A tour around the geometry of a cyclic quadrilateral

7/28/2019 A tour around the geometry of a cyclic quadrilateral

1/31

A Tour around the

Geometry of the

Cyclic Quadrilateral

School of Science, Mathematics and Technology Education

Faculty of Education

University of KwaZulu Natal

Durban

12 April 2013

Dr Chris Pritchardd

m2

m3m4

m1

ZY

X

W

P


2/31

Alternate angles have equal sums


3/31

Interior angle bisectors of any quadrilateral

meet at the vertices of a cyclic quadrilateral

360

360 180

180

EHG GFE a b c d


4/31

Exterior angle

bisectors of an

arbitrary quadrilateralmeet at the vertices of

a cyclic quadrilateral

(and at the centres of

the escribed circles)


5/31

Proof


6/31

Perpendicular Bisectors

Perpendicular bisectorsare concurrent at the

circumcentre and divide

the cyclic quad into four

smaller cyclic quads

The circumcentres ofthe smaller cyclic

quads form the vertices

of another cyclic quad

This central cyclic quadis similar to the original.

In fact, STUVis an

enlargement ofABCD

(factor , centre O)


7/31

Four Beer Mats Theorem

Four beer mats are

placed so that their

circumferences all

pass through aparticular point, P.

A dinner plate can

be positioned to fit

over the beer matsexactly.


8/31

Four Beer Mats Theorem

Draw in four

equal diameters

to prove it!


9/31

Five Beer Mats Theorem

Four beer mats are

placed so that their

circumferences all

pass through a

particular point, P.Common tangents

are drawn pairwise to

produce another

quadrilateral. An evenlarger dinner plate

can be positioned

over the quadrilateral

exactly.


10/31

Revealing the Fifth Beer Mat

The fifth beer

mat has the

centres of the

first four on itscircumference.


11/31

Miquels Six

Circle Theorem

ABCD is a cyclic quadrilateral.

Four circles, with centres outside

the quadrilateral, are drawn sothat the sides of the quadrilateral

are chords.

The second set of circle intersections,

E ,F, G, Hform the vertices of

another cyclic quadrilateral.

F

E

G

H

C

D

A

B


12/31

Pairwise Incentres of a Cyclic Quadrilateral

ABCD is divided into four triangles, either side of a diagonal, twice.

Incircles are drawn.

The incentres form the vertices of a rectangle, JKLM.


13/31

Pairwise Incentres of a Cyclic Kite

Now stir in some

bilateral symmetry

and the rectangle

becomes a square.

And a square has

equal diagonals.


14/31

Pairs of Radii

Pairs of radii have

equal sum.

This is still true

if the symmetryis dropped, i.e.

for the general

cyclic quad.


15/31

A Sangaku Cyclic Polygon Theorem (c. 1800)

F

F F

F

A

B

D

C

E

A

B

D

C

E

A

B

D

C

E

A

B

D

C

E

Divide a cyclic

polygon into

triangles arbitrarily.

The incircle radii

have the same total

length, regardless

of the configurationchosen.


16/31

. . .AC BD AB CD BC DA If is any point on

the minor arc ,

then

B

ABC

BD AB CB

Ptolemys Theorem Van Schootens Theorem(special case of Ptolemys Theorem

whenACD is equilateral)


17/31


18/31

Length of Diagonal of Regular

Pentagon of Unit Side

2 .1 1.1a a 2

1a a 2 1 0a a

1.618a

.


19/31

Diagonal of Regular Heptagon of Unit Side

.

Applying Ptolemys Theorem toRVWXyields ab b a

.

2 1a b

3 2Eliminate to give 2 1 0.b a a a

Applying Ptolemys Theorem toRTWXyields


20/31

3 2 2 1f a a a a

Need the root which is greater than 1.

Solution: 1.802, 2.247.a b


21/31

Addition Formula for Sine:

Defining the Lengths of the Sides

Consider a circle with

unit diameter; then sin

cossin

cossin ( + )

+W

Z

OY

X

sin , sin ,

cos , cos ,

XY YZ

WX WZ

The full version of the Sine

Rule for triangle WXZis that:

sin=

sin=

sin= 2 = 1 if =

1

2

w= sinW

XZ = sin (+)


22/31

Addition Formula for Sine:

Applying Ptolemys Theorem

By Ptolemys Theorem

. . .1.sin sin cos cos sin

sin sin cos cos sin

WY XZ XY WZ WX YZ

sin

cossin

cossin ( + )

+W

Z

OY

X


23/31

cos

sin Z'

-

cossin

sin

sin ( - )

cos

WO

Y

X

90 -

sin (90 - )

sin (90 - ) sin

sin

sin (90 - ( - ))

90 -

90 - ( - )W

Z

OY

X

90 - ( + )90 - 90 -

Z'

sin (90 - )

sin (90 - )sin

sin (90 - ( - ))sin

WO

Y

X

Equivalent

Diagrams for other

Addition Laws

cos


24/31

Double Angle Formulas

X'

sin

sin

cos

cos

sin 22

W

OY

X

1.sin2 sin cos cos sin

Let WY= 1

2 2

2 2

2 2

sin 90 1.sin 90 2 sin

cos cos 2 sin

cos 2 cos sin

90 -

X'

90 - 2sin

sin (90 - )

sin

W

O

Y

X


25/31

Brahmagupta

(1) If the elements of two Pythagorean triples (k, l, m) and (K,L,M) arecombined to form products kM, mL, lMand mKrepresenting the

lengths of the sides of a quadrilateral, then the quadrilateral is cyclic.

