Download Proof - Jeremy Cue

Math 166: Advanced Euclidean Geometry
Jeremy Cue
Problem 1: Given a circle and its center, construct a regular octagon inscribed in the
circle. Prove that your construction works.
Construction: Consider the given circle with center O.
F
E
G
D
O
H
C
A
B
Construction Steps:
1) Construct a line through O. Get the points A and E as the intersection of this line and
the circle (so line segment AE is any diameter of the circle).
2 – 4) Construct a line through O perpendicular to AE. Get the points C and G as the
intersection of this line and the circle (so GC is also a diameter of the circle).
5 – 8) Construct the angle bisector of angle AOC. Get the point B as the intersection of
this bisector with the circle.
9 – 12) Construct the angle bisector of angle COE. Get the point D as the intersection of
this bisector with the circle.
13 – 16) Construct the angle bisector of angle EOG. Get the point F as the intersection of
this bisector with the circle.
17 – 20) Construct the angle bisector of angle GOA. Get the point H as the intersection of
this bisector and the circle.
21 – 28) Construct line segments AB, BC, CD, DE, EF, FG, GH, HA.
Claim: Octagon ABCDEFGH is both equilateral and equiangular, meaning it is a regular
octagon.
Proof: By construction, we know that the angles AOC, COE, EOG, and GOA are right
angles (because CG was constructed perpendicular to AE). Therefore, the 8 angles AOB,
BOC, COD, DOE, EOF, FOG, GOH, HOA are each equal to one half of a right angle
(because they were constructed by bisecting right angles). Therefore, these 8 angles are
all congruent to one another.
Also, OA, OB, OC, OD, OE, OF, OG, and OH are all congruent to one another because
they are radii of the same circle. Therefore, the 8 triangles in the above construction are
all congruent to one another by I.4 (SAS). Thus, all the corresponding sides and angles of
these 8 triangles are congruent, so AB, BC, CD, DE, EF, FG, GH, and HA are all
congruent to one another, and all the corresponding interior angles are congruent.
Therefore, all the sides of this octagon are congruent, so it is equilateral. Also, all the
angles of this octagon are congruent, so it is equiangular. Hence, octagon ABCDEFGH is
regular.
Problem 2: Settle the following conjecture:
Conjecture: Let ABCD be a quadrilateral. Then ABCD is circumscriptable if, and only
if, AB + CD is in the same segment class as AD + BC.
Proof: This conjecture needs to be proven in both logical directions:
First, we will assume that ABCD is circumscriptable and show that AB + CD is in the
same segment class as AD + BC.
Consider a circumscriptable quadrilateral ABCD and the circle with center O tangent to
each of the four sides at points E, F, G, and H:
B
E
F
A
O
C
H
G
D
Mr. Winder’s Theorem 21 tells us that two tangent lines to a circle from a point outside of
the circle are congruent. Therefore, AH is congruent to AE, BE is congruent to BF, DG is
congruent to DH, and CG is congruent to CF. This means that
AE + BE + DG + CG is in the same segment class as AH + BF + DH + CF,
(AE + BE) + (DG + CG) is in the same segment class as (AH + DH) + (BF + CF),
so
(AB) + (CD) is in the same segment class as (AD) + (BC),
because AB is in the same segment class as AE + BE, CD is in the same segment class as
DG + CG, AD is in the same segment class as AH + DH, and BC is in the same segment
class as BF + CF.
The first logical direction has been proven. In order to prove the second direction, I will
first prove a short lemma.
Lemma (Problem 2): In an isosceles triangle, the angle bisector of the angle contained
by the two congruent sides is the perpendicular bisector of the third side.
Proof: Consider the isosceles triangle ABC with AB congruent to BC and the angle
bisector of angle ABC meeting side AC at point D.
B
A
C
D
We know angle CAB is congruent to angle ACB by I.5, AB is congruent to BC, and
angle ABD is congruent to angle CBD. Therefore, by I.26 (ASA), triangle ABD is
congruent to triangle CBD. Thus, AD is congruent to DC and angle ADB is congruent to
angle CDB because corresponding parts of congruent triangles are congruent. So, by
Book 1, Definition 10, angles ADB and CDB are right angles. Therefore, the angle
bisector of angle ABC is the perpendicular bisector of AC.
