Transcript Math 258 - Webpages - Individual Webpages on Homepage

Chapter 7: Triangles and Circles
Circumcircles
• Our main goal of this section is to show that for any
triangle ABC there exists a unique circle that contains the
three vertices, A, B, and C. We say that this circle is
circumscribed about ABC and call it the circumcircle
of ABC . (We also say that ABC is inscribed in the circle.)
• The center of the circumcircle is called the circumcenter,
and it will be denoted O. The radius of the circumcircle is
called the circumradius, and it will be denoted R.
• The key step in the proof is the following lemma:
• Lemma. Given a line segment AB and a point P, the point P
will be on the perpendicular bisector of AB if and only if P
is equidistant from A and B (i.e., PA  PB).
• Proof. Let M be the midpoint of AB . We will show that PM
is perpendicular to AB if and only if PA  PB.
• Since we are proving an if and only if statement, we have
two statements to prove. First, we assume that PA  PB.
In this case PAM  PBM by SSS. So AMP  BMP

and since the sum AMP  BMP is 180 , each must be

.90. Next, assume that PM is perpendicular to AB . In this
case sinceAMP  BMP we use SAS to conclude that

. PAM  PBM . So PA  PB .
• We now prove our main theorem.
• Theorem. Given any triangle ABC, there exists a unique
circle circumscribed about it.
• Proof. Let l1 be the perpendicular bisector of BC and let
be the perpendicular bisector of AC. Let O be the
intersection point of l1 and l 2 .
• Apply the lemma to see that, since O is on
OB = OC,
and, since O is on l ,
2
OA = OC.
l1 ,
l2
• Combining these two equations, we see that OA = OB =
OC. Hence, O is equidistant from the three points A, B,
and C and there is a circle with center O containing these
three points.
• Now suppose that O is the center of a circle that passes
through each of A, B, and C. Because OA  OB  OC and
O lie on both
l1
another application of the lemma,
must
and l 2 ,and hence O   O. Therefore, only one circle
contains A, B, and C and the circumscribed circle we have
constructed is unique.
• As a corollary of the proof we get the next result.
• Corollary. In any triangle ABC the three perpendicular
bisectors of the sides AB , BC , and AC meet in a point.
• We remark that this proof is constructive. Given ABC we
not only know that a circumcircle exists but we also have a
technique to find O and we may draw the circle.
• Although O is the center of ABC in the sense that it is the
unique point equidistant from A, B, and C, we point out
that O does not have to lie inside the triengle. O lies inside
of ABC if and only if ABC is acute.
A Theorem of Brahmagupta
• When you first learned about areas you were told that the
area of a rectangle was length times width.
• It might have occurred to you that the area of a triangle
should be the product of the three sides, abc.
• Of course it isn’t, but in this section we will prove that the
product abc is related to area.
• We start off with a consideration of the product of two
sides of a triangle.
• Theorem. Given ABC , let AC = b, AB = c, let R be the
circumradius, and let h = the length of the altitude AD
from A to BC . Then bc = 2Rh.
• Proof. Let A be the point such that AA is a diameter of
the circumcircle and consider the triangles ADB and AC A
in the Figure.
• The angles Aand  B each subtend the arc AC and are
congruent, A  B . Also, ACA is inscribed in a
semicircle, so it is a right angle.
• Therefore, AC A  ADB. By AA, ADBis similar to
.AC A.So
AD AB
AC

