Geometry and the Common Core Standards

This document is intended for those who, like me, are newcomers to the transformational

geometry of the new Common Core Standards. We teachers of geometry have been asked (told,

really) to radically revise our courses, and I suspect that many are as yet unclear just how radical

the revision will be. I think the revision good on balance. But that issue to the side, I hope to

make clear what the revision entails.

The Common Core Standards (CCS) require that we teach geometry in a way very different than

before. What before were theorems become postulates and vice versa. What before was true by

definition must now be proven and vice versa. Geometry has been turned on its head.

Before the CCS, we were followers of David Hilbert and George Birkhoff.1 We defined polygon

congruence in terms of angle and side congruence; we defined polygon similarity in terms of

angle congruence and side proportionality. With the new CCS, we are followers of Hilbert

Birkhoff no more. Now we say that polygons are congruent if they can be superposed by a

sequence of so-called Rigid Transformations; we say that they are similar if they can be

superposed by a sequence of Similarity Transformations. (More on these, and on superposition,

in just a moment.) It is still true that polygons with all angles and sides congruent are themselves

congruent. But it is no longer true by definition. On the contrary, it must now be proven.

When we were followers of Hilbert and Birkhoff, we had a SAS triangle congruence postulate.

(For Birkhoff, that’s not quite the truth. He actually gave us a SAS triangle similarity postulate,

of which SAS congruence is a special case.) No more. SAS congruence (and SAS similarity too)

is now a theorem, proven by the Principle of Superposition.

When we were followers of Hilbert and Birkhoff, geometrical objects were static, forever locked

in place, forever changeless. No more, and from their transformations, the theory of congruence

and similarity follows.2 The transformations - both the Rigid and the Similarity - now become

axiomatic. They’re the foundation on which we erect our geometry. Before the transformations

were a curious little side topic, of no greater importance than, say, the properties of a trapezoid.

With the CCS, they could not be more important.3

What follows is my attempt to get clear about the new geometry. I begin with a syllabus for a

year-long course. After is a fragment of the new system of geometry. It includes only that part of

1 See George Hilbert’s Foundations of Geometry and Birkhoff’s Basic Geometry. The postulate sets in the two

works are most certainly not identical. The most obvious difference is that Birkhoff gives a pair of metric postulates

– the so-called Ruler and Protractor Postulates. But modern textbook geometry is based upon the School

Mathematics Study Group (SMSG) postulate set, and that is an amalgam of Hilbert and Birkhoff. (The original, and

still the best, SMSG text – the ur-text, as it were, for the modern crop – is Edwin Moise’s Geometry. I am of the

opinion that every teacher of geometry should have Foundations of Geometry, Basic Geometry and Geometry on

their shelf.)

2 Mathematicians do not conceive of the transformations as motions. Instead they are functions, and functions are

sets of pairs of points. But I think it a deep error to import this into the high school classroom. Students should be

taught that transformations are motions in accordance with certain rules, for this will accord most closely with their

intuitions.

3 The reader curious about the historical roots of transformational geometry should do a bit of research into Felix

Klein and the so-called Erlangen Program.

the new that differs from the old (and by “old” I mean the system of the current crop of mass-

market texts, which as I said is based on the SMSG postulate set). This comprises the

development of (a portion of) the theories of triangle congruence, triangle similarity and

parallels.

