Download Hints for Exam Two

Hints for Exam Two
1.
A. (a.) This is the statement of SAS congruence, which is always valid regardless of α.
B. (c.) This is a statement of AAS congruence, which can fail for “large” triangles when α < ∞.
C. (d.) This is a statement of SSA congruence, which always might fail if the remaining base angles
are supplementary.
D. (d.) This is a statement of AAA congruence, which is not a valid congruence criteria in most
geometries.
E. (c.) On a sphere, large enough triangles can have an obtuse angle and a right angle.
F. (a.) This can be proven from SSS congruence or the isosceles triangle theorem, both of which are
valid generally.
G. (d.) Even when α = ∞, the angle sum might be less than β.
H. (a.) This is the triangle inequality, which is valid generally (though more difficult to prove when
α < ∞).
I. (d.) Although this looks like the scalene inequality, the two angles are not in the same triangle and
therefore do not have to have much to do with each other.
J. (b.) We have done this construction on a sphere (using the foot of a point of distance α/2 from a
line), but if α = ∞ the exterior angle inequality would prevent this from occurring.
K. (c.) You can always find such a line, but on a sphere it will not be unique.
L. (a.) This is a simple application of the angle construction postulate, which is true generally.
M. (a.) Recall that we defined circles on a sphere precisely in a way which guaranteed this property.
N. (d.) ♦ABCD is a Saccheri quadrilateral. If α = ∞ it may or may not be a rectangle depending on
whether we are in Euclidean or Hyperbolic geometry.
O. (b.) This is related to the notion that if a line contains an antipodal point, it must contain the other
antipodal point. However when α < ∞, we can actually explicitly construct parallel lines.
2. The simplest way to prove this is to use the perpendicular bisector theorem, since X and M must
be on the perpendicular bisector.
3. First establish that ∠ABC and ∠CBD are actually adjacent and not overlapping (this will happen
whenever there are four distinct points which make two equal angles). Then construct the opposite ray
−−→
−→
−−→
to BC and calculate the angles that BA and BD make with this ray. Using betweenness, you should
be able to prove the correct measure.
4.
If you can prove that ∠ACB is the only non-acute angle in the triangle, the Scalene inequality
will imply the desired result. But it is possible that the other two angles are right. Suppose that one of
them is right, and derive a contradiction.
5. Draw the angles and see if you can find a pair of congruent triangles. Recall that you know that
vertical angles are congruent.
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6.
You are trying to prove that the diagonals are congruent. Try to find two triangles in the
quadrilateral which have the diagonals as one of their sides, and which contain the congruent angles.
From here you should be able to find a congruence between triangles.
7. If there is a point of intersection, then you can draw a triangle 4XBD. What do you notice about
the angles of this triangle?
8. Suppose that D ∈ Int 4ABC. What can you say about the sides and angles of 4DBC relative
to 4ABC? You can prove that the triangles cannot be congruent either by finding a angles which
are not congruent or a pair of sides which are not congruent, but the angle approach is probably more
immediate.
9. The quickest approach is to use the Isosceles triangle theorem. If ∠ABC ∼
= ∠BCA ∼
= ∠CAB then
∼
in particular ∠ABC = ∠BCA, so what can you say about AC and BA. From there, what can you say
about BA and CB?
10. The three triangles 4AOB, 4BOC, 4COA make up the triangle 4ABC. If you can prove that
these three smaller triangles are congruent, then you can prove that the larger triangle is equilateral. The
three triangles have congruent angles by assumption, but how can you show that they have congruent
sides?
11. Consider the triangles 4ABC and 4ADC. Can you find a congruence between them?
12. Find a point D so that (BCD) is true and BD = α/2. Then what can you say about the triangles
4ABC and 4ADC? Since you have a triangle with multiple right angles, what can you say about the
size of ∠BAD?
13. Since the sides are of length less than α/2, you can say that AF < α/2 and therefore the exterior
angle inequality applies to any triangle that you draw. Therefore in particular m∠AF C < m∠ABC.
But since 4ABC is equilateral you can prove that all of its angles are congruent, so that ∠ABC ∼
=
∠ACB. Then using the scalene inequality, you should be able to prove the desired result.
14. If you draw the triangles connecting D to either of its feet on the sides of the angles, you will find
a SSA correspondence. But this is generally not a congruence condