Download Developing Neutral Geometry We continue proving basic theorems

Developing Neutral Geometry
We continue proving basic theorems in neutral geometry, demonstrating how much mileage
we can get out of our very modest beginnings!
We continue the development of neutral geometry.
Theorem 5 Let 4ABC be isosceles, with AB = AC. Then angles ∠ABC and ∠ACB
are equal. Additionally, if we extend AB past point B to point D, and extend AC past C
to E, then angles ∠DBC and ∠ECB are also equal.
Proof: Since AB = AC and ∠BAC is congurent to itself, the triangles 4ABC and
4ACB are congruent. Thus, corresponding parts ∠ABC and ∠ACB are equal. Additionally, ∠ABD and ∠ACE are equal because both are striaght angles, so we can subtract by
common notion 3 to get the last pair of angles we desire.
Are we really sure all straight angles are equal? Does it need proof? It does, but
fortunately it’s easy. Since there is only one straight line through two given points, if we set
two straight angles to coincide at their vertex with one side of one along a side of the other,
the remaining sides must also be along each other.
Theorem 6 Given triangle 4ABC with ∠B = ∠C, then sides AB = AC. (This is the
converse to the previous theorem.)
Proof: Assume that AB > AC. Then (common notion 5) there is a point D on AB such
that BD = AC. By SAS, 4DBC ∼
= 4ACB. But this implies that ∠DCB = ∠B = ∠C
which is impossible because the part is less than the whole.
←→
Theorem 7 Given segment AB, there is not more than one point on each side of AB with
←→
given distances to each of A and B. That is, given AB and point C not on the line AB,
there is no other point D on the same side of the line as C and such that CA = DA and
CB = DB.
Proof: Assume there is a second such point D. Draw the line segment CD. Now 4CAD
is isosceles, so angles ∠ACD and ∠ADC are equal. Similarly, 4CBD is isosceles, with equal
angles ∠BCD and ∠BDC. Now B is in the interior of ∠ACD and A is in the interior of
∠BDC, so we have ∠BCD < ∠ACD = ∠ADC < ∠BDC which is a contradiction because
∠BCD = ∠BDC.
Note there might not be any point with given distances to A and B, if the triangle
inequality is not satisfied. This ability to pinpoint objects based on their distances to two
given points is used to improve the accurace of radar systems.
Theorem 8 (side-side-side) If triangles 4ABC and 4DEF have sides AB = DE,
AC = DF , and BC = EF , then they are congruent.
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Proof: Apply 4ABC to 4DEF so that A coincides with D, AB is along DE and C
and F are on the same side of the common line. Because AB = DE points B and E must
coincide. Then, since A = D, we have DC = AC = DF and EC = BC = EF . Applying
theorem 7 then tells us that points C and F coincide, so we are done.
Theorem 9 The bisector of any angle may be constructed.
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Proof: Let our ange be ∠BAC. Choose any point D on AB and construct point E on
−→
AC so that AE = AD. (This is possible by theorem 3.) Now construct equilateral triangle
←→
4DEF by theorem 1. Then AF is the bisector of ∠A. This is because, by construction,
AD = AE and EF = DF , while of course AF = AF . So triangles 4ADF and 4AEF are
congruent (SSS) so angles ∠F AE and ∠F AD are equal and together make up angle A.
Can an angle be divided equally in 3, 4, 5, or more parts? We’ll see later that it is
actually quite easy to divide a length into any fixed number of equal parts—in Euclidean
geometry once we have parallel lines. Other geometies may present a challenge! Angles are
a much more difficult beast. It is known that it is impossible (with the geometry as we have
defined it!) to trisect an angle. Although trisections are possible with slight rule changes, as
we’ll hopefully see later. The proof that trisections are impossible is probably beyond our
ability as a class to understand. But in any geometry we can cut a line segment into two
equal pieces:
Theorem 10 The midpoint of a line segment can be constructed.
Proof: Given segment AB construct equilateral triangle 4ABC. Then bisect angle C
and let D be the intersection of the bisector with AB. (Why do they necessarily intersect?)
Then AC = BC, CD = CD and ∠ACD = ∠BCD. So by SAS we have 4ACD ∼
= 4BCD
so BD = AD and D is the midpoint of AB.
Now we prove that right angles actually exist.
←→
Theorem 11 Given point C on line l = AB it is possible to construct the perpendicular
line to l through C.
−→
Proof: Let D be any point on CA other than C. We may construct E on the other ray
of l from C so that CD = CE. Construct equilateral triangle 4DEF . Then by construction
CD = CE, DF = EF , and CF = CF . By SSS triangles 4F CD and 4F CE are congruent,
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so angles ∠F CD and ∠F CE are equal. Since these are adjacent angles, line F C must be
perpendicular to l.
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