AA
• Or, by cross multiplication, AA. AD  AB. AC . But AA  2 R
AD = h, AB = c, AC = b, and the theorem follows.
• As an immediate consequence we get the following elegant
theorem due to Brahmagupta. We use the letter K to denote
the area of triangle ABC .
• Theorem. abc = 4RK.
• Proof. By the previous theorem, bc = 2Rh. Now multiply
both sides by a to get
abc = 2Rah.
1
But K = 2 ah and the theorem follows.
Inscribed Circles
• In this section we will prove that given any triangle ABC
there is a unique circle inscribed in ABC called the
incircle of ABC .
• A circle is said to be inscribed in a triangle if each of the
three sides is tangent to the circle. The triangle is said to be
circumscribed about the circle.
• The center of the incircle (inscribed circle) is called the
incenter and is denoted I. The radius is called the inradius
and is denoted r.
• In order to prove the existence and uniqueness of incircles,
we need a lemma on angle bisectors.
• Lemma. Let  A be any angle and let P be a point in the
interior of  A . Then P is on the bisector of  A if and
only if P is equidistant from the sides of  A .
• Proof. Let PB and PC be the perpendiculars from P to the
sides of  A . We need to show that PB  PCif and only
if PAB  PAC.
• First assume that PB  PC . Then PAB  PAC by SSA
for right triangles. Since PAB and PAC are
corresponding parts, they must be congruent.
• Next, assume that PAB  PAC. Then by SAA, PAB
and PAC are congruent for they have two angles
congruent and share a side. Hence PB  PC .
• Theorem. Given any triangle ABC , there is a unique
circle inscribed in it.
• Proof. Let l1 be the bisector of  A and let l 2 be the
bisector of  B . Let l1 and l 2 intersect at the point I and
apply the lemma:
• Since I is on the bisector of  A , the distance from I to AB
equals the distance from I to AC. Similarly, since I is on the
bisector of
equals the
BC
, the
B distance from I to
distance from I to AB .
• Combining these, we see that I is equidistant from the three
sides AB , BC , and AC . So there is a circle with center
I that will be tangent to all three.
• To show that this circle is the unique incircle, suppose that
P is the center of a circle D inscribed in ABC. Drop
segments from P to E and F, the points of tangency of the
lines AB and AC , respectively, with D.
• We know from previous work that PE  AB and PF  AC;
moreover, PE = PF since both are radii of D. Applying our
lemma, we see that P is on l , the bisector of  A .
1
Similarly, P is also on l 2 , which implies that P = I and
that the first circle we constructed is unique.
• As a corollary of the proof we get the next result.
• Corollary. Given any triangle ABC , the bisectors of the
three angles meet in one point.
• We point out that our proof here is also constructive. Given
, we canABC
construct I by finding the meeting point of the
angle bisectors and we can use I to construct the incircle by
dropping perpendiculars.
• One way in which I is better behaved than O is that I
always lies inside the triangle. This fact has an interesting
implication for area.
• Theorem. If K = the area of ABC , r = the inradius, and
s = 12 times the perimeter, then K = sr.
• Proof. Draw IA, IB, and IC as in the Figure. It is clear that
the area of ABC is the sum of the areas of ABI , BCI ,
and ACI . Consider ABI . The base is AB, which has
length c, and the height is the distance from I to AB, which
is r. So area ABI  (1 / 2)cr. Likewise, BCI has area (1 / 2)ar
and ACI has area (1 / 2)br . By addition,
1
1
1
1
K  ar  br  cr  (a  b  c)r  sr.
2
2
2
2
An Old Chestnut
(The Steiner-Lehmus Theorem)
• Theorem 1. In ABC let the bisector of  B meet AC at D
and let the bisector of C meet AB at E. If BD  CE ,
then B  C.
• Theorem 2. In ABC let the altitude of the side ACmeet it
at D and let the altitude of the side AB meet it at E. If BD  CE
, then B  C.
• Theorem 2. In ABC let the median of the side AC meet it
at D and let the median of the side AB meet it at E. If BD  CE
, then B  C.
• Proof of Theorem 1. The proof will be by contradiction.
We will assume that BD  CE but that  B is larger than C
• Since B is greater than C ,ABD , which is half of  B, is
greater than half of C . So we can find a point F strictly
1

DBF

C . Let G be the
between A and D such that
2
BF
intersection point of CE and
.
• Now consider BDF and CGF . Each has BFC for one
angle. Also, FCG  FBD , since each is 1 C . Hence,
CG CF
by AA, BDF ~ CGF and thus BD  BF . 2
• By hypothesis CE = BD, which forces CG  BD . This
implies that 1  CG and, by the preceding equation, that
BD
BF  CF .
• Now that we have shown that BF  CF .we can get a
contradiction, because we have another line of reasoning
that will show the opposite.
1
1

