Transcript 3.2 More Neutral Theorems

3.2 More Neutral Theorems
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Pasch’s Axiom-If a line l intersects Δ PQR at
point S such that P-S-Q, then l intersects
PR or RQ
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Crossbar Theorem-If X is a point in the
interior of triangle Δ UVW, then UX
intersects WV at a point Y such that W-Y-V
   
3.3 Congruence Conditions
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We already had SAS congruence for
triangles (postulate 15)
Can also have
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ASA
AAS
SSS
Isosceles Triangle Theorem
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If 2 sides of a triangle are congruent then the
angles opposite them are congruent.
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Converse Isosceles Theorem:
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If 2 angles of a triangle are congruent then the
sides opposite them are congruent.
Converse Isosceles Proof
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Suppose we have ΔABC with <A ≡ <C. Now
draw BD as the bisector of <ABC. Let D be
the intersection of BD with AC
Want to Show that ΔABD ≡ ΔCBD.
Since <BAC ≡ <BCA, <ABD ≡ <CBD, and
BD ≡ BD , we know that ΔABD ≡ ΔCBD by
AAS, therefore BA ≡ BC
Hence, the sides opposite the angles are
congruent.
Inverse Isosceles Theorem
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If 2 sides of a triangle are not congruent, then the
angles opposite those sides are not congruent, and
the larger angle is opposite the larger side.
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Similarly, If two angles of a triangle are not
congruent, then the sides opposite them are not
congruent and the larger side is opposite the larger
angle.
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Proof
Triangle Inequality Theorem
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Triangle Inequality Theorem:
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The sum of the measures of any 2 sides of a
triangle is greater than the measure of the third.
The Hinge Theorem:
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Draw picture
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Do #6 and #10
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Assign #4, #7
3.4 The Place of Parallels
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We have not talked about parallel lines
because we have only assumed SMSG
postulates 1-15
However, in Neutral Geometry it can be
shown that at least one line can be drawn
parallel to a given line through a point not on
that line.
Alternate Interior Angle Thrm
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If two lines are intersected by a transversal such
that a pair of alternate interior angles formed is
congruent, then the lines are parallel.
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Read Proof
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Note: the converse (what you were probably given in high
school geometry), which states that if two parallel lines are
intersected by a transversal then the pairs of alternate
interior angles are congruent is NOT a theorem in neutral
geometry.
Corollary 3.4.2
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Two lines perpendicular to the same line are
parallel.
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Proof: Given lines l, m, n such that l ┴ n, m ┴ n. Then wts: l
is parallel to m.
By definition, the intersection of perpendicular lines
produces four right angles.
Since l and m are both intersected by n at right angles, the
alternate interior angles are congruent.
By Alt. Interior Angle Thrm, l is parallel to m.
Therefore two lines perpendicular to the same line are
parallel.
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Corollary 3.4.3
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l
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m
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n
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If two lines are intersected by a transversal such that
a pair of corresponding angles formed is congruent,
then the lines are parallel .
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Proof: Given n is a transversal of l and m, and
corresponding angles <2 and <6 are congruent.
wts: l is parallel to m
Since <2 and <4 are vertical angles, <2 ≡ <4.
Since <2 ≡ <4 and <2 ≡ <6, by transitivity of congruence,
<4 ≡ <6.
Since <4 and <6 are alternate interior angles, by alt. int.
angle thrm. we have that l is parallel to m.
Corollary 3.4.4
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If two lines are intersected by a transversal
such that a pair of interior angles on the
same side of the transversal is
supplementary, then the lines are parallel.
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Proof
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l
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Given n is a transversal of l and m and angles 1 and 6
are interior angles on the same side that are
supplementary.
wts: l is parallel to m
Since <1 and <6 are supplementary, m<1+m<6 =180.
n
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We know that <6 and <7 are supplementary by
definition, so m<6+m<7=180.
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This implies that m<1=m<7 and so <1 ≡ <7.
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Since <1 and <7 are congruent alt. interior angles, by
alt. int. angle theorem, I is parallel to m
#3
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l
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Prove: If two lines are intersected by a transversal
such that a pair of alternate interior angles is
congruent, then the lines have a common
perpendicular.
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Proof: Given n is a transversal of l and m and <1 ≡<7 , and
<4 ≡ <6.
wts: l and m have common perp, i.e. let n be perpendicular to
l then show n is perpendicular to m
Since n is perp to l, then m<1=m<2=m<3=m<4=90
Since <1 ≡<7 , and <4 ≡ <6 , m<6=m<7=90.
So m<8=m<5=90 since they are vertical angles.
Therefore n is perpendicular to m.
So they share a common perpendicular.
Equivalences to Euclidean
Parallel Postulate
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Theorem (3.4.5) Euclid’s 5th postulate is
equivalent to the Euclidean parallel postulate.
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Euclid’s 5th: That, if a straight line falling on
two straight lines makes the interior angles
on the same side less than two right angles,
the straight lines, if produced indefinitely,
meet on that side on which are the angles
less than the two right angles.
Proof
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(→) Assume Euclid’s 5th and prove EPP.
Given line m and point A not on m, draw line t perp to m through
A which intersects m and B.
Let l be a line through A perp to t.
By Cor 3.4.2, l || m
wts: there is not another line through A || to m
Let n be another line through A || to m
Since n ≠ l then m∩n = Ø and E is interior to <DAB
So m<EAB < m<DAB and so <EAB is acute.
Therefore m<EAB + m<ABC < 180 and m∩n ≠ Ø which is a
contradiction.
So there is only one parallel line through A.
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(←) Assume EPP and prove Euclid’s 5th.
Assume m<1 + m<2 < 180.
Prove l intersects m on the same side of t.
Draw ray AD such that <BAD ≡ <3.
So ray AD || m (Alt. Int. Angle Thrm)
So since l ≠ AD, the EPP guarantees l∩m ≠ Ø
In order to show I and m meet on the same side of m (on the
RHS), we will assume l∩m on the LHS of t.
Then <1 is exterior angle of ΔABC.
This implies <1 < <DAB which implies <1 < <3
This is a contradiction, so l∩m on the RHS.
Other equivalences to EPP
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The EPP is equivalent to
The converse of the Alt. Int. Angle Theorem
If a line intersects one of two parallel lines,
then it intersects the other.
If a line is perpendicular to one of two
parallel lines, then it is perp to the other.
Given ΔPQR and any line segment AB,
there exists ΔABC having a side ≡ to AB
and ΔABC is similar to ΔPQR.
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Assign #5, 6, 10, 11
#5 and #10 for turn in (due Monday Sep 24)
3.5 The Saccheri-Legendre
Theorem
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Saccheri studied a specific class of
quadrilaterals now called Saccheri
quadrilaterals.
‫ם‬ABCD where AB ≡ CD, <A ≡ <D =90
So either <B and <C are right angles, acute
angles or obtuse angles
He tried to show that if they were acute/obtuse
you get contradictions
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Saccheri-Legendre Theorem:
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Lemma 3.5.2
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The angles sum of any triangle is less than or
equal to 180
The sum of any 2 angles of a triangle is less than
180
Lemma 3.5.3
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For any ΔABC, there exists ΔA’B’C’ having the
same angle sum but m<A’ ≤ ½ m<A
To prove theorem, reused this lemma repeatedly
Cor 3.5.4
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The angle sum of any convex quadrilateral is
less than or equal to 360
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Convex: ‫ם‬ABCD is convex if a pair of
opposite sides exist AB and CD such that CD
is contained in one of the half-planes
bounded by line AB and AB is in one of the
half planes bounded by line CD
Proof
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wts: m<A + m<B +m<C + m<D ≤ 360
Assume we have convex ‫ם‬ABCD.
Draw line segment BD.
It separates ‫ם‬ABCD into ΔABD and ΔBDC
Thrm 3.5.1 says sΔABD ≤ 180 and
sΔBDC ≤
180
So sΔABD + sΔBCD ≤ 360
From angle addition postulate <ADB + <BDC=<D
and <ABD + <CBD =<B
So m<A + m<B +m<C + m<D ≤ 360
#3
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State and prove the converse of Euclid’s 5th.
If 2 straight lines meet on a particular side of
a transversal, then the sum of the interior
angle on that side is less than 2 right angles
(180)
Proof
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Let l and m be lines such that l∩m = A and let
t be a transversal of I and m.
wts: <1 + <2 < 180
Let B = l∩t and C = t∩m and consider ΔABD
By lemma 3.5.2 m<C +m<B < 180
So the sum of the interior angles on the same
side of a transversal is < 180.
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From this Saccheri easily showed that in the
quadrilateral <B and <C were not obtuse
A contradiction to acute angles still hasn’t
been found.
Homework: #2
3.6 The Search for a Rectangle
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For SQ ‫ם‬ABCD:
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AD, BC called sides or legs
AB called base
DC called summit
<C, <D called summit angles
His goal: show <C=<D=90
So he looked at ways to maybe construct a
rectangle in neutral geometry
rectangle: quadrilateral with 4 right angles
Theorem 3.6.1
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The diagonals of a Saccheri quadrilateral are
congruent
Proof:
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Let ‫ם‬ABCD be a SQ. So then AD ≡ BC, <A ≡ <B
=90
Draw diagonals AC and DB and consider ΔABC
and ΔBAD
Since AB ≡ AB, and <A ≡ <B, AD ≡ BC
Then ΔABC ≡ ΔBAD by SAS
So DB ≡ AC.
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Theorem 3.6.2
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Theorem 3.6.3
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The summit angles of a SQ are congruent
The summit angles of a SQ are not obtuse, thus
are both acute or both right
Theorem 3.3.4
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Trying to prove two quadrilaterals are congruent,
need to show SASAS
Theorem 3.6.4
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The line joining the midpoints of both the
summit and the base of a SQ is
perpendicular to both.
Proof:
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Given SQ ‫ם‬ABCD, let MN join the midpoints of AB
and DC
wts: MN ┴ AB and MN ┴ CD
By def, AD ≡ BC and <A ≡ <B =90
Also, <C ≡ <D by theorem 3.6.2
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By def of MN, AM=MB and DN=NC
So ‫ם‬AMND ≡ ‫ם‬BMNC by SASAS theorem
This implies <NMB ≡ <NMA or m<NMB=m<NMA
We know m<NMA + m<NMB =180
So then m<NMB =½(180) and m<NMA=½(180)
Similarly, m<MND=½(180) and m<MNC=½(180)
Therefore, MN ┴ AB and MN ┴ CD
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Cor. 3.6.5
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The summit and base of a SQ are parallel
Theorem 3.6.6
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In any SQ, the length of the summit is greater
than or equal to the length of the base.
Lambert
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Lambert Quadrilateral: quadrilateral with at
least 3 right angles
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Lambert’s Theorem 3.6.7
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The fourth angle of a LQ is not obtuse, thus is
either acute or right.
Proof
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Let ‫ם‬PQRS be a LQ
Then <P = <Q = <R =90
wts: <S ≤ 90
Since <P + <Q + <R + <S ≤ 360 by Cor 3.5.4
and <P + <Q + <R =270
Then <S ≤ 90
So it is either acute or right.
Theorem 3.6.8
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The measure of the side included between 2 right
angles of a LQ is less than or equal to the measure
of the side opposite it.
Proof:
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Let ‫ם‬PQRS be a LQ. Then <P=<Q=<R=90 and <S ≤ 90
wts: PQ ≤ SR and QR ≤ PS
Assume PQ > SR and choose P’ such that P’Q = SR.
Then ‫ם‬P’QRS is a SQ
So then <1 ≡ <2 and m<1 = m<2 ≤ 90
This is a contradiction by exterior angle theorem to ΔPP’S
So PQ ≤ SR
Similarly QR ≤ PS (you try to finish this one)
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Theorem 3.6.9
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The measure of the line joining the midpoints of
the base and summit of at SQ is less than or
equal to the measure of it sides.
Proof:
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in book
Lots of theorems and corollaries about when
triangles and rectangles exist that you should
look at
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Theorem:
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Theorem:
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The EPP is equivalent to saying the angle sum of
every triangle is 180
The Pythagorean Theorem implies the existence
of a rectangle.
Theorem:
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The Pythagorean Theorem is equivalent to EPP
#25 If a quadrilateral is both SQ
and LQ then it is a rectangle.
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Proof:
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Given ‫ם‬ABCD such that AB ≡ BC and
<A=<B=<C=90
wts: ‫ם‬ABCD is a rectangle.
By SQ theorem, <C ≡ <D
So then <D=90
So ‫ם‬ABCD is a rectangle.
Homework: #1, 3, 4, 6, 7, 12, 14, 26
Turn in #3, 6, 14