhombus	a parallelogram with all four sides equal
square	a rectangle with all four sides equal (a rhombus with every angle 90∘°)

So every rectangle, rhombus and square is a parallelogram, and everything this unit proves about parallelograms holds for them too.

The half turn is the secret of the whole unit. Turn any parallelogram half way round its middle point and it lands exactly on itself: each corner on the opposite corner, each side on the opposite side, each half of a diagonal on the other half. Everything in 14.1 is that one fact, proved step by step.

Here is the family with all its properties in one table. The first row holds for all four; each row below adds something.

Figure	Sides	Angles	Diagonals
parallelo­gram	opposite ones parallel and equal	opposite ones equal; neighbors add to 180∘°	bisect each other
rectangle	opposite ones equal	all 90∘°	bisect each other; equal
rhombus	all four equal	opposite ones equal	bisect each other at 90∘° and bisect the angles
square	all four equal	all 90∘°	equal; bisect each other at 90∘°

Proving them: the seven theorems

Measuring shows a property for the figures you drew. A proof shows it for every parallelogram there could ever be. Each proof below is a table: a statement on the left, the reason it is true on the right, each reason something given, something already proved, or a fact you know (alternate angles on parallel lines, vertically opposite angles, the congruence tests S.A.S., A.S.A., S.S.S.). Almost every one finds two congruent triangles, and then their matching sides and angles are equal.

Theorem 1

The line segments joining the ends on the same side of two equal and parallel line segments are also equal and parallel.

Given: AB = CD and $AB \parallel CD$; A is joined to C, and B to D.

To prove: AC = BD and $AC \parallel BD$.

Construction: join B and C.

	Statements	Reasons
1.	In △ABC and △DCB: i. AB = DC (S) ii. ∠ABC = ∠DCB (A) iii. BC = CB (S)	i.given ii. alternate angles, $AB \parallel CD$ iii. common side
2.	$\triangle ABC \cong \triangle DCB$	by S.A.S.
3.	AC = DB	corresponding sides of congruent triangles
4.	∠ACB = ∠DBC	corresponding angles of congruent triangles
5.	$AC \parallel BD$	from 4: the alternate angles are equal
6.	AC = BD and $AC \parallel BD$	from 3 and 5

Proved. What it means: take any stick, slide it without turning, and the two ends trace out equal, parallel paths. So one pair of opposite sides equal and parallel is enough to make a parallelogram.

Theorem 2

The line segments joining the opposite ends of two equal and parallel line segments bisect each other.

Given: PQ = RS and $PQ \parallel RS$; the opposite ends are joined, P to S and Q to R, and PS and QR cross at O.

To prove: PO = OS and QO = OR, that is, PS and QR bisect each other.

	Statements	Reasons
1.	In △POQ and △SOR: i. ∠OPQ = ∠OSR (A) ii. PQ = SR (S) iii. ∠PQO = ∠SRO (A)	i.alternate angles, $PQ \parallel RS$ ii. given iii. alternate angles, $PQ \parallel RS$
2.	$\triangle POQ \cong \triangle SOR$	by A.S.A.
3.	PO = SO and QO = RO	corresponding sides of congruent triangles

Proved.

Theorem 3

The opposite sides and the opposite angles of a parallelogram are equal.

Given: ABCD is a parallelogram: $AB \parallel DC$ and $AD \parallel BC$.

To prove: 1. AB = DC and AD = BC; 2. ∠ABC = ∠ADC and ∠DAB = ∠BCD.

Construction: join B and D.