FBC


B

C  C ,
• Consider BFC . The angle
2
2
because we assumed thatB  C. But, in any triangle, a
larger angle must be opposite a larger side. So CF  BF .
• This is a contradiction and so our initial assumption is
impossible. Assuming C  B will yield the same
contradiction. Thus the theorem is proved.
Excribed Circles
• Before discussing escribed circles, it is worthwhile to
define external bisectors of angles.
• Let BAC be any angle with bisector AD .Extend one of
the sides of the angle, say AB , in the other direction to
form BAC , the supplementary angle. Then the line AE
which bisects this angle is called the external bisector of
BAC .
• External bisectors have two simple properties that will be
important to us.
• Lemma. Let BAC have internal bisector AD and external
bisector AE. Then
(1) AD is perpendicular to AE .
(2) The points of AE are equidistant from AB and AC .
DAE  DAC  CAE
• Proof. (1)
1
1
BAC  CAB
2
2
1
1
 BAC  (180   BAC )
2
2

 90
• (2) The points of AE are on the bisector of BAC and so
they are equidistant from AC and AB  AB.
• Theorem. Let ABC be any triangle . Then the external
bisector of  B , the external bisector of C , and the
internal bisector of  A all meet in a point I a . Moreover,
there is a circle with center I tangent to the three lines AB
a
, BC , and AC .
• The circle we constructed in this manner is said to be an
excribed circle for ABC , the point I a is called an
excenter, and the radius of the circle, which we denote ra
, is called an exradius.
• The triangle ABC has two more excircles, one touching AC
and one touching AB . We denote their centers by Ib and I c
their radii as rb and rc , respectively.
• Our next theorem generalizes the area formula to the case
of excircles.
• Theorem. Let K = the area of ABC , ra = the radius of
the excribed circle opposite  A , BC = a, and s = the
semiperimeter. Then K  (s  a)ra .
• Proof. The area of ABC = the area of ABIa + the area
of ACIa  the area of BCIa (See the Figure). Taking AC
1
as the base, we see that ACIa has area bra , taking BC as
2
1
base BCIa has area ara and taking AB as base, ABIa has
2
area 1 cr . Hence the area of
1
1
1
a
cra  bra  ara 
2
2
2
1
1
1
1
ra (b  c  a)  ra (a  b  c  2a)  ra .(2s  2a)  ra .2.(s  a)  ra ( s  a).
2
2
2
2
2
ABC 
• Corollary. If r is the inradius of ABC and K is its area,
then rra rb rc  K 2 .
K
K
r

r

• Proof. By the theorem, a
. Likewise b
s

a
s b
K
K
and rc 
. Also, since K  rs , r 
. Hence
sc
s
K4
K4
rra rb rc 
 2  K2
s( s  a)(s  b)(s  c) K
by Heron’s formula.
• Corollary. 1  1  1  1
r ra rb rc
• Proof.
1 1 1 s  a s  b s  c 3s  (a  b  c) 3s  2s s 1
  


 
ra rb rc
K K K
K
K
K r
Euler’s Theorem
• In this section we will prove a celebrated theorem due to
Euler that calculates the distance from the incenter I to the
circumcenter O.
2
• Theorem. OI  R( R  2r ).
• The proof will depend upon a two part lemma. In the
circumcircle of ABC pictured in the Figure we will denote
by M the midpoint of arc BC .
• Lemma. (1) M lies on the perpendicular bisector of BC and
the bisector of  A .
(2) There is a circle with center M that contains the points
I, B, C, and I .
a
• Proof. (1) Since CM  MB , CM  MB. But since M is
equidistant from B and C, it must lie on the perpendicular
bisector of BC . Also, BAM  CAM since they
subtend equal arcs.
• (2) First notice that IB bisects  B and that BIa bisects
, where D is a point on AB extended. Since ABD is a
straight angle, the angle IBIa is a right angle. Similarly,
. ICIa is a right angle.

• Hence, if we draw a circle with IIa as diameter it will
contain the points B and C. M must be the center of this
circle because it lies on the intersection of a diameter with
the perpendicular bisector of a chord.
• Proof of Euler’s Theorem. Extend OI to a chord PQof
the circumcircle. The line AI extends to the chord AM .
Hence PI.IQ  AI.IM .
• The rest of the proof will consist of an examination of each
side of the above equation.
• We first consider the left-hand side. PI = OP – OI = R – OI
and IQ = IO + OQ = (R+OI).
2
2
R

OI
.
Hence PI.IQ = (R-OI)(R+OI) =
• To evaluate the right-hand side we draw IZ perpendicular
to AB and extend MO to a diameter MM .


