Download Q1: Which of the following would be considered a "line" by Euclid`s

Experiment with hyperbolic geometry (solutions):
Plotting and measuring: applet at
http://www.math.psu.edu/dlittle/java/geometry/spherical/toolbox.html
You can set the background to white under View>>Background. This may work better for copying/printing.
Q1: Start off by drawing and measuring a few triangles.
 Select Constructions >> Plot Point, and plot 3 points by clicking.
 Select Constructions >> Draw Segment, and click endpoints to connect (A to B, B to C, C to A).
 Select Measurements >> Measure Triangle, and it will give all three side lengths and angle measures.
Select Edit >> Move Point, drag the vertices of your triangle around, and watch the measurements
change. The angle sum appears to always be less than 180˚, as expected.
Q2: Draw a pair of parallel lines by constructing alternate interior angles equal. You can select File >> New to clear the
previous construction
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Constructions >> Draw Infinite Line. Click on two spots
on the disk, and you'll get points and a line.
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Constructions>>Draw Ray at Specific Angle.
Enter a value for the angle (default is 45)
Click B, then A, then somewhere on the disk.
You'll get ray AC and ray AB forming an angle.
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Constructions>>Draw Ray at Specific Angle.
Keep the same value as above.
Click A, then C, then somewhere on the disk, to the side of
ray AC which is opposite B. (You want alternate angles).
You should get a ray CD.
You may need to extend ray CD into an infinite line by
choosing Constructions >> Draw Infinite Line, and
clicking points C and D.
The lines should be non-intersecting; i.e. parallel.
Measurements >> Measure Angle and measure angles
BAC and DCA to verify that alternate interior angles are
equal.
Now, Edit >> Move point
Slide point C around. The angles should stay fixed by
construction. Can you ever get CD to intersect AB, or are
they always parallel? (Being the same line doesn't count!)
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They will always be parallel – the only time you'll intersect is at the
instant they're the same line.
Q3: A parallel to a parallel may not be parallel!
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Draw infinite line AB.
Constructions >> Draw Perpendicular to line AB through
point B by first clicking point B, then the line.
Same thing through AB through point B.
You may need to Edit>>Move Point a bit to bring points C
and D closer. Get something that looks basically like the
image to the right.
Draw a perpendicular to line AD through point D (which
will give you a point E). Because alternate interior angles
are equal (both 90), line AB and line DE are parallel.
Also draw a perpendicular to line BC through point C
(which will give you a point F). Because alternate interior
angles are equal (both 90), line AB and line CF are
parallel.
Go to Edit >> Move Point, and slide points C and D
around. You should be able to get things so that line DE
and line CF intersect. This means that (1) you have two
distinct lines through the same point (the point of
intersection) that are parallel to a given line, and therefore
(2) lines which are parallel to a third line are not
necessarily parallel to each other. (These are intersecting at
point G.)
Q4: Triangle area: (answers will vary, of course)
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Plot 3 points (A, B, and C).
Draw segments A to B, B to C, C to A.
Measure triangle ABC (gives all sides and all angles).
Record segment lengths:
AB = 1.836
BC = 1.893
CA = 2.699
Draw perpendicular to segment AB through C.
Measure distance CD
CD = 1.831
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Calculate value of .5(base)(height), using AB as the base,
and CD as the height:
.5(AB)(CD) = .5(1.836)(1.831) = 1.689
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Draw perpendicular to segment BC through A.
Measure distance AE
AE = 1.775
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Calculate value of .5(base)(height), using BC as the base,
and AE as the height:
.5(BC)(AE) = .5(1.893)(1.775) = 1.680
You should recognize .5(base)(height) as the formula for the area of a triangle in Euclidean geometry. Does it work as the
area formula in hyperbolic geometry? The value changes depending on what side you use for the base (this does not happen
in Euclidean geometry).
Q5: SAS congruence
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Start by constructing two segments of the same length:
Constructions>> Draw Segment of Specific Length.
Then, with vertices at points A (on one segment) and C (on
the other), Constructions>> Draw Ray at a Specific Angle
(pick an angle). [ Click B to A to somewhere, then D to C
to somewhere.]
By default, it should put point on those rays that are the
same length as the original segments. (You can measure
distance to check, and move point if needed.)
This means SAS = SAS by construction.
Form triangles by drawing segment EB and segment FD.
Measure triangle ABE and triangle CDF.
Are the triangles congruent? YEP.
AB = DE = 2 constructed, then rays at angle A and angle E. Length
AC = EF = 2 as well. Triangle ABC congruent to triangle FDE.
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Since ASA congruence can be proven from SAS
congruence, we should have that as well. Construct a pair
of triangles with ASA = ASA (construct a matching side
for each, then construct matching pairs of angles at each
endpoint). Measure. Congruent? (Hope so!)
For this one, first segments AB = CD = 3. Then, angles at A and C
(45), then angles at B and D (5). Find intersection point was used
to get vertices I and J. Triangle ABI congruent to triangle CDJ