Download 1 Assignment 5 – due on 3/4

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Assignment 5 – due on 3/4
I would like you to go through the following theorems in your own time, but before next class.
You do not have to submit these proofs. The homework is listed below the last theorem .
In proving that the sum of the interior angles in any triangle (in NEUTRAL GEOMETRY)
is smaller than or equal to 180◦ in class, we moved backwards. I will list the steps in order
we did them in class:
• Saccheri wants to show that a quadrilateral with 3 congruent sides and two right base
angles is a rectangle. He showed that the summit angles are congruent.
• Need to rule out that summit angles are obtuse.
• That will be achieved if the sum of interior angles in a quadrilateral is smaller than
360◦ .
• Previous will be true if the the sum of interior angles in a triangle is smaller than 180◦ .
• We will suppose that there is a triangle with angle sum bigger than 180 deg. In fact,
let the angle sum be 180◦ + p.
• If we can construct a triangle with the same angle sum, but whose one angle is smaller
than p, then we would have two angles in a triangle that add up to more than 180◦ .
• If we knew that any two angles in every triangle add up to less than 180◦ we would
have a contradiction.
Now let’s list all the things we proved today, but we will list them in order we have to
prove them: so that everything follows neatly and nicely from each other and we rule out
the obtuse angles.
Theorem 1.1 (AIA - Alternate interior angle theorem). If two distinct lines cut by a
transversal have a pair of congruent alternate interior angles, then the two lines are parallel.
Proof. By contradiction: Suppose the two lines ℓ and m intersect, there will be a triangle
formed. Construct a triangle congruent to it (on the other side of the transversal) and prove
that the two lines ℓ and m must in fact intersect in another point, which is a contradiction
(with what?).
Theorem 1.2 (Exterior angle theorem). An exterior angle of a triangle is greater than either
remote interior angle.
Proof. Contradiction again: if the exterior angle were not greater than a remote interior
angle, then it’s either congruent to it, or smaller than it. In either case, use AIA to deduce
that two lines are parallel, but those two lines contained a vertex of the original triangle,
therefore were intersecting. Kaboom.
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Proposition 1.3. Any two angles in a triangle add up to less than 180◦ .
Proof. An angle α and adjacent exterior angle add up to 180◦ ; an interior angle β remote to
the aforementioned exterior is smaller than it, so it must be α + β, 180◦ . You may need to
write down some equations.
Theorem 1.4. Angle sum of any triangle is less than or equal to 180◦ .
Proof. Contradiction again!
Suppose there is a triangle △ABC whose angle sum is larger than 180◦ , say 180◦ + p. I
will construct a new triangle with the same angle sum of 180◦ + p, but one of whose angles
is smaller than p. That would, however, mean that its other two angles add up to MORE
than 180◦ , which contradicts the Proposition1.3.
First thing we will do is construct a new triangle △ABC ′ with the same angle sum of
180◦ + p, but whose one angle (either one at A or one at C ′ ) is at most half of the angle at
A.
Then we will construct another triangle with the same angle sum of 180◦ + p, but whose
one angle is at most half of the angle that was at most half of the one at A, which would
make this one at most a fourth of the angle at A.
After n repetitions of this process we will arrive at a triangle whose angle sum is 180◦ + p,
and one of whose angles is at most 21n m(A). We will stop the process when 21n m(A) < p
– how do you know that such n exists?
So there is only one piece of the proof missing: construct a new triangle △ABC ′ with
the same angle sum of 180◦ + p, but whose one angle (either one at A or one at C ′ ) is at
most half of the angle at A:
−→
Let E be the midpoint of BC. Let a point C ′ be the point on AE such that A ∗ E ∗ C ′
and AE ∼
= EC ′ .
What can you say about triangles △ECA and △EBD?
Give all angles inside these triangles a name, and write down the sum of the angles in:
1. △ABC
2. △ABC ′
Is this triangle △ABC ′ one we were looking for? Why? I’m done, are you?
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Homework
We can now prove the following theorems:
Question 2.1. AAS: If AC ∼
= DF , A ∼
= D, and B ∼
= E, then △ABC ∼
= △DEF .
Question 2.2. The sum of the lengths of any two sides of a triangle is greater than the
length of the third side.
To prove them, you may use the following propositions without proving them here. However, you should be able to do it if asked.
Proposition 2.3. Given two non-congruent sides in a triangle, the angle opposite the longer
side is greater than the angle opposite the shorter side.
Proposition 2.4. Given two non-congruent angles in a triangle, the side opposite the larger
angle is longer than the side opposite the smaller angle.
Question 2.5. Let’s do a problem. Three squares are given side by side. Prove that
α + β + γ = 90◦ .
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