Download 14 Neutral Geometry VI

The Rest of Neutral Geometry
We complete our development of neutral geometry—and take a look at what’s missing.
Theorem 12 Given a line and a point not on it, the perpendicular to the line through the
point can be constructed.
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Proof: Let the line be AB and the point be C. Choose a point D not on the line and
not on the same side of the line as C. Then the segment CD crosses the line. Draw the
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circle of radius CD centered at C. Because the intersection of CD and AB must be inside
the circle, and there are points outside the circle on the line, we know the line intersects the
circle. In fact, it does so at least twice—once “on its way in” and once “on its way out.”
Call these points E and F . Let their midpoint be G. Then SSS shows that angles ∠CGE
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and ∠CGF are equal, so line CG is perpendicular to AB.
Theorem 13 The sum of any two supplementary angles is equal to two right angles.
Conversely, if angles ∠DCA and ∠DCB are supplementary, then A, B, and C are collinear.
Proof: The first part is obvious, since both supplementary and pairs of right angles form
a straight angle. The converse is by application of 4DCA to 4ECA where E is any point
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on the perpendicular to AC through C. Since the angles are supplementary, the ray CB
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must lie on the continuation os line AC.
Theorem 14 Vertical angles are equal.
Proof: Since they have a common supplement, each together with that supplement adds
to two right angles, and equals minus equals are equal.
Theorem 15 Any exterior angle of a triangle is greater than either of the opposite angles.
Proof: We are given 4ABC with exterior angle ∠ACD. Let E be the midpoint of AC.
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Draw line BE and find the point on this line, on the opposite side of AC and equidistanct
from E as B is. Call this point F . By SAS 4AEB ∼
= 4CEF . Now ∠BAE = ∠ECF <
∠ECD.
This theorem is not quite what we want. In Euclidean geometry, the exterior angle is
equal to the sum of the other two angles. But we need parallel lines to prove that, and we
don’t have parallel lines in general geometry!
Also, think about spherical geometry. This isn’t quite going to work! For instance, the
90-90-120 triangle has an exterior angle of 90◦ which is not larger than the opposite angle!
Corollary The sum of any two angles of a triangle is less than two right angles.
Proof: An angle plus its exterior angle equal two right angles (they are supplements)
and either of the other angles is less than the exterior angle.
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This is even more trouble for spherical geometry, where we can have an equilateral triangle
with three 179◦ angles if we want!
Theorem 16 In any triangle, the larger side is opposite the larger angle.
Proof: In 4ABC let AC > AB. Locate D on AC with AD = AC. Then triangle
4ABD is isosceles, so ∠ABD = ∠ADB. But because D is in the interior of ∠ABC, we
have ∠ABD < ∠ABC. Meanwhile, ∠ADB > ∠C because of the exterior angle theorem
above. So in the original triangle, ∠B > ∠C.
Corollary The converse to this is also true—the greater angle is opposite the greater side.
For if the corresponding sides were equal, the angles would be equal (isosceles triangle) and
the if wrong side were longer we’d have a contradiction to the theorem above.
Lemma Of all the segments joining a given point to a given line, the shortest is the
perpendicular. From this, we define its length to be the distance from the point to the line.
Proof: Let A be the point, B the foot of the perpendicular, and C any other point
on the line. Then ∠ABC is a right angle, as is the exterior angle at B by the definition of
perpendicular. So ∠C is less than a right angle, and by the theorems just proved, AC > AB.
Theorem17 (Triangle inequality) In any triangle, the sum of two sides is always larger
than the third side.
Proof: Consider triangle 4ABC. Extend side BA past A to D so that AC = AD.
Then triangle 4ACD is isosceles, so ∠ADC = ∠CDA. But ∠ACD < ∠BCD so in 4BCD
we have BD > BC. But BD = BA + AC, so we’re done.
Theorem 18 Let P be a point in the interior of 4ABC. Then AP + P B < AC + CB
and ∠AP B > ∠ACB.
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Proof: Let ray CP continue to point Q. Then by exterior angles we have ∠AP Q >
∠ACQ and ∠BP Q > ∠BCQ. Adding these gives the desired angle result. To prove the
distance result, extend AP until it meets BC at Q. Then by two applications of the triangle
inequality, AC + CQ > AQ = AP + P Q and P Q + QC > P C. Adding these and cancelling
P Q from both sides gives the desired result.
Theorem 19 Let three lengths a, b, c be given with a ≥ b ≥ c, and b + c > a. Then a
triangle may be constructed whose side lengths are the given lengths.
Proof: On any line, constuct segment BC with length a. Then construct circles with
center B and radius c and center C with radius b. Since a ≥ b we know that B is outside
the circle centered at C and similarly C is outside the circle centered at B. But because
b + c > a the circles intersect. Call this point A and 4ABC is the needed figure.
Theorem 20 Angles may be copied. That is, given segment AB and angle ∠XY Z we can
find C so that ∠CAB = ∠XY Z.
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−→
Proof: Draw segment XZ. Construct point P on AB so that AP = Y X. Then, by the
previous theorem, construct Q so that AQ = Y Z and P Q = XZ. By SSS, ∠P AQ is the
needed angle.
Theorem 21 (side-angle-side) Given triangles 4ABC and 4DEF with ∠A = ∠D,
∠B = ∠E, and AB = DE then the triangles are congruent.
Proof: First, if AC < DF , extend AC to point G with AG = DF . Then by SAS we
have 4GAB ∼
= 4EDF . In particular, ∠GBA = ∠F ED = ∠CBA but this last contradicts
common notion 5. A similar contradiction shows that we can’t have AC > DF either. So
they are equal and the triangles are congruent by SAS.
Theorem 22 (side-side-angle) Given triangles 4ABC and 4DEF with ∠A = ∠D,
∠B = ∠E, and BC = EF then the triangles are congruent.
Proof: In Euclidean geometry, we would know the third angles are also equal and be
able to appeal to ASA, but we don’t know that in general geometries. Also, note that this
theorem is not true in spherical geometry! So we’re going to have to work a little.
Assume AB > DE. Then let P on AB so that BP = DE. By SAS we have 4BP C ∼
=
4EDF . But in 4AP C, ∠CP B is external and is thus larger than ∠A, so couldn’t be
equal to ∠D. Contradiction. A similar argument shows AB 6< DE. So AB = DE and the
triangles are congruent by ASA.
Theorem 23 If a line l intersects two other lines p and q so that alternate interior angles
are equal, then p and q are parallel—they never intersect.
Proof: For the sake of the proof, let l cross p and B and cross q and C. Let ehe given
alternate interior angles be ∠ABC and ∠BCD. Then A and D are on opposite sides of l.
Assume p and q meet at E, on the same side of l as A. Then ∠BCD is exterior to
triangle 4BCE and is thus greater than ∠ABC, a contradiction.
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