Transcript Chapter 4-One Way to Go: Euclidean Geometry of the Plane

Chapter 4-One Way to Go:
Euclidean Geometry of the Plane
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Neutral Geometry: all theorems derived
without making any assumptions about
parallel lines
Geometers were trying to extend neutral
geometry to include all of Euclidean but none
ever succeeded
Because several people were able to show
Euclid’s 5th postulate is independent
4.2 The Parallel Postulate and
Some Implications
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Assume Postulate 16 (or Playfair’s Postulate)
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Through a given point there is at most one line
parallel to a given line.
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Alternate def: Two lines are parallel if they are
coplanar and do not intersect.
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Now that we have assume the parallel postulate,
we can add theorems we mentioned before (as
being equivalent to the EPP)
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Theorem 4.2.1: If two parallel lines are
crossed by a transversal, the alternate
interior angles are congruent
Theorem 4.2.2: The sum of the measures of
the interior angles of a triangle is 180
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Read proof on page 129
Cor 4.2.3: The measure of an exterior angle
of a triangle is equal to the sum of the
measures of the two remote interior angles.
Proof:
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Let ΔABC exist such that D is a point on AC
such that A-C-D
Know that sΔABC=180, so m<1+m<2+m<3=180
By def of linear pair, m<2+m<4=180
By substitution, m<1+180-m<4+m<3=180
So m<1-m<4+m<3=0
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Or m<1+m<3=m<4.
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More results of EPP
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Parallelogram: A quadrilateral is a
parallelogram if and only if both pairs of
opposite sides are parallel
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Theorem 4.2.4: The opposite sides of a
parallelogram are congruent
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Read proof, page 130
More results
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Theorem 4.2.5: If a transversal intersects three
parallel lines in such a way as to make congruent
segments between the parallels, then every
transversal intersecting these parallel lines will do
likewise.
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See picture and read Proof page 131
Cor 4.2.6: If a transversal crosses three or more
parallel lines in such a way as to result in congruent
segments between the parallels, then every
transversal will do likewise.
Induction
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Proof by induction:
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Need to show true for n=1
Assume true for n=k
Then show true for n=k+1
Proof of Cor 4.2.6
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Shown true for n=3 (theorem 4.2.5)
Assume true for n=k, which implies
A1A2 = A2A3 = … = Ak-1Ak and
B1B2 = B2B3 = … = Bk-1Bk
Now show true for line lk+1
Well if A1A2 = A2A3 = … = Ak-1Ak = AkAk+1 then
B1B2 = B2B3 = … = Bk-1Bk = BkBk+1 by
definition of distance.
More Results
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Def: A median of a triangle is a line segment that has
as its endpoints a vertex of the triangle and the
midpoint of the side opposite that vertex.
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Theorem 4.2.2: Median Concurrence Theorem
(neutral)
The three medians of a triangle are concurrent at a
point called the centroid
Cor 4.2.8: Any two medians of a triangle intersect at a
point that is 2/3 the distance from any vertex to the
midpoint of the opposite side.
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More Results
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Theorem 4.2.9: Two lines parallel to the same line are
parallel to each other.
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Proof:
Let l, m, n be lines such that l || m, m||n
wts: l||n
Let t be a transversal of n,m,l
Then <1 and <7 are congruent as well as <5 and <11
Since vertical angles are congruent, <1 ≡<7 ≡ <5 ≡<11
and <1 ≡ <11
And <1 and <11 are congruent alt. Int. angles, l || n
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More Results
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Property: Parallel lines are everywhere
equidistant
Proof:
Let l || m. Draw PS ┴ m at S and QR┴m at R.
Then ‫ם‬PQRS is a parallelogram.
By theorem 4.2.4, PS ≡QR
So parallel lines are equidistant everywhere.
More Results
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There are 12 more simple theorems listed on
page 134-135 (read)
#8 Prove Theorem 4.2.11
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Each diagonal of a parallelogram partitions
the parallelogram into a pair of congruent
triangles.