Z


C

90

We claim that AIZ ~ M MC. This is because
1
and
ZAI  CM M  A . Hence AI  M M .
2
IZ
CM
• What can we say about these lengths? By the lemma
CM = IM, since they are radii of the same circle. MM  is
a diameter of the circumcircle and has length 2R.IZ is the
distance from the incenter to a side of the triangle, so it is r.
Making these three substitutions we see that AI  2 R
r
IM
, so AI . IM =2Rr.
• Now we substitute the results of each of the preceeding
paragraphs into the equation to get R 2  OI 2  2Rr or
OI 2  R2  2Rr  R( R  2r ).
• Corollary. R  2r .
Center of Gravity
• In ABC let the midpoints of sides BC, AC , and AB be A
, B , and C  , respectively. The line segments AA , BB ,
and CC connecting the three vertices to the midpoints of
the opposite sides are called the medians of ABC .
• Our main theorem states that the three medians meet in a
point G called the centroid of gravity of ABC .
• Lemma. Let B be the midpoint of AC and C  be the
midpoint of AB . Then BC is parallel to BC and is half as
long.
• Proof. The triangles ABC and AC B are similar but the
SAS theorem for similar triangles because BAC  C AB
, AC   1 AB , and AB  1 AC .
2
2
• HenceAC B  ABC and BC BC , since they make
equal corresponding angles with the transversal AC .
Finally,
BC  AC  1

 .
BC
AB 2
• Let us label the intersection point of BB and CCas G. We
don’t yet know that G is also on
. AWe
A first prove, as an
intermediate step, that G is the trisection point of BB
.
• Lemma. GB 
1
BB.
3
•
• Proof. Since BC  is parallel to BC , BBC  BBC and
. BCC   BC C since they are alternate interior angles.

Hence, by AA, GBC ~ GBC . So
GB
BC
GB 

BC 
2
• Hence GB  2GB. If we add GB to both sides we see
that BB  3GB .
• Theorem. In ABC the three medians AA , BB , and CC
meet in a point.
• Proof. Let G be the point of intersection of BB and CC .
We need to show that G also lies on AA . In order to do
this we let G1  the intersection point of AA and BB ,
and we will show that G  G1 .
BG 
1
BB 
3
• By the lemma
. However, we can also apply the
lemma to G1 . G1 is the intersection point of two medians
1
in a triangle and so it must also trisect them. Hence, BG1  BB.
3
But BB only has one trisection point closer to B so
G  G1.
• Corollary. The center of gravity trisects each of the
medians.
Length Formulas
• In this section we will calculate the lengths of the medians
of ABC . We will denote these lengths by AA  ma, BB  mb
and CC  mc .
• Theorem. 2ma  b 2  c 2 
2
1 2
a .
2
• Proof. Let AD be the altitude. We assume for convenience
that D is between B and
, as shownAin the Figure. Now
apply the geometric law of cosines to the triangles,
AC A

AB
and
to Aobtain
AA2  AB2  BA2  2BD.BA
• and
• or
• and
AA2  AC 2  CA2  2CD.CA
1
1
2
ma  c 2  ( a) 2  2( BD )( a)
2
2
1
1
2
ma  b 2  ( a) 2  2(CD )( a).
2
2
• Adding, we get
1
1
1
2
2ma  b 2  c 2  a 2  a(CD  BD )  b 2  c 2  a 2  a 2  b 2  c 2  a 2 .
2
2
2
as claimed.
• Of course, there are similar formulas for the other two
medians:
1 2
2
2
2
2mb  a  c  b
2
and
1 2
2
2
2
2mc  a  b  c .
2
• With a bit of algebra we can deduce a number of striking
formulas from these.
3 2
• Corollary. (a) ma  mb  mc  (a  b 2  c 2 )
4
1
(b) GA 2  GB 2  GC 2  (a 2  b 2  c 2 )
3
1 2
2
2
2
2
2