Syllabus

I. The Language of Geometry

A. Introduction to Mathematical Systems: Definition, Postulate and Theorem

B. Point, Line and Plane Definitions

C. Point, Line and Plane Postulates

D. Congruence, Betweenness and Midpoints

E. The Ruler Postulate

F. Angle Definitions and the Protractor Postulate

G. Angle Relationships

H. Polygon Definitions

I. The Algebraic Postulates

II. The Language of Logic

A. Conjecture and Counterexample

B. Conditional Propositions

C. Related Conditionals and Biconditionals

D. Paragraph Proof

E. Indirect Proof

F. The Fallacies

G. Angle Relationship Proofs

III. The Rigid Transformations

A. Slides

B. Turns

C. Flips

D. The Principle of Superposition

IV. Triangle Congruence

A. Triangle Categorization

B. SAS Triangle Congruence and the Isosceles Triangle Theorem

C. SSS and ASA Triangle Congruence

D. Applications of Triangle Congruence

V. Parallels

A. Definitions

B. The Alternate Interior Angles Theorem (AIAT) and its Converse (CAIAT)

C. The Corollaries of AIAT and CAIAT

D. The Triangle Angle Sum Theorem

E. Angles of Polygons

VI. Quadrilaterals

A. Properties of Parallelograms

B. The Parallelogram Tests

C. Properties of Rectangles

D. Properties of Rhombuses

E. Properties of Trapezoids

F. Properties of Kites

VII. Similarity

A. Ratios and Proportions

B. Similarity Transformations

C. AA Similarity

D. SAS and SSS Similarity

E. Parallel Lines and Proportional Parts

VIII. Right Triangles and Trigonometry

A. Right Triangles and the Dunlap Table4

B. The Pythagorean Theorem

C. The Converse of the Pythagorean Theorem and the Pythagorean Inequalities

D. The Special Right Triangles

E. The Trigonometric Ratios: Sine, Cosine and Tangent

F. The Inverse Trigonometric Functions

G. The Law of Sines, The Law of Cosines

H. Applications

IX. Circles

A. Definitions

B. Pi

C. Angles and Arcs

D. Arcs and Chords

E. Inscribed Angles

F. Tangents

G. Secant-Tangent Theorems

H. Special Segment Theorems

X. Points of Concurrency

A. Perpendicular Bisectors and the Circumcenter

B. Angle Bisectors and the Incenter

C. Medians and the Centroid

D. Altitudes and the Orthocenter

4 Dunlap Tables are named after a particularly insightful student of mine who created a clear, clean way to display

the side and angle correlations in the three right triangles present when the altitude is dropped to the hypotenuse in a

right triangle. The idea is to make a table in which the elements of any row are the three sides of a triangle arranged

so that adjacent elements in two rows are sides which correspond. Proportions are then easily constructed.

XI. Area and Volume

A. The Rectangle Area Postulate and the Areas of Parallelograms and Triangles

B. Areas of Rhombuses, Trapezoids and Kites

C. Areas of Regular Polygons

D. The Uniform Cross Section Principle and the Volumes and Prisms and Cylinders

E. Cavalieri’s Principle and the Volumes of Pyramids and Cones

F. Frusta

G. Volume of Spheres

H. Surface Area of Spheres

A Fragment of the New System of Geometry

No doubt what follows will not please professional mathematicians. It is not up to their usual

standards of rigor. But perfect rigor is not my goal. The fragment below is for use in a high-

school classroom. What’s the standard of rigor appropriate there? Euclid nailed it. My goal is to

match the rigor of Euclid’s Elements.

Definitions

D1. Preimage and Image The preimage of a transformation is a figure prior to its

transformation. The image of a transformation is the figure that results from the transformation.

When we have two names identical except that one ends with a prime and the other does not –

for example, F and F' – the one without the prime suffix designates the preimage of a

transformation and the one with the prime suffix designates the image.

D2. Slide For a slide, we must have a figure F and a vector v. Choose a point P on F, any point at

all. Place the tail of the vector on P; the image of P then lies at the tip of the vector. Thus all

points of F move exactly the same distance in exactly the same direction. Slides are also called

translations.

D3. Turn For a turn, we must have a figure F, an angle measure g and a center of rotation O.

Choose a point P on F, any point at all. If P = O, P stays put in the turn. Otherwise, the turn

leaves P' the same distance from O as P but swings it through an angle of measure g. It is as if we

draw the circle with center O which passes through P. We then sweep out an angle of measure g

from P and then mark P'. We observe the convention that a positive angle measure means to

sweep out the angle counterclockwise. Turns are also called rotations.