Proof:
Let ‫ם‬ABCD be a parallelogram
wts: ΔABC ≡ ΔCDA and ΔBAD ≡ ΔDBC
By theorem 4.2.4, AB ≡ DC and BC ≡ AD
Draw diagonals BD and AC
By SSS, ΔABC ≡ ΔCDA and ΔBAD ≡ ΔDBC
#10 Prove Theorem 4.2.13
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If the diagonals of a quadrilateral bisect each other, then the
quadrilateral is a parallelogram
Proof:
Let ‫ם‬ABCD be a quadrilateral such that AC bisects BD and BD
bisects AC
wts: ‫ם‬ABCD is a parallelogram
Know BE ≡ ED, AE ≡ EC, and <1 ≡ <3 and <2 ≡ <4 (vertical
angles)
So ΔCED ≡ ΔAEB and ΔCEB ≡ ΔAED by SAS
Thus, AB ≡ CD and BC ≡ AD
Since opposite sides of ‫ם‬ABCD are congruent, by theorem 4.2.4,
‫ם‬ABCD is a parallelogram
#15 Prove Theorem 4.2.18
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The median to the hypotenuse of a right triangle is one-half the
length of the hypotenuse
Proof:
Let ΔABC be a right triangle with hypotenuse BC
Draw AD such that D is the midpoint of CB. Then AD is a median
of ΔABC.
wts: AD =1/2 BC
Draw line AE such that A-D-E and AD ≡ DE
Draw DE and BE
Since AE and BC bisect each other ‫ם‬ABEC is a parallelogram
(theorem 4.2.12)
So AB ≡ CE and CA ≡ EB (theorem 4.2.4)
cont.
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So by theorem 4.2.12, we have ΔABC ≡ ΔECB and
ΔBAE ≡ ΔCEA
which implies <A ≡ <E and <B ≡ <C
This implies m<A=m<E=90 and <B=m<C
Since the sum of the angles of ‫ם‬ABEC=360,
m<B=m<C=90
So by SAS, ΔABE ≡ ΔBAC
Then CB ≡ EA which implies CB=EA which implies
AD= ½ CB since they bisect each other.
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Assign 4, 13, 22, 23
4.3 Congruence and Area
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Postulates 17-20
17 (area): Every polygon has a unique area
18 (area congruence): Congruent triangles
(congruent polygons) have the same area
19 (area addition): Area, as a quantity, is
additive in nature
20 (formula): We can find the area of a
rectangle as length*width
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Theorem 4.3.1-Parallelograms that share a
common base and that have sides opposite
this base contained in the same (parallel) line
are equal (in area)
Every parallelogram has two altitudes and 2
bases.
Theorem 4.3.2-The area of a parallelogram is
the product of the lengths of its base and its
height.
Definitions
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The height of a triangle is the measure of the
perpendicular line segment that is drawn from
the base line to the opposite vertex
trapezoid: quadrilateral having two parallel
and two non-parallel sides and the height of
the trapezoid is the measure of the
perpendicular drawn between the parallel
sides
rhombus: parallelogram in which all 4 sides
are congruent
Theorem 4.3.3-The area of a right
triangle is ½ the product of the lengths
of its legs
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Proof:
Given right ΔABC with m<C=90
wts: A= ½ ab
Construct l ┴BC at B
Construct m┴ l from A
Then ABCD is a rectangle.
So, area of ACBD=ab
Also ΔABC ≡ ΔBAD, so area ΔABC = area ΔBAD
And area of ACBD= aΔABC + aΔBAD
Then aΔABC= ½ area ACBD = ½ ab
Theorem 4.3.4-The area of a triangle is
½ the product of any base and the
corresponding height
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Set up proof:
In any triangle, a= ½ bh
Have to show three cases:
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If right, each leg is an altitude to the other and leg
is the base
If acute, drop the altitude to form two right
triangles
If obtuse, drop the altitude down to the extension
of one of the sides
Case 1-right triangle
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Let ΔABC is right triangle with m<C=90
wts: a= ½ bh
Since the triangle is right, each leg is a height
to the other leg, which is a base.
So for ΔABC, by theorem 4.3.3,
aΔABC= ½ ab = ½ bh
Case 2- Acute
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Consider acute ΔABC and let CB be the base
Draw AD ┴ CB such that C-D-B
This separates the triangle into 2 right Δ ‘s
Which implies aΔADB= ½ (AD)(DB) and
aΔADC= ½ (DC)(AD)
which implies aΔACB= aΔADB + aΔADC =
½ (AD)(DB) + ½ (DC)(AD) = (½AD)(DB+DC)
= ½ (AD)(CB) = ½ bh
Theorem 4.3.5- The area of a trapezoid
is the product of its height and the
arithmetic mean of its bases
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Proof:
Draw trapezoid ABCD, such that BC || AD and BC < AD
Since AD > BC, there are two possible cases:
Case 1: AB ┴ AD
Which implies AB ┴ BC
Draw CE ┴ AD such that A-E-D
Then ‫ם‬ABCE is a rectangle and ΔCED is right
So, a‫ם‬ABCE = (AB)(AE) and aΔCED= ½ (ED)(CE)
Then aABCD= a‫ם‬ABCE + aΔCED (postulate 19)
= (AB)(AE) + ½ (ED)(CE)
Since AB ≡ CE, aABCD = (AB)(AE + ½ ED)= (AB)(BC+AD)/2
= h(b1 + b2)/2
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Definition/notations: n-gon
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Assign: #2, 3, (both we did in class), 4, 5, 6,
7, 13, 17, 18.
Turn in 5, 6, 7 Wed October 24
Also due Monday October 22: One page
description of paper
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4.4 Similarity
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Similarity: Two polygons ABCD…Z and
A’B’C’D’…Z’ are similar if and only if
m<A=m<A’, m<B=m<B’, … , m<Z=m<Z’
 AB = BC = CD = … = ZA
A’B’ B’C’ C’D’
Z’A’
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Theorem 4.4.1: Similarity is an equivalence
relation
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Theorem 4.4.2: If a line parallel to one side of a
triangle intersects the other two sides in two different
points, then it divides these sides into segments that
are proportional.
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Read Proof pages 146-148
Read Cor 4.4.3
Theorem 4.4.4: If a line l intersects two sides of a
triangle in different points so that it cuts off
segments that are proportional to the sides, then the
line is parallel to the third side.
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Read Proof page 148-149
Theorem 4.4.5- AAA Similarity: If for two
triangles, all angles are congruent, then
the triangles are similar.
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Read proof
Finish proof:
Now assume AC ≠ DF, so assume DF > AC
Locate point J between D and F such that DJ=AC
Likewise, locate I on DE s.t. DI = AB.
Since m<A =m<D, ΔABC ≡ ΔDIJ (by SAS)
So then m<IJD = m<C =m<F and thus IJ || EF
Then by Cor 4.4.3: DI/DE = DJ/DF
Since DJ=AC, DI=AB then we have AB/DE = AC/DF
Similarity Theorems
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4.4.6- If for two triangles, an angles of one
triangle is congruent to the corresponding
angle from the other triangle, and if the
corresponding sides that surround this angle
are proportional, then the triangles are
similar.
4.4.7- If for two triangles, the lengths of the 3
sides of one triangle are proportional to the
lengths of the corresponding 3 sides of the
other triangle, then the triangles are similar.