G
A

G
B

G
C

(
a

b

c
)
(c)
12
2
2
2
Complementary and Anticomplementary
Triangles
• Let ABC be a triangle and let the midpoints of the sides
be A , B  , and C  . Then the triangle with these three
points for vertices is called the complementary or median
triangle of ABC .
• Theorem. Given ABC with complementary triangleABC 
,
 parallel to the sides of ABC .
(1) The sides of ABC are
1
(2) ABC and ABC  are similar with ratio 2 .
(3) ABC and ABC  have the same centroid G.
• Proof. (1) We showed before that BC and BC are parallel.
The same proof shows that AB and AB are parallel and AC
and AC  are parallel.
1
• (2) we also proved that BC   .BC . The same proof
2
1
1




shows that A B  2 . AB and A C  2 . AC .
• Hence ABC ~ ABC  by SSS for similar triangles.
• (3) Let AA intersect BC at the point M. Consider the
triangle AAB and AMC . Since BC BC  ,C   B by
the corresponding angles theorem.
• Likewise, AM C   AAB . So the two triangles must be
similar by AA. Hence
C M AC  1
BA
AB

2
1

B
A

BC
But
2
1
thatC M  BC 
2
1
C M  BA.
2
• Which implies that
Putting these together we see
the midpoint of BC  .

so BA  BC .
and that M is
• With this observation in hand the rest of the proof follows
easily. AA is a median of ABC , AM is a median of
. ABC and AA  AM .

• Similar statements are true for the other two medians.
Hence, the point of intersection of the three medians of ABC
is the same as the point of intersection of the three medians
of ABC .
• Next, given a triangleABC , we define the
anticomplementary triangle of ABC to be the unique
triangle ABC  with the properties that BC contains
the point A and is parallel to BC , AC contains the point
B and is parallel to AC , and ABcontains the point C and
is parallel to AB .
• Theorem. If a triangleABC has anticomplementary
triangle ABC , then ABC is the complementary
triangle of ABC .
• Proof. To prove this theorem it is worthwhile to recall the
definition of a complementary triangle. In order to prove
that ABC is the complementary triangle of ABC  we
need to show (see the Figure) that A is the midpoint of BC
and C is the midpoint of BA. We will only prove that A
is the midpoint of BC since the other two cases are
similar.
• Because AB BC and AB BC , the quadrilateral BC BA
must be a parallelogram. We know that opposite sides of a
parallelogram are congruent, so AB  BC.
• Likewise, since AC  BC and BC  AC, the quadrilateral
BCA
C  is also a parallelogram and consequently AC  BC.
.
• Comparing these two congruences, we see that CA  AB,
which means that A must be the midpoint of BC. This is
what we wanted to show.
• Corollary.
1. ABC and ABC are similar with ratio 2.
2. ABC and ABC  have the same centroid.
The Orthocenter
• Previously, we proved that in any triangle ABC the three
medians meet in a point G, the three angle bisectors meet
in a point I, and the three perpendicular bisectors of the
sides meet in a point O.
• Here we will prove the corresponding theorem for the
altitudes, namely that the three altitudes meet in a point.
The point is called the orthocenter of ABC , and we
will denote it by H.
• We point out that H does not necessarily lie inside the
triangle: If ABC is a right triangle, then H is at the
vertex, and if ABC is obtuse, then H will lie outside.
• Theorem. If a triangle ABC has altitudes AD , BE ,
and CF , then AD , BE , and CF meet at a point.
• Proof. Let ABC  be the anticomplementary triangle
of ABC . Since A is the midpoint of BCand since AD
is perpendicular to BC , we conclude that AD is the
perpendicular bisector of BC.
• Similarly, BE is the perpendicular bisector of AC
and CF is the perpendicular bisector of AB. But we
know that the three perpendicular bisectors of the sides of
any triangle meet in a point. Hence, AD , BE , and CF
meet in a point, as claimed.
• In addition to being short, this proof yields two corollaries.
Their proofs follow from the fact that the perpendicular
bisectors of the sides of a triangle meet at the circumcenter
of that triangle.
• Corollary. Let ABC be a triangle with anticomplementary
triangle ABC . Then the orthocenter of ABC is the
circumcenter of ABC .
• Corollary. Let ABC be a triangle with complementary
triangle ABC  . Then the circumcenter of ABC is the
orthocenter of ABC  .
Fagnano’s Problem
• Here we will solve a geometric optimization problem due
to G. C. Fagnano in 1775; Fagnano also solved it. But the
proof we shall give is due to L. Fejer in 1990.
• Problem. Given an acute triangle ABC, find points X
on BC , Y on AC , and Z on AB that minimize the
perimeter of the resulting triangle XYZ .
•
We claim that FEG is the required triangle.
• First Step: Given an acute angled triangle ABC and given
a fixed point X on BC , then find points Y on AC and Z
on AB that minimize the perimeter of XYZ .
X2
A
Y
X1
Z
F
E
B
C
X
SAS
XYF  YFX2 
 XY  YX 2
SAS XZ  ZX
XZE  ZEX1 