Now, we will assume that AB + CD is in the same segment class as AD + BC and show
that ABCD is circumscriptable.
There are three cases to consider: 1) All four sides of ABCD are congruent (a rhombus),
2) Two pairs of adjacent sides of ABCD are congruent (a kite), 3) No sides of ABCD are
congruent.
Case 1: All four sides of ABCD are congruent (rhombus)
Consider the rhombus ABCD with diagonals AC and BD:
G
B
C
F
O
H
A
E
D
We know that the diagonals bisect the angles at A, B, C, and D. We also know that the
diagonals of a rhombus meet at point O by Mr. Flugum’s proof to Problem 2. Now,
construct lines perpendicular to each side through point O (lines OE, OH, OG, and OF).
Consider triangles AEO and AFO. They share a side and have two corresponding
congruent angles (the angles bisected at angle DAB and the right angles), so they are
congruent by I.26 (AAS). Therefore, OE is congruent to OF. Similarly, OF is congruent
to OG, and OG is congruent to OH, so all four of these perpendicular segments are
congruent. So, since E, F, G, and H are all the same distance from O, a circle with center
O containing these points exists. Furthermore, because the four sides of the rhombus
meet radii of this circle at right angles, by III.16, these four sides are tangent to the circle.
Therefore, ABCD is circumscriptable.
Case 2: Two Pairs of Adjacent Sides of ABCD are congruent (kite)
Consider the kite ABCD with angle bisectors at angles A, B, C, and D.
B
C
E
A
O
D
Consider triangles ABC and ADC; they are congruent by 1.8 (SSS), so the corresponding
angles are congruent. Therefore, AC is the angle bisector of both angle A and angle C,
and angles B and D are congruent. We also know that the angle bisector of angle D (DO)
must meet AC because the angle bisector of angle D must cross the opposite side of its
corresponding triangle (triangle ADC), which is AC. Similarly, the angle bisector of B
must meet AC. I want to show that all four angle bisectors meet at the same point,
namely point O.
We know the angle bisector of angle D meets the angle bisector of angle A and angle C at
point O (because point O lies on line segment AC, which is the angle bisector of both
angle A and angle C). Therefore, the only thing left to show is that the angle bisector at
angle B also meets at point O.
Assume the angle bisector of angle B meets AC at a point E different from point O, and
assume AE is greater than AO (because one of them must be greater). Because angles B
and D are congruent (shown in the first paragraph), angle ADO is congruent to angle
ABE because DO and BE bisect angles D and O, respectively. Also, side AD is
congruent to side AB because ABCD is a kite. Also, angle DAO is congruent to angle
BAE because line segment AC bisects angle A. Therefore, by I.26 (ASA), triangle ADO
is congruent to triangle ABE, so AE must be congruent to AO (CPCTC). This is a
contradiction, because AE is greater than AO. Therefore, E must be the same point as O.
Thus, the angle bisector at angle B meets the other three angle bisectors at point O.
Now, consider kite ABCD with perpendiculars drawn from each side to point O:
B
G
F
C
A
O
E
H
D
Consider triangles DEO and DHO. They share a side and have two corresponding
congruent angles (the angles bisected at angle D and the right angles), so they are
congruent by I.26 (AAS). Therefore, OE is congruent to OH. Similarly, OH is congruent
to OG, and OG is congruent to OF, so all four of these perpendicular segments are
congruent. So, since E, H, G, and F are all the same distance from O, a circle with center
O containing these points exists. Furthermore, because the four sides of the kite meet
radii of this circle at right angles, by III.16, these four sides are tangent to the circle.
Therefore, ABCD is circumscriptable.
Case 3: No sides of ABCD are congruent.