Proof for SAS Similarity
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Given ΔABC and ΔXYZ
Let <A ≡ <X and AB/XY = AC/XZ
wts: ΔABC ~ ΔXYZ
Assume m<C > m<X
Then by angle construction postulate, there exists
<YZW such that <YZW ≡ <C and Y-W-X
By AAA, ΔABC ≡ ΔWYZ which implies
AB/WY = AC/WZ which is a contradiction unless
W=X
Proof for SSS Similarity
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Given ΔABC and ΔDEF such that AB/DE = BC/EF= AC/DF
wts: ΔABC ~ ΔDEF
Find E’ on AB such that AE’=DE and find F’ on AC such that
AF’=DF
Then AB/AE’ =AC/AF’ by substitution
So ΔABC ~ ΔAE’F’ since <A ≡ <A (SAS)
So E’F’/BC=AE’/AB
So E’F’ = (BC)(AE’)/AB = (BC)(DE)/AB
And EF = (BC)(DE)/AB
So EF= E’F’
So ΔAE’F’ ≡ ΔDEF by SAS
Hence <A = <D and
ΔABC ~ ΔDEF by SAS
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Similarity is also use to prove the
Pythagorean Theorem (4.4.8)
Do number 11
Assign 1, 3, 10, 12, 13, 14
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Turn in 1, 10, 14 Monday October 29
4.5-Euclidean Results About
Circles
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Definition: A circle is the set of all points equidistant
from a given point
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The given point is called the center and the common
distance is the radius
Theorem 4.5.1- In the Euclidean plane, three
distinct, non-collinear points determine a unique
circle
Theorem 4.5.2-Every triangle can be circumscribed,
and the center of the circumscribing circle is the
concurrence point of the perpendicular bisectors of
two of the sides of the triangle.
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Def: A chord of a circle is a line segment
joining two points on the circle
Def: A diameter of a circle is a chord that
contains the center of the circle
Theorem 4.5.3: If AB is a diameter of a circle
and if CD is another chord of the same circle
that is not the diameter then AB > CD
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Read Proof Page 160
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Theorem 4.5.4: If a diameter of a circle is
perpendicular to a chord of the circle, then
the diameter bisects the chord.
Theorem 4.5.5: If a diameter of a circle
bisects a chord of the circle (which is not a
diameter), then the diameter is perpendicular
to the chord.
Theorem 4.5.6: The perpendicular bisector of
a chord of a circle contains a diameter of the
circle.
Proof Theorem 4.5.5
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Let AB be a diameter of C(O, OA)
Let CD be a chord, not the diameter on circle
Let AB bisect CD at point P
wts: AB ┴CD
Draw OC and OD. Since C and D are on the circle,
then OC ≡ OD
We know OP ≡ OP and we are given CP≡PD
So by SSS ΔCPO ≡ ΔDPO
which implies <OPC ≡ <OPD and since they form a
linear pair both must be 90.
So AB ┴CD
Other Definitions
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A secant of a circle is a line that contains exactly 2
points of the circle
A tangent of a circle is a line that contains exactly
one point of the circle
Theorem 4.5.7: If a line is tangent to a circle, then it
is perpendicular to the radius drawn to the point of
tangency.
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Read proof page 162
Any angle whose vertex is the center of a circle is
called a central angle for the circle
#7
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Given C(O,OA) and P in the exterior of circle
Draw tangents PT and PS such that T and S lie on
circle C(O,OA)
wts: PT ≡ PS
Draw OS and OT
By theorem 4.5.7, <PSO and <PTO are right angles.
Draw OP
Since OP ≡ OP and OS ≡ OT, then by hypotenuseleg congruence condition ΔOPT ≡ ΔOPS
So PT ≡ PS
#8
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Suppose t is tangent to C(O,OP) at P
Also suppose we have secant l, which is parallel to t, intersects
C(O,OP) at the two points A and B
wts: AP ≡ BP
By theorem 4.5.7 OP ┴ t which implies diameter DP ┴ t
Let the intersection of DP and AB be M
By the converse of the Alt. Int. Angle theorem, <BMP ≡ <CPM,
which implies m<BMP=90
So then m<PMA=90 since it is supplementary
And by Theorem 4.5.4, BM ≡ MA
So by SAS ΔBMP ≡ ΔAMP
Therefore BP ≡ AP
#9
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Given C(O, OA)
Let DA be a diameter of the circle
Draw line t perpendicular to AD at A
Let C be any other point on t
Draw OC
Since the perp segment from a pt to a line is the
shortest segment from the pt to a line, OC > OA
Therefore, C lies in the exterior of the circle
Hence, t is tangent to C(O,OA).