D4. Flip For a flip, we must have a figure F and a line of reflection t. Choose a point P on F, any

point at all. If P is on t, the flip leaves it where it is. Otherwise we place P' so that t is the

perpendicular bisector of PP'. Flips are also called reflections.

D5. Dilatation For a dilation, we must have a figure F, a center of dilation O and a scale factor r.

Choose a point P on F, any point at all. If P = O – that is, if P is at the center of the dilation – the

image of P is simply P; that is, if P = O, then P' = P. Otherwise, P ' lies on the ray OP such that

OP' = r · OP. A dilation is (one variety of) a similarity transformation.

D6. Congruent Figures are congruent when one can be carried onto the other by a sequence of

slides, turns and flips. This is the Principle of Superposition; it tells us that figures which can be

superposed by a sequence of flips, turns and slides are congruent. We call flips, turns and slides

together the Rigid or Congruence Transformations. Note that we do not define polygon

congruence in terms of side congruence and angle congruence. On the contrary, the link between

congruence of polygons and congruence of sides and angles must be proven. This is done in

Theorems 2 and 3 below.

D7. Similar Figures are similar when one can be carried onto the other by a sequence of slides,

turns, flips and dilations. We call these four transformations the Similarity Transformations. Thus

the Congruence Transformations are a subclass of the Similarity Transformations; that is, all

congruent figures are also similar. Note that, as before, we do not define polygon similarity in

terms of relationships of sides and angles. On the contrary, the link between similarity of

polygons and relationships between sides and angles must be proven.

D8. Parallel We say that lines are parallel either when they are coplanar and fail to intersect or

when they coincide. (The second clause might come as a surprise, but it simplifies a proof

below.)

Postulates

I include in the set of postulates below only those that are likely new to the reader.

P1. The Crossbar Postulate Every ray that begins at the vertex of a convex angle intersects any

segment whose endpoints are on opposite sides of that angle.

P2. The Congruence Postulate Slides, turns and flips take lines to lines, rays to rays and line

segments to line segments. Moreover, slides, turns and flips preserve both distance and angle

measure. This means: if points P and Q lie on the preimage, and points P' and Q' on the image

correspond to P and Q respectively, then PQ = P'Q'; and if angle A in the preimage corresponds

to angle A' in the image, then the measure of angle A equals the measure of angle A'.

P3. The Similarity Postulate Dilations take lines to lines, rays to rays and line segments to line

segments.

P4. The Playfair Postulate Through a point not on a given line, there's at most one line parallel

to the given line.

Theorems

The theorems below are divided into three sets. The topic of the first is congruence, of the second

parallels, and of the third similarity.

Congruence

T1. The Rigid Transformations preserve segment length.

Proof The length of a segment is the distance between its endpoints. But by P2, the Congruence

Transformations preserve distance measure. Thus they preserve segment length.

T2. When polygons are congruent, sides and angles which correspond are congruent.

Proof Assume that the pair of figures F and G are congruent. By D6, this means that one can be

superposed on the other by a sequence of Rigid Transformations. But by P2, those

transformations preserve angle measure; and by T1, they preserve side length. Thus the sides and

angles of F and G which correspond – the one superposed after transformation – are congruent.

Commentary In the Hilbert/Birkhoff amalgam that we find in the SMSG system, T2 is no

theorem at all. Instead it is a definition. Is the new approach superior? It has this advantage at

least: it gives us a single, unambiguous account of congruence that applies to all figures, whether

they are polygons or not. The traditional approach fails in this regard.

T3. If the sides and angles of two polygons can be paired off in such a way that sides and angles

which correspond are congruent, then those polygons are congruent.

Proof In the proof below (and in later proofs too), I won’t always note that a rigid transformation

preserves distance and angle measure. To do so quickly becomes tedious.

Let F and G be two polygons such as are described in the hypothesis. To prove that F and G are

congruent, we must prove that a sequence of Rigid Transformations can carry one onto the other.

We do this for the particular case of triangles ABC and RST below. The same method can be

applied to any pair of congruent polygons.