A s a s b s c s d

(2) If the sides of a cyclic quadrilateral have lengths a, b, c, dand semi-

perimeters, then its area is given by:

(3) The lengths of the diagonals of a cyclic quadrilateral are given by:

ad bc ac bd xab cd

ab cd ac bd yad bc


26/31

Brahmaguptas Trapezium

If the elements of two Pythagorean

triples (k, l, m) and (K,L,M) are

combined to form products kM, mL, lMand mKrepresenting the lengths of the

sides of a quadrilateral, then the

quadrilateral is cyclic.

(3, 4, 5) and (8, 15, 17) give a (40, 51, 68, 75) cyclic quadrilateral.


27/31

Area and Diagonals of the

(40,51,69,75) Cyclic Quad

3234A s a s b s c s d

84

ad bc ac bd

x ab cd

77

ab cd ac bd y

ad bc

y

77

84x

421/2R

68

40

51

75

d

c

a

b

O

12

42

4

ab cd ac bd ad bcR

A

and by a formula of Paramesvara:


28/31

Brahmaguptas Theorem and Corollary

For a cyclic quadrilateral withperpendicular diagonals meeting

at P(known as the anticentre), lines

drawn perpendicular to the four sides

through Pbisect the opposite sides.

P

T

M

B

DA

C

34

20251/2

371/2

68

40

51

75

O and the distance of thecircumcentre from each side is half

the length of the opposite side


29/31

References 1 Alexander Bogomolnys Cut-the-knot website

Antonio Gutierrez website, Geometry Step by Step from the

Land of the Incas Eric Weissteins Wolfram MathWorld

H S M Coxeter & S L Greitzer, Geometry Revisited, MAA,1967.

Honsberger, R. More Mathematical Morsels. Washington, DC:

MAA (1991), 36-37; also Episodes in Nineteenth andTwentieth Century Euclidean Geometry. Washington, DC:MAA (1995), 35-40.

Hidetoshi Fukagawa & Daniel Pedoe, Japanese TempleGeometry Problems, Winnipeg: Charles Babbage ResearchFoundation, 1989.

Roger A Johnson, Modern Geometry: An Elementary Treatiseon the Geometry of the Triangle and the Circle. Boston, MA:Houghton Mifflin, 1929; also,Advanced Euclidean Geometry,New York: Dover, 1960.

Eli Maor, Trigonometric Delights, Princeton University Press,1998.


30/31

References 2

Michael de VilliersSome Adventures in Euclidean

Geometry(2009)

Rethinking Proof with Geometers

Sketchpad, Key CurriculumPress, (2003)

Chris Pritchard (ed.)

The Changing Shape of Geometry,

Cambridge University Press / MAA (2003)
http://www.google.co.uk/url?sa=i&rct=j&q=michael+de+villiers+geometry&source=images&cd=&cad=rja&docid=N1EYEwVrrW6GRM&tbnid=ovTV-dpi6N9EsM:&ved=0CAUQjRw&url=http://www.amazon.com/Some-Adventures-In-Euclidean-Geometry/dp/images/0557102952&ei=AtJCUb8BwezSBdD8gJgI&psig=AFQjCNHXQD_bAhwkra659mEj6sbsFlyh6Q&ust=1363420032059339http://www.google.co.uk/url?sa=i&rct=j&q=chris+pritchard+geometry&source=images&cd=&cad=rja&docid=GEX30UBHsawffM&tbnid=RCpjeD_yeCxhjM:&ved=0CAUQjRw&url=http://www.mei.org.uk/?section=resources&page=books2&ei=cNFCUf6HLKbX0QXNsYDIBQ&bvm=bv.43828540,d.d2k&psig=AFQjCNFxxxQOf7SdJkzavwDtGc4ChIZxZA&ust=1363419815475540


31/31

A Tour around the

Geometry of the

Cyclic Quadrilateral

School of Science, Mathematics and Technology Education

Faculty of Education

University of KwaZulu Natal

Durban

12 April 2013

Dr Chris Pritchard

[email protected]

d

m2

m3m4

m1

ZY

X

W

P
mailto:[email protected]:[email protected]:[email protected]:[email protected]

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A tour around the geometry of a cyclic quadrilateral