#11
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Given C(O,OA), such that OX┴AB, OY┴CD, and
OX=OY
Draw OD, OC, OA, and OB
we know that OD≡OC≡OA≡OB since they are all
radii
Then ΔOCY≡ΔODY≡ΔOAX≡ΔOBX by the
hypotenuse leg condition
Therefore, YC≡YD≡XA≡XB or YC=YD=XA=XB
so YC+YD=XA+XB
which implies CD=AB, so CD≡AB
Definitions
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Draw example of semicircle, minor arc and
major arc
Draw example of inscribed angles and
intercepted arcs and measure of angles.
Theorem 4.5.8:If two chords of a circle are
congruent, then their corresponding minor arcs
have the same measure (see pic page 165)
Converse: If two minor arcs are congruent, then
so are the corresponding chords
#17 proof
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Let arcAB ≡ arcCD
wts: AB ≡ CD
We know by definition that <AOB ≡ <COD
Draw OA, OB, OC, and OD
Since they are radii, OA≡OB≡OC≡OD
Draw AB and CD
Then by SAS ΔCOD ≡ ΔAOB
Therefore, AB≡CD
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Theorem 4.5.11: The measure of an angle
inscribed in an arc is ½ the measure of its
intercepted arc.
Proof:
Draw OA, then m<AOC=m arcAC
m<1=m arcAC
m<1=m<2+m<3
So, m<1=2m<3
Then, m<3= ½ m<1= ½ m arcAC
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Theorem 4.5.14: If two chords intersect in the
interior of a circle to determine an angle, the
measure of the angle is the average of the
measure of the arcs intercepted by the angle
and its vertical angles.
Theorem 4.5.15: If two secants intersect at a
point in the exterior of a circle, the measure
of the angle at the point of intersection is ½
the positive difference of the two intercepted
arcs.
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Suppose we are given line AB is tangent to
C(O,OA) at point A and AC is a chord such
that m arcAPC=xº and O is an interior point
of <BAC. Then what is m<BAC?
Well m<BAC=90+ ½ m arcCD
= ½ m arcAPD + ½ m arcCD
= ½ (m arcAPD + arcCD)
= ½ m arcAPC
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Theorem 4.5.16: If line AB is tangent to C(O,OA) at
point A and if AC is a chord such that m arcAPC=x,
then m<BAC= ½ xº
Proof: (#25) when AC contains the center:
wts: xº= ½ m arcAPC
It has already been shown that m<BAP = ½ m
arcAP
m<PAC= ½ m arcPC
xº= m <BAP +m<PAC = ½ m arcAP + ½ m arc PC
= ½ (m arcAP + arcPC)= ½ m arcAPC
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Homework: 10, 12, 14, 22, 24, 26, 27, 30, 31
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Look at #24 if one of the secants is the center of the
circle:
Given line AC a secant of C(O,OC) such that A-O-C
and AD a secant of C(O,OC) such that AC and AD
intersect on the exterior of circle at A
Draw EC. Then m<BCE= ½ arcBE and m<CED= ½
arcCD.
We know m<CAD+m<ACE=m<CED
So m<CAD=m<CED- m<ACE
By substitution, m<CAD= ½ (arcCD – arcBE)
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4.6 Euclidean Results of
Triangles
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Already have:
The three medians of a triangle are concurrent (at a
point called the centroid)
The three perpendicular bisectors of a triangle are
concurrent (at a point called the circumcenter)
Shortest distance from a point to a line is the
perpendicular dropped from the point to the line.
From this we have: A point is on the bisector of an
angle if and only if it is equidistant from the sides of
the angle.
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Theorem 4.6.3: The 3 bisectors of the interior
angles of a triangle are concurrent (at a point
called the incenter).
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Definition: An altitude of a triangle is a
perpendicular line segment from a vertex to
the side opposite it
Theorem 4.6.4: The lines containing the 3
altitudes of a triangle are concurrent.
(orthocenter)
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Read through 4.6 proofs and read 4.7 by
Wednesday.
Assign: 4.6: #7,