Begin with a pair of vertices which correspond, say A and R. Slide ΔABC so that A and R

coincide. Since by T1 turns leave side length unchanged, we may rotate ΔABC about A until B

coincides with S. Do so. Now, either C and T lie on the same side of segment AB or they do not.

If they do not, flip ΔRST over RS.

Thus segments AB and RS coincide and T and C lie on the same side of AB. By P2, angles A, B,

R and S did not change measure as the triangles were rigidly transformed, and so the sides of

angle A coincide with those of angle R and the sides of angle B coincide with those of angle S.

Moreover, by T1, no side length changed as the triangles were transformed. Thus segments AC

and BC coincide with RT and ST respectively. But this means the ΔABC coincides with ΔRST.

D6 then entails that the triangles are congruent.

Commentary As before, in the SMSG system, T3 is in fact a definition, not a theorem.

T4. SAS Triangle Congruence If two sides and an included angle of one triangle are congruent

to two sides and an included angle of second triangle, then those triangles are congruent.

Proof Consider triangles ABC and RST below. RS = AB, RT = AC and angle A = angle R. We must

prove that the triangles can be carried onto one another by a series of Rigid Transformations. We'll

let ΔRST stay put.

Slide ΔABC so that vertex A comes to coincide with vertex R. Since by T1 turns leave side length

unchanged, we may rotate ΔABC about A until B coincides with S. Do so.

Now, vertices C and T are either on the same side of segment AB or they are not. If they are not,

flip ΔABC over AB.

Thus sides AB and RS coincide and vertices C and T are on the same side of segment AB. Since

by P2 angle measures remain unchanged throughout a sequence of Rigid Transformations, the

sides of angle A coincide with those of angle R. Moreover, since by T1 side lengths also remain

unchanged, sides RT and AC also coincide.

Think carefully about what’s happened. Triangle ABC has been rigidly transformed so that

segments AB and AC coincide with RS and RT respectively. This means that vertices A, B and C

coincide with vertices R, S and R respectively. That is: after rigid transformation, A = R, B = S

and C = T. But from three points, only one triangle may be constructed. But this means that

ΔABC coincides with ΔRST. By D6, they are congruent.

T5. The Isosceles Triangle Theorem In a triangle with a pair of congruent sides, the angles

opposite those congruent sides are themselves congruent.

Proof I'll only sketch the proof. Drop in an angle bisector from the vertex angle B of isosceles

triangle ABC. Let M be the point of intersection of the angle bisector with AC. (The Crossbar

Postulate guarantees that this point of intersection exists.) The two triangles that result – ABM

and CBM – are congruent by SAS. But T2 tells us that when figures are congruent, all sides and

angles which correspond are congruent. Thus angles A and C are congruent, as was to be proven.

Alternatively, we can flip triangle CBM over segment BM. It is then easily proven that the image

of CBM then coincides with triangle ABM, from which it follows that triangles CBM and ABM

are congruent; and then by T2, we may say that angles A and C are congruent.

T6. SSS Triangle Congruence If the three sides of one triangle are congruent to the three sides

of a second triangle, then those triangles are congruent.

Proof We begin with triangles ABC and RST in which AB = RS, BC = ST and AC = RT. We

rigidly transform one or both so that AB and RS coincide and C and T lie on opposite sides of

segment AB. (Rigid transformations of this sort have been described many times. The details are

omitted.) We then connect C to T.

This gives rise to three cases. First, segment CT might pass through the interior of segment AB.

Second, it might pass through an endpoint of AB. Third, it might fail to intersect AB altogether.

The three cases are pictured below.

I'll consider only the first case. Since BC = ST, ΔCBT is isosceles with vertex B. Likewise, since

AC = RT, ΔACT is isosceles with vertex A. By the Isosceles Triangle Theorem, angles ACT and

ATC are congruent, as are angles BCT and BTC. Angle ACB is then congruent to angle ATB. So,

then, triangles ABC and RST fit the SAS pattern – two sides and an included angle of one are

congruent to two sides and an included angle of the other. Thus by T4 the triangle are congruent,

as was to be proven.

