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Geometry
EXAM II
Solutions
1. (a) Assume that 2ABCD is a Saccheri quadrilateral, that is ∢A, and
∢B are right angles and that AD ∼
= BC. Prove that ∢C ∼
= ∢D.
Proof: Consider the triangles △DAB and △CBA. By SAS, these
are congruent since AB ∼
= CB (by hypothesis) and
= BA, DA ∼
∼
∢DAB = ∢CBA (both are right angles).
Therefore, DB ∼
= CA.
Now consider the triangles: △DAC and △CBD. By SSS, these
are congruent.
Therefore, ∢CDA ∼
= ∢DCB.
(b) Let M be the midpoint of AB and M ′ the midpoint of DC. Prove
that MM ′ is perpendicular to AB and to DC.
Proof: By SAS, △DAM ∼
= △CBM
∼
Therefore DM = CM.
By SSS, △DM ′ M ∼
= △CM ′ M,
so ∢DM ′ M ∼
= ∢CM ′ M.
So MM ′ ⊥DC.
By part (a) ∢ADC ∼
= ∢BCD, so we can apply the same argument
′
to show that MM ⊥AB.
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2. In Theorem 4.1 it was proved in neutral geometry that if alternate interior angles are congruent, then the lines are parallel. Prove that in
hyperbolic geometry, such two parallel lines have a common perpendicular. In other words, let m and l be two parallel lines. Let P be a
point on m and Q a point on l, such that the alternate interior angles
at P and Q are congruent (see the picture). Prove that l and m have
a common perpendicular.
Hint: Let M be the midpoint of P Q and drop perpendiculars M N and M L
to the lines m and l, respectively. Prove that L, M and N are collinear.
Proof: By AAS, △NP M ∼
= △LQM.
Therefore, ∢P MN ∼
= ∢QML.
Then the supplementary angle of ∢P MN: ∢QMN ∼
= ∢LMP , which
is supplementary to ∢QML.
Therefore (∢LMP )o + (∢P MN)o = 180o .
Therefore L, M, N have to be collinear.
3. Prove that every segment has a unique midpoint.
Hint: Recall the construction of the midpoint, or exercise 12 in Chapter 4.
Proof: In the proof of exercise 12 in chapter 4, we chose a point C on a
←→
side of AB, and then found the point C ′ on the other side of the line,
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so that: ∢BAC ∼
= BC ′ . Then M was defined as
= ∢ABC ′ and AC ∼
the intersection of CC ′ with AB (such point exists because C and C ′
are on opposite sides of AB). Then we showed that AM ∼
= MB.
By RAA, assume that there is another point M ′ such that AM ′ ∼
= M ′ B.
By SAS: △ACM ′ ∼
= △BC ′ M ′ .
Then ∢CM ′ A ∼
= ∢C ′ M ′ B and their supplementary angles are also
congruent: ∢BM ′ C ∼
= ∢C ′ M ′ A.
′
′
Therefore (∢C M B)o + (∢BM ′ C)o = 180o ,
i.e. C, M ′ , C ′ are collinear.
Therefore M and M ′ have to be the same point.
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