Assume AB is greater than BC (because one of them must be greater – if they were
congruent, the assumption that AB + CD = AD + BC would force CD to be congruent to
AD, which would be case 1 or case 2). So, since AB is greater than BC, this forces AD to
be greater than CD. Construct BE congruent to BC on AB, and construct DF congruent to
CD on AD. Also, construct the angle bisectors of angles A, B, and D. Consider the
following diagram with these conditions:
A
F
D
E
O
C
B
(AB) + CD is in the same segment class as (AD) + BC, so
(AE + EB) + CD is in the same segment class as (AF + FD) + BC,
and since EB is congruent to BC and CD is congruent to FD, we know that AE is
congruent to AF. Therefore, we have three isosceles triangles, namely AEF, BCE, and
DCF. We constructed the angle bisectors of angles A, B, and D, and by Lemma (Problem
2), the angle bisector of angle A is the perpendicular bisector of EF, the angle bisector of
angle B is the perpendicular bisector of CE, and the angle bisector of angle D is the
perpendicular bisector of CF. These three perpendicular bisectors are the perpendicular
bisectors of triangle CEF, and by Ms. Gilles’ proof to Problem 18, they all meet at the
same point, namely point O.
Now, construct perpendiculars from each of the four sides to point O:
A
I
D
O
H
J
C
G
B
Consider triangles BGO and BHO. They share a side and have two corresponding
congruent angles (the angles bisected at angle B and the right angles), so they are
congruent by I.26 (AAS). Therefore, OG is congruent to OH. Similarly, OH is congruent
to OI, and OI is congruent to OJ, so all four of these perpendicular segments are
congruent. So, since G, H, I, and J are all the same distance from O, a circle with center
O containing these points exists. Furthermore, because the four sides of ABCD meet radii
of this circle at right angles, by III.16, these four sides are tangent to the circle. Therefore,
ABCD is circumscriptable.
The second logical direction has been proven, so the if and only if conjecture has been
proven.
Problem 3: Given a triangle ABC and a segment DE, construct a rectangle with the same
content as the triangle ABC and with DE as one side.
Consider the given triangle ABC:
B
G
F
A
C
As a part of problem 55, Mr. Andera constructed (and proved that the construction
worked) a rectangle with the same content as triangle ABC with AC as one of its sides.
Using Mr. Andera’s construction, we obtain rectangle ACFG, so this rectangle and
triangle ABC have equal content. Now, consider the following construction:
D
A
F
E, C
H
•
•
•
•
Construct line segment AC + CF.
Place the given line segment DE so that point E and point C are the same point.
Construct a circle through points A, D, and F (by Euclid IV.5).
Extend DE (the same segment as DC) to the other side of the circle and get point
H.
We know the rectangle on AC and CF has the same content as the rectangle on DC and
CH (equivalently, the rectangle on DE and EH) by Euclid III.35. So, rectangle ACFG has
the same content as the rectangle with sides DE and EH.
Now, construct a rectangle with sides DE and EH by following the steps below:
Y
H
I
X
D
•
•
•
E
Construct EH perpendicular to DE.
Construct a line segment HX perpendicular to EH.
Construct a line segment DY perpendicular to DE. Get the point I as the
intersection of these two line segments.
Claim: DEHI is a rectangle with the same content as triangle ABC.
Proof: First, we will show that DEHI is a rectangle. By construction, the angles at D, E,
and H are right angles. By I.29, DE and IH are parallel (because straight line EH falls on
them and the interior angles on the same side add up to two right angles). Similarly by
I.29, EH and DI are parallel. Therefore, DEHI is a parallelogram, so the opposite angles
are congruent to one another by I.34. Thus, DEHI has four right angles, so it is a
rectangle.
As shown earlier, the rectangle with sides DE and EH (DEHI) has the same content as the
rectangle with sides AC and CF (ACFG). But, as shown earlier, rectangle ACFG has the
same content as triangle ABC. Therefore, by axiom E.C. 2, rectangle DEHI has the same
content as triangle ABC.
Therefore, we have constructed a rectangle DEHI with the same content as triangle ABC
and with DE as one side.