T7. ASA Triangle Congruence If two angles and an included side of one triangle are congruent

to two angles and an included side of a second, then those triangles are congruent.

Proof Since we've seen examples of this sort of proof before, I'll move quickly. Assume that, in

triangles ABC and RST, angles A and C are congruent to angles R and T respectively and that

AC = RT. Transform the triangles so that segment AC coincides with segment RT, and if

necessary flip triangle RST over RT so that angles B and S lie on the same side of segment AC.

Angles A and C will then coincide with angles R and T respectively.

Since the lengths of AB, BC, RS and ST were unchanged throughout the sequence of

transformations, AB and RS must coincide and BC and TS must coincide. This means that the

triangles themselves coincide and so are congruent.

Parallels

T8. If we rotate a line 180° about a point not on it, the line that results is parallel to the original.

Proof We assume in the proof below that if a point P is rotated 180° about point O and P ≠ O,

then P, O and P' are collinear. How to prove this assumption? It says, in effect, that if we rotate a

ray 180° about its endpoint, the result in a ray opposite the original. Perhaps this will be a

postulate. (In some systems, it is.) Perhaps it will be proven. But no matter which, I assume that

the reader is already quite familiar with it.

Below, the line we rotate is t. O is center of rotation. t' is the image of t after a turn of 180°.

The proof is indirect. Thus we assume that t' does in fact intersect t. Call the point of intersection

P.

Like all points on t, P was rotated 180° about O. But this implies that P, O and P' are collinear.

(This is the assumption mentioned above.) However, both P and P' lie on t'. Thus O lies on t' too,

for through two points only one line may be drawn. But if O lies on t', it must lie on t as well; for

every point on t' is the image of a point on t under the rotation, and by definition a rotation of a

figure leaves every point on it the same distance from the center of rotation.

But we were given that O does not lie on t. Thus the assumption that t' intersects t is false. That

is, t and t' are parallel.

Do notice the role played by the assumption that the rotation carried t (and every point on it)

through 180°. This is what allowed us to conclude that P, O and P' are collinear and that as a

result O must lie on t'. If the rotation had not been 180°, no such conclusion would have been

possible.

T9. When a transversal cuts a pair of lines and creates congruent alternate interior angles, those

lines are parallel.

Proof Consider lines r, s and t below. t cuts r and s at different points and thus is a transversal for

them. The points of intersection are P and Q. A and B are points on lines r and s respectively that

lie on opposite sides of t. Angles OPA and OQB are congruent alternate interior angles. O is the

midpoint of segment PQ. We are to prove that lines r and s are parallel.

Let us rotate angle OQB about O a total of 180°. We will prove this rotation will superpose ray

Q'B' on ray PA. From this it will follow that lines r and s are parallel.

Since O and Q are collinear and OQ = OP, the rotation will take Q to P; that is, Q' = P. From this

it follows that ray OQ' coincides with ray OP.

Consider now the angles OPA and OQ'B'. Since rays OQ' and OP are equal, angles OPA and

OQ'B' share a side on common. Moreover, since rotations leave angle measure unchanged,

angles OPA and OQ'B' are congruent. This implies that their second sides must coincide as well.

We conclude, then, that ray PA coincides with ray Q'B'.

Think, now, of what this means for lines r and s', where s' is the result of a rotation of s 180°

about O. If rays coincide, so too do the lines which contain them. But line r includes ray RA and

line s' includes ray Q'B'. Thus r = s'!

Thus whatever is true of r is true of s' also, for they are in fact the same line. But by T8, lines s

and s' are parallel. Thus lines r and s are parallel, as was to be shown.

What’s happened is this. The congruence of the alternate interior angles insures that, when line s

is rotated 180° about O, it must come to coincide with line r. But the rotation of a line by 180°

yields an image that is parallel to the preimage. Thus lines r and s are parallel.