	Statements	Reasons
1.	In △ABD and △CDB: i. ∠ABD = ∠CDB (A) ii. BD = DB (S) iii. ∠ADB = ∠CBD (A)	i.alternate angles, $AB \parallel DC$ ii. common side iii. alternate angles, $AD \parallel BC$
2.	$\triangle ABD \cong \triangle CDB$	by A.S.A.
3.	AB = CD and AD = CB	corresponding sides of congruent triangles
4.	∠DAB = ∠BCD	corresponding angles of congruent triangles
5.	∠ABD + ∠DBC = ∠CDB + ∠BDA	equals added to equals (from 1 i and iii)
6.	∠ABC = ∠ADC	from 5: each side is one whole angle
7.	AB = DC, AD = BC, ∠ABC = ∠ADC, ∠DAB = ∠BCD	from 3, 4 and 6

Proved. On the way it showed something more: a diagonal cuts a parallelogram into two congruent triangles (step 2).

Theorem 4

A quadrilateral whose opposite sides are equal is a parallelogram.

This is Theorem 3 turned round, its converse: Theorem 3 starts from a parallelogram and finds equal sides; Theorem 4 starts from equal sides and finds a parallelogram. It is the theorem in Lean the frame.

Given: in quadrilateral ABCD, AB = CD and AD = BC.

To prove: ABCD is a parallelogram, that is, $AB \parallel CD$ and $AD \parallel BC$.

Construction: join A and C.

	Statements	Reasons
1.	In △ABC and △CDA: i. AB = CD (S) ii. BC = DA (S) iii. AC = CA (S)	i.given ii. given iii. common side
2.	$\triangle ABC \cong \triangle CDA$	by S.S.S.
3.	∠BAC = ∠DCA and ∠BCA = ∠DAC	corresponding angles of congruent triangles
4.	$AB \parallel CD$ and $BC \parallel AD$	from 3: the alternate angles are equal

Proved.

Theorem 5

A quadrilateral whose opposite angles are equal is a parallelogram.

Given: in quadrilateral ABCD, ∠ABC = ∠ADC and ∠DAB = ∠BCD.

To prove: ABCD is a parallelogram.

	Statements	Reasons
1.	$\angle ABC + \angle BCD + \angle CDA + \angle DAB = 360^\circ$	the angles of a quadrilateral add up to 360∘°
2.	$\angle ABC + \angle BCD + \angle ABC + \angle BCD = 360^\circ$, so $2\angle ABC + 2\angle BCD = 360^\circ$ and $\angle ABC + \angle BCD = 180^\circ$	putting ∠CDA = ∠ABC and ∠DAB = ∠BCD into 1 (given)
3.	$AB \parallel DC$	from 2: co-interior angles adding up to 180∘°
4.	$\angle BCD + \angle CDA = 180^\circ$	in the same way as 1 and 2
5.	$BC \parallel AD$	from 4: co-interior angles adding up to 180∘°
6.	$AB \parallel DC$ and $BC \parallel AD$	from 3 and 5
7.	ABCD is a parallelogram	from 6

Proved.

Theorem 6

The diagonals of a parallelogram bisect each other.

Given: PQRS is a parallelogram whose diagonals PR and QS intersect at O.

To prove: PO = OR and SO = OQ.

	Statements	Reasons
1.	In △POQ and △ROS: i. ∠OPQ = ∠ORS (A) ii. PQ = RS (S) iii. ∠OQP = ∠OSR (A)	i.alternate angles, $PQ \parallel SR$ ii. opposite sides of a parallelogram (Theorem 3) iii. alternate angles, $PQ \parallel SR$
2.	$\triangle POQ \cong \triangle ROS$	by A.S.A.
3.	PO = RO and QO = SO	corresponding sides of congruent triangles

Proved. It is Theorem 2 again: PQ and SR are equal and parallel, and the diagonals join their opposite ends.

Theorem 7 (the converse of Theorem 6)

If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.

Given: in quadrilateral ABCD, the diagonals cross at O, with AO = OC and DO = OB.

To prove: ABCD is a parallelogram.

