Quadrilaterals
MA 341 – Topics in Geometry
Lecture 21
Ptolemy’ss Theorem
Ptolemy
Let a, b, c, and d be the
l
lengths
h of
f consecutive
sides of a cyclic
quadrilateral and let x
and y be the lengths of
the diagonals. Then
ac + bd = xy.
17-Oct-2011
MA 341 001
b
y
a
x
c
d
2
Ptolemy’ss Theorem
Ptolemy
We have: ∆ABP ~ ∆CDP &
∆
∆BCP
~ ∆DAP.
∆
So
a u r
b u s
 
and
 
c s v
d r v
C
b
B
s
u
a r
P
v
c
A
d
as = uc, br = ud, and uv = rs.
D
sa2 + rb2 = uac+ubd = u(ac+bd)
xu2+xrs = xu2+xuv=xu(u+v)=uxy
By Stewart’s Theorem
u(ac
(  bd)  ssa2  rb
b2  xu2  xrss  uxy
ac  bd  xy MA 341 001
17-Oct-2011
3
The Converse of Ptolemy’s
Theorem
Let a, b, c, and d be the lengths of consecutive
sides of a quadrilateral and let x and y be the
lengths of the diagonals. If
ac + bd = xy,
then the quadrilateral is a cyclic quadrilateral.
quadrilateral
17-Oct-2011
MA 341 001
4
Euler’ss Theorem
Euler
Let a, b, c, and d be the
l
lengths
h of
f consecutive
sides of a quadrilateral,
m and n lengths of
diagonals, and x the
distance between
midpoints of diagonals.
Then
b
p
a
x
c
q
d
a2 + b2 + c2 + d2 = m2 + n2 + 4x2
17-Oct-2011
MA 341 001
5
Brahmagupta’ss Theorem
Brahmagupta
There is an analog of
H
Heron’s
’s Formula
F
l for
f
special quadrilaterals.
b
a
Let a, b, c, and d be
lengths
g
of consecutive
sides of cyclic
quadrilateral, then
c
d
K  ( s  a)( s  b)( s  c )( s  d )
17-Oct-2011
MA 341 001
6
Area of a Quadrilateral
C
c
D
Using triangle trigonometry
you can show that
q
d
θ
p
b
A
1
1 2
A  pq sin   (b  d 2  a 2  c 2 ) tan  a
2
4
B
1

4 p 2 q 2  (b 2  d 2  a 2  c 2 ) 2
4
 ( s  a )( s  b)( s  c)( s  d )  abcd cos 12 ( A  C )
17-Oct-2011
MA 341 001
7
Maltitudes
For a quadrilateral the maltitude
( id i t altitude)
(midpoint
ltit d ) iis a perpendicular
di l
through the midpoint of one side
perpendicular
di l tto th
the opposite
it side.
id
17-Oct-2011
MA 341 001
8
Maltitudes
Rectangle
Square – same
Parallelogram
17-Oct-2011
MA 341 001
9
Maltitudes
Rhombus
Kite
17-Oct-2011
MA 341 001
10
Maltitudes
Trapezoid
Isosceles trapezoid
17-Oct-2011
MA 341 001
11
Maltitudes
Cyclic quadrilaterals
17-Oct-2011
MA 341 001
12
Quadrilaterals and Circles
• For a cyclic quadrilateral,
quadrilateral area is easy
and there are nice relationships
• Maybe maltitudes of cyclic quadrilateral
are concurrent
• Can
C we tell
t ll when
h a quadrilateral
d il t
l is
i
cyclic?
• Can we tell when a quadrilateral has an
inscribed circle?
17-Oct-2011
MA 341 001
13
Theorem
For a cyclic quadrilateral the maltitudes
intersect in a single point, called the anticenter.
center
17-Oct-2011
MA 341 001
14
Proof
Let O = center of circle
G = centroid,
intersection of midlines
Let T = point on ray
OG so that OG = GT
17-Oct-2011
MA 341 001
15
Proof
PG = RG
OG = GT
PGT = RG0
∆PGT = ∆RG0
PTG = ROG
PT || OR
17-Oct-2011
MA 341 001
16
Proof
OR  CD
Thus, PT  CD
T lies
li on maltitude
ltit d
through P
Use ∆RGT = ∆PG0
to show T lies on
g R
maltitude through
17-Oct-2011
MA 341 001
17
Proof
Using other midline we show T lies on
maltitudes through Q and S.
17-Oct-2011
MA 341 001
18
Anticenter
T = anticenter =
i t
intersection
ti of
f
maltitudes
G = midpoint
O = circumcenter
OG = GT
17-Oct-2011
MA 341 001
19
Other Anticenter Properties
Perpendiculars from
midpoint
id i t of
f one
diagonal to other
intersect
T.
i t
t att T
17-Oct-2011
MA 341 001
20
Other Anticenter Properties
Construct 9-point
circles
i l of
f the
th f
four
triangles ∆ABD,
∆BCD,
∆ABC,, and
∆ADC.
The 4 circles
intersect at the
anticenter.
anticenter
17-Oct-2011
MA 341 001
21
Other Anticenter Properties
The centers of the
9 i t circles
9-point
i l are
concyclic with
center
T.
t T
17-Oct-2011
MA 341 001
22