1
 PerXYZ   X 1ZYX 2
SAS
AYX  AYX2 
A1  A2
SAS
AX1Z  AZX 

A3  A4  X 1 AX2  2A
X1 X 2  2 AX . SINA
• Second Step: We have shown that how to choose Y and Z
if we are given X. Our next goal is to determine a choice of
X. As before, let X be on BC , X  be the reflection of X
through AC , and X  the reflection of X through AB .
• We have shown that the minimum perimeter of a triangle XYZ
inscribed in ABC with X as a vertex is the length of X X .
What we have to do now is find that particular X that
minimizes the length X X  .
• Since A is on the perpendicular bisector of XX, AX  AX 
and since A is on the perpendicular bisector of XX 
,
BC
, AX
for any
.
 Achoice
X  of X on
• Also, using the fact that AX X  is isosceles and has
congruent base angles, X AC  CAX ; and since AX X 
is isosceles, we have X AB  BAX .
• Adding, we see that
X AX   X AB  BAX  XAC  CAX 
 2BAX  2XAC  2BAC
• Now consider AX X . This triangle has three properties:
1. X AX   2BAC , a relation that does not depend
upon the choice of X.
2. The sides AX and AX  are congruent to each other and
to AX .
3. The length of the base is the perimeter of XYZ , which
we want to minimize.
• Since AX X  is an isosceles triangle with a fixed summit
angle, we minimize the base by minimizing the length of
the legs AX   AX . But this is the length AX.
• Now we are as good as done. We know that of all the line
segments AX joining A to a point on BC the shortest one
is the perpendicular, AD . Therefore X = D, will be a
vertex of the triangle that solves Fagnano’s problem.
• By similar arguments the other two vertices will be E and
F and the orthic triangle is the one of smallest perimeter.
• We remark that DEF has perimeter 2 K
,where
R
= area and R = circumradius.
K
The Euler Line
• Recall the notation O = the circumcenter, G = the centroid,
and H = the orthocenter of ABC . We now have the
follwing remarkable theorem, due to Euler.
• Theorem. The three points O, G, and H lie on a straight
line, with G between O and H. Moreover, GH = 2GO.
• Proof. As usual, we let A be the midpoint of BC and AD
the altitude to BC as in the Figure. Then AA is a median,
it contains G, and AO is the perpendicular bisector of BC.
• Connect O to G and let the resulting line intersect AD at
the point P. Since AD and OA are each perpendicular to
, they
BCmust be parallel.
• Hence, by the alternate interior angle theorem APG  GO A
and GAP  GAO . By AA, we conclude that APG ~ AOG
So
OG AG

GP GA
1
• But the centroid G trisects the median AA , so AG
 .
GA 2
Hence
2OG  GP.
• This equation is the key to the proof. Consider what we
just proved from the point of view of O and G.
• We have shown that if you extend the line segment OG to
a new point P such that G is between O and P and GP  2OG
, then this point P must be on the altitude AD , since these
two conditions determine P uniquely.
• However, the same manner of proof would show that this
point P will be on the altitude BE and the altitude CF .
Thus, P is the orthocenter H and the theorem is proved.
• The line joining O, G, and H is called the Euler Line of