	Statements	Reasons
1.	In △AOB and △COD: i. AO = CO (S) ii. ∠AOB = ∠COD (A) iii. OB = OD (S)	i.given ii. vertically opposite angles iii. given
2.	$\triangle AOB \cong \triangle COD$	by S.A.S.
3.	AB = CD	corresponding sides of congruent triangles
4.	∠OBA = ∠ODC	corresponding angles of congruent triangles
5.	$AB \parallel DC$	from 4: the alternate angles are equal
6.	$AD \parallel BC$ and AD = BC	from 3 and 5, by Theorem 1 (AB and DC are equal and parallel)
7.	ABCD is a parallelogram	its opposite sides are equal and parallel

Proved. This is the quickest way to draw a parallelogram with no set square: draw two lines crossing at any angle, mark off equal halves on each from the crossing, and join the four ends.

Rectangles, rhombuses and squares

Each special parallelogram has a property of its diagonals that an ordinary one doesn’t. The two proofs below are extra: they are not among the seven theorems you are asked to prove. But the exam asks you to use both properties (see How the exam asks it), and a short proof is the surest way to remember them. Both work in the same way as the theorems.

The diagonals of a rectangle are equal. In rectangle ABCD, compare △DAB and △CBA: AD = BC (opposite sides), $\angle DAB = \angle CBA = 90^\circ$, and AB is common. So $\triangle DAB \cong \triangle CBA$ by S.A.S., and DB = CA. Since the diagonals also bisect each other, all four halves are equal: OA = OB = OC = OD.

The diagonals of a rhombus bisect each other at right angles. In rhombus ABCD the diagonals meet at O. A rhombus is a parallelogram, so AO = CO (Theorem 6). In △AOB and △COB: AB = CB (sides of a rhombus), AO = CO, and OB is common, so $\triangle AOB \cong \triangle COB$ by S.S.S. Then ∠AOB = ∠COB; together they make a straight angle, so each is 90∘°. The same congruence gives ∠ABO = ∠CBO: the diagonal BD bisects the angle B. In the same way each diagonal of a rhombus bisects the angles at both its ends.

A square is both a rectangle and a rhombus, so its diagonals are equal and bisect each other at right angles.

Worked examples

Back to the mason

This proof is extra too, not one of the book’s theorems, but it answers the question the unit began with. Narayan’s strings make a quadrilateral ABCD with AB = DC = 4 m and AD = BC = 3 m. By Theorem 4 it is a parallelogram, whatever its corners. Now he pulls the frame until the two diagonals are equal, AC = BD. Then:

	Statements	Reasons
1.	In △DAB and △CBA: i. AD = BC (S) ii. AB = BA (S) iii. DB = CA (S)	i.opposite sides (his strings) ii. common side iii. the diagonals he made equal
2.	$\triangle DAB \cong \triangle CBA$	by S.S.S.
3.	∠DAB = ∠CBA	corresponding angles of congruent triangles
4.	$\angle DAB + \angle CBA = 180^\circ$	co-interior angles, $AD \parallel BC$
5.	$\angle DAB = \angle CBA = 90^\circ$	from 3 and 4
6.	every angle of ABCD is 90∘°	opposite angles of a parallelogram are equal

A parallelogram with equal diagonals is a rectangle, so the mason’s check works: equal opposite sides and equal diagonals mean square corners. For his 4 m by 3 m room, both diagonals must come to 5 m (by Pythagoras’ rule, $\sqrt{4^2 + 3^2} = 5$), exactly what Lean the frame shows at 90∘°. Both checks are needed: a trapezium whose two slanting sides are equal also has equal diagonals without being a rectangle, which is why he makes the opposite sides equal first.

Grade 9’s final exam is set to CDC’s specification grid for Compulsory Mathematics, Grades 9–10. There, Geometry is 13 marks in three questions. In the grid’s model paper, this unit is question 12, about a rhombus:

Part	Level	Marks	What it asks
a.	knowledge	1	a property of its diagonals
b.	understanding	1	find an unknown from two equal sides
c.	application	2	construct the rhombus from its diagonals (Unit 15)

Part a wants one sentence; part b one equation from a property; part c a construction with its steps. Here is one of our own in that shape.

Question 1

What must x and y be for each figure to be a parallelogram?

Question 2

Find the values of x and y.

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9