The other similar results – that when consecutive interior angles are supplementary lines are

parallel, that when alternate exterior angles are congruent lines are parallel, etc. – follow easily

from T9.

T10. When a pair of parallel lines is cut by a transversal, alternate interior angles are congruent.

Proof Consider lines a, b and t pictured below. a and b are parallel. a and b are cut by t; the

points of intersection are A and B. We are to prove that angles 1 and 2 are congruent.

Construct line a' through A so that alternate interior angles are congruent. (At present, we remain

agnostic about whether a and a' are the same or different lines.) By T9, a' is parallel to b.

Assume now that a' and a are distinct lines. We thus have two lines through A, each of which is

parallel to b. But by the Playfair Postulate, this is impossible. Thus a' = a.

By construction, a' makes alternate interior angles congruent. Thus a does so as well. We

conclude that angles 1 and 2 are congruent, as was to be proven.

As before, the other similar results about parallels – that when lines are parallel, angles which

correspond are congruent, etc. - can now easily be proven.

The last result about parallels will prove quite useful in the proofs that follow.

T11. If the sides of polygons F and G can be paired off so that each side of G is parallel to the

side of F to which it corresponds, then each angle of G is congruent to the angle of F to which it

corresponds.

Proof It’s really quite easy. Just construct lines through pairs of vertices which correspond and

identify congruent angles. (Ones which correspond will work nicely.) I leave it to the reader.

Similarity

T12. If three or more parallel lines cut congruent segments on one transversal, then they cut

congruent segments on all transversals.

Proof In the diagram below, we assume that lines AD, BE and CF are parallel and that AB = BC.

We wish to prove that DE = EF. We have constructed segments DG and EH as shown so that

each is parallel to line AB.

ADGB is a parallelogram, as is BEHC. Thus AB = DG and BC = EH. (I assume that the reader is

quite familiar with properties of parallelograms.) But AB = BC. Thus DG = EH. Moreover, since

the sides of triangle DGE are parallel to those of EHF - in particular, DG || EH, DE || EF and GE

|| HF - their angles are congruent. Thus those triangles are congruent. (Take your pick here: ASA

or AAS.) We now conclude that DE = EF, as was to be proven.

T13. If a segment begins in the interior of one side of a triangle and ends in the interior of another

and is parallel to the third side, then it cuts the sides it intersects proportionally.

In triangle ABC below, we assume that segment MN is parallel to side AC. We wish to prove that

BM:MA::BN:NC.

We assume that segments BM and MA are commensurate. We assume, that is, that there exists a

quantity u (for ‘unit’) such that BM = k ∙ u and MA = p ∙ u, where p and k are positive integers.

(We have in effect assumed that the ratio of the lengths of BM and MA is rational.)

We next divide segment BM into k equal parts and segment MA into p equal parts. Each of these

divisions will of course be u long. Moreover, from each point of division along BA, we construct

a segment parallel to AC that ends on BC.

By T12, that sequence of segments parallel to AC cuts off congruent pieces of BC. Thus BN = k ·

u' and NC = p · u', for some positive quantity u'.

If we form the ratios BM:MA and BN:NC, we find that each is simply k:p. Thus they are equal,

as was to be proven.

T14. The dilation of line segment yields a second line segment parallel to the first.5

Proof Assume that we scale line segment AB. We wish to prove that image A'B' is parallel to the

preimage AB.

If AB passes through the center of the dilation, it will lie on the same line as A'B', in which case

it is trivially parallel to AB. (This is where my strange definition of parallel pays off.) So let us

assume that this is not so. Likewise, if the scale factor of the dilation is 1, AB = A'B'; as before, it

is then trivial that AB and A'B' are parallel. So let us assume that the scale factor is not 1.

In the diagram below, O is the center of the dilation of segment AB.

Since the transformation is of the Similarity sort, OA:AA'::OB:BB'.

The proof is now indirect. Assume that segments AB and A'B' are not parallel. Let T be the point

on OB' such that AT || A'B'. By assumption, T ≠ B.

