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Math 102B Hw 4
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P.191 (2 points)
Incorrect
Incorrect
Correct (p. 242)
Correct (p. 245)
Correct (p. 245 and 247)
Correct
Correct (Def of square is 4 equal sides and 4 equal angles)
Correct (Thm 6.1)
Correct (SAS still holds)
Incorrect
P.262
1 (1 point)
- Every triangle has a circumscribed circle.
- Wallis’ postulate on the existance of similar triangles.
- A rectangle exist.
- Clavius’ axiom that the equidistant locus on one side of a line is the set of points on a line.
- Some triangle has an angle sum equal to 180◦
- An angle inscribed in a semicircle is a right angle.
- The Pythagorean equation holds for right triangles.
- A line cannot lie entirely in the interior of an angle.
- Any point in the interior of an angle lies on a segment with endpoints on the sides of the angle.
- Areas of triangles are unbounded.
- Hilberts Parallel Axiom: For every line l and every point P not lying on l there is at most one line m
through P such that m is parallel to l.
- If a line intersects one of two parallel lines, then it also intersects the other.
- Converse to the alternate interior angle.
- If t is a transversal to l and m, l||m, and t⊥l, then t⊥m.
- If k||l, m⊥k, and n⊥l, then either m = n or m||n.
- The angle sum of every triangle is 180◦ .
- The sum of the angles is the same for every triangle.
- There exist a pair of similar, not congruent triangles.
- Two lines that are parallel to the same line are also parallel to each other.
- Every triangle can be circumscribed
- If three angles of a quadrilateral are right angles, then the fourth angle is also a right angle.
- There exists a pair of straight lines that are at constant distance from each other.
- The summit angles of the Saccheri quadrilateral are right angles.
2 a (1 point)
We are following Figure 6.15 p. 270. Since J is the midpoint of AC, we have that AJ ∼
= CJ. Also the
vertical angles at J are congruent, namely ∠AJD ∼
= ∠CJF . Since the angles at D and at F are right angles
we have by AAS that ∆ADJ ∼
= ∆CF J. From this we get that AD ∼
= CF . Similarly, I is the midpoint of
BC, so we have that BI ∼
CI.
Also
the
vertical
angles
at
I
are
congruent,
namely ∠BIE ∼
=
= ∠CIF . Since
the angles at E and at F are right angles we have by AAS that ∆BEI ∼
∆CF
I.
From
this
we get that
=
∼
∼
BE ∼
CF
.
Giving
us
that
AD
CF
BE,
and
hence
EDAB
is
a
Saccheri
quadrilateral
with
base ED
=
=
=
and summit AB. The content in the polygon ABIJ is common to both the ∆ABC and the quadrilateral
EDAB. Since ∆ADJ ∼
= ∆CF J their content are equal, likewise for ∆BEI ∼
= ∆CF I.
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So we have (content ∆ABC) = (content polygon ABIJ) + (content ∆CF I) + (content ∆CF J) =
(content polygon ABIJ) + (content ∆BEI) + (content ∆ADJ) = (content EDAB).
b (1 point)
By Prop. 4.3 every segment has a unique midpoint. Let M be the midpoint of ED (it might be that
←
→
M = F or not, but in any case midpoint exist). The midpoint also lies on the medial line IJ by defenition of midpoint and the medial line. By Prop. 12b, the line joining the mipoints of the summit and
←
→
←→
the base is perpendicular to both the summit and the base, i.e. midpoints are K and M . IJ lies on ED
←
→
by the construction of part (a) hence (perpendicular bisector to AB) = KM and KM is perpendicular to IJ .
c (1 point)
From part (a) we have that ∆BEI ∼
= ∆CF I and ∆ADJ ∼
= ∆CF J, from this we get that EI ∼
= IF
∼
and F J = JD. Since we are in a real plane lengths are assigned. So ED = EI + IF + F J + JD =
IF + IF + F J + F J = 2(IF + F J) = 2IJ.
If the plane is hyperbolic then we can say that the sumit angles are accute by thm. 6.1. Since ED is the
base, by corollary 4 to prop. 4.13 we get that AB > ED, and by the above we get the desired AB > 2IJ.
If the plane is euclidean then Thm. 6.1 gives us that the summit angles are right angles and we get again
by corollary 4 to prop. 4.13 that AB = ED. Hence AB = 2IJ.
d (2 points)
(⇐) Suppose that AC ∼
= BC. Then ∆ABC is an isosceles triangle and we have that ∠A ∼
= ∠B. Since K
is the midpoint of AB we have that BK ∼
= AK, then ∆BCK ∼
= ∆ACK by SAS. That is CK is a bisector
of angle C. Also CF is common to both ∆CF I and ∆CF J, angle F is a right angle for both triangles and
CJ ∼
= CI since both are half of congruent sides of isosceles triangle. Thus ∆CF I ∼
= ∆CF J by SAS. That
←→ ←→
is CF is a bisector of angle C. But by prop. 4.4 the angle bisector is unique. Thus CF = CK and so C, F
and K are colinear as desired.
(⇒) Suppose that C, F and K are colinear. By construction CF is perpendicular to IJ, and ∠KF D is a
right angle (since it is supplementary to ∠DF C). Again let M be the midpoint of ED guarantied by prop.
4.3. Suppose that M 6= F . Then we can form ∆KM F , we know by prop. 4.12 that ∠KM F is a right
angle. Then ∆KM F has two base angles that are right angles, namely ∠KF M and ∠KM F , contrary to
Saccheri-Legendre (otherwise (∠F KM )◦ is zero or negetive!). So F = M and in particular ∠CKA is a right
angle. Since K is the midpoint of AB we have AK ∼
= BK. Since ∆CKA and ∆CKB share common side
CK, have a right angle and AK ∼
= BK, by SAS we have ∆CKA ∼
= ∆CKB. This last steps gives the desired
conclusion of AC ∼
= BC.
Assume that F is the midpoint of CK. Then CF ∼
= F K, but by part (a) we know that BE ∼
= CF , so
that BEF K is a lambert quadrilateral. By corollary 3 to prop. 4.13 ∠B is a right angle, and therefore
we are in an Euclidean plane. Assume we are in an euclidean plane, then ABED is a rectangle. Thus
FK ∼
= AD, but again by part (a) we have that AD ∼
= CF . So we have that CF ∼
= F K, and since we assume
C, F and K are colinear, we get that F is the midpoint of CK.
Suppose C, F and K are not colinear, and we are not in an euclidean plane. Then the EDAB Saccheri
−−→
quadrilateral has acute summit angles. Furthermore ray CF intersects AB at some point by the crossbar
theorem, call that point G. But CG is not an altitude of the triangle if the plane is not euclidean. Thus CF
is not perpendicular to AB.
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e (2 points)
Suppose that the pythagorean equation holds for all right triangles and ∠C is a right angle. Since ∆CIJ is
a right triangle we have the equation (IJ)2 = (CI)2 + (CJ)2 . Similarly ∆ABC is a right triangle so again
we have equation (AB)2 = (AC)2 + (BC)2 . But from part (a) we have that BC = 2CI and AC = 2CJ. So
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by substitution we have: (AB)2 = (AC)2 + (BC)2 = (2CJ)2 + (2CI)2 . So (AB)2 = 4(CJ + CI ) = 4(IJ)2 .
From wich we get the desired result AB = 2IJ. Since we get AB = 2IJ part (c) gives us that the plane
must be Euclidean.
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