5 One way in which to simplify the system herein presented is to load T11 into the Similarity Postulate. I won’t for

my Honors classes. Likely I will for non-Honors.

By T13, OA:AA'::OT:TB'. Thus the ratios OT:TB' and OB:BB' are equal. But this implies that T

= B! (Only a bit of algebra is required to reach this conclusion. I leave it to the reader.)

From the contradiction that T = B and T ≠ B, we infer that the assumption that AB and A'B' are

not parallel is false. AB and A'B' are then parallel, as was to be proven.

T15. The dilation of a line segment by a certain scale factor yields a second segment whose ratio

with the first is that very scale factor; that is, the scale factor for the segments is the scale factor

of the dilation.

Proof The proof isn’t difficult. We need the diagram from T13 again. Here it is.

Segment MN was scaled to yield segment AC. We wish to prove that the scale factor from MN to

AC is the same as that from BM to BA. I’ll only sketch the proof.

The idea is this: from each of the marked points on AB, drop segments to AC parallel to BC. We

will thus divide MN into k equal parts and AC into p + k equal parts. (Recall that the number of

divisions of BM is k and the number of divisions of AM is p.) Why equal? T13 applies; the ratio

of the parts of AM and BM is 1:1, and thus the ratio of the pieces of MN and of AC must also be

1:1. Moreover, we can also prove that the pieces of MN are equal to those of AC. How? I’ll give

a little hint: parallelograms have congruent opposite sides.

Thus for some u'', MN = k ∙ u'' and AC = (p + k) ∙ u''. (u'' is the length of each piece of AC and

MN.) The ratio of AC to MN is thus k:(p + k).

However, the scale factor from BM to BA is also k:(p + k). Thus the scale factors are one and the

same, as was to be proven.

T16. If polygons are similar, then sides which correspond are proportional.

Proof If polygons are similar, the two can be superposed by a sequence of Similarity

Transformations. But by T15, each Similarity Transformation scales the sides by a constant scale

factor. (If Rigid, that scale factor is 1. If non-Rigid, it is some value greater or less than 1.) Let

k1, k2, k3 . . . be the scale factors of the members of the sequence of transformations. The scale

factor from first preimage to last image is thus k1 ∙ k2 ∙ k3 ∙ . . .. This product is the scale factor of

the sequence, i.e. each side is scaled by this product. Thus sides are proportional.

T17. If polygons are similar, then angles which correspond are congruent.

Proof This follows immediately from the conjunction of T11 and T14. By definition, polygons are

similar when they can be superposed by a sequence of Similarity Transformations. Thus if we

can prove that every sort of Similarity Transformation (slides, turns, flips and dilations) leaves

angle measures intact, we are done.

By postulate, slides, turns and flips do not change angle measure. So that leaves only dilations.

By T14, the dilation of a line segment yields a second parallel to the first. Thus if we dilate a

polygon, each side of the image is parallel to the side of the preimage to which it corresponds.

By T11, if the sides of two polygons are pair-wise parallel, their angles are congruent. Thus the

dilation of a polygon leaves angle measures intact, as was to be proven.

T18. AAA Similarity If the angles of one triangle are congruent to the angles of another, then

those triangles are similar.

Proof We are given triangles ABC and RST with angles A, B and C congruent to angles R, S and

T respectively. We are to prove the triangles similar, and to do this, we must prove that they can

be superposed by a sequence of Similarity Transformations.

If AB = RS, the triangles are congruent and thus trivially similar. We thus assume that AB ≠ RS.

Indeed we assume that AB < RS, but the proof would be essentially the same if AB > RS.

Since angles A and R are congruent, we may rigidly transform triangle ABC so that sides AB and

AC come to lie along sides RS and RT respectively. This is shown below. B' is the image of B. C'

is the image of C. A' and R coincide.

Now we will perform a dilation of A'B'C'. Its center is R; it takes B' to S.

Consider segment B''C'' – where B'' and C'' are the images of B' and C' under the dilation. By T14,

B''C'' is parallel to B'C'. But segment ST is also parallel to B'C', for angles TSR and C'B'R are

congruent. Thus, since both ST and B''C'' pass through S, the Playfair Postulate entails that they

are collinear. Moreover, since both terminate on the line RT, they are in fact the same segment.

We conclude that RST and A'B''C'' are the same triangle. But the latter resulted from triangle

ABC by a sequence of Similarity Transformations. Thus triangles ABC and RST can be

superposed by a sequence of Similarity Transformations, as was to be shown.

T19. SAS Similarity If two sides of one triangle are proportional to two sides of another and

included angles are congruent, then those triangles are similar.

Proof We are given triangles ABC and RST such that AB:RS::AC:RT and angles A and R are

congruent. We must prove that by a sequence of Similarity Transformations, triangles ABC and

RST can be made to coincide. It will then follow by definition that they are similar.

I assume that the scale factor from ABC to RST is greater than 1. (If it was 1, the triangles would

be congruent and hence trivially similar. If less than 1, the proof would differ in only minor

detail.)

We first rigidly transform ABC so that A' and R coincide and B' and C' lie on segments RS and

RT respectively. (The possibility of this is guaranteed by the congruence of angles A and R.)

Now we perform a dilation of A'B'C'. Its center is R; it takes B' to S. We will prove that this

dilation takes C' to T.

We were given at the start that AB:RS::AC:RT. After the dilation, A'B':A'B''::A'C':A'C''. But A' =

R, B'' = S, AB = A'B' and AC = A'C'. Thus the latter proportion becomes AB:RS::AC:RC''. The

first and third proportions together imply that RT = RC''.

We know, then, that triangles RST and A'B''C'' coincide. But the latter resulted from triangle

ABC by a sequence of Similarity Transformations. Thus triangles ABC and RST are similar.

T20. SSS Similarity If the sides of one triangle are proportional to the sides of another, then

those triangles are similar.

Proof I adopt what seems the simplest strategy. I make use of AA Similarity. (Here and in what

follows, I speak not of AAA Similarity but of AA Similarity. The Triangle Angle Sum Theorem

implies that, if two angles of one triangle are congruent to two angles of a second, then their third

angles are congruent too.)

We begin with triangles ABC and RST. We assume that sides are proportional, i.e. for some

constant k, RS = k ∙ AB, ST = k ∙ BC and RT = k ∙ AC. We must prove that triangles ABC and

RST are similar.

To this end, we construct triangle FGH such that FG = k ∙ AB and angles F and G are congruent

to angles A and B respectively. Of course, then, RS = FG.

By AA Similarity, triangles ABC and FGH are similar. By T16, their sides are proportional. Thus

GH = k ∙ BC and FH = k ∙ AC. From this it follows that triangles RST and FGH have congruent

sides and so are themselves congruent. Triangle RST is thus congruent to a triangle that is itself

similar to triangle ABC. We conclude that triangles ABC and RST are similar. (We assume that if

one triangle is similar to a second and that second is congruent to a third, then first and third are

similar. This is obvious and easily proven.)

2nd amendment essay outline (Pro 2nd amendment)

Introduction:

The second amendment in the United States Constitution states:

Some people feel that the second amendment should be taken away and guns should be restricted:

I, however, am in favor of the second amendment and feel that it is the right of the American people to own guns. I feel this way because: (three reasons why)

1.

2.

3.

Alternatively people would argue that guns are dangerous and people should not be able to own them. They would argue that,

Back up your reasons with claims from readings…

In (name of article)

(author) claims, (or fill in with an alternative to says)

I disagree

with what (author) states because, (or I feel that this is not true

because)

Conclusion for anti 2nd amendment:

Many people will say that it is a right of American citizens to own guns. They claim

However, I would argue…

It is because of each of the above mentioned reasons that the second amendment needs to be changed and gun use needs to be restricted in America.

(List your three reasons from the introduction)

1.

2.

3.

Argument for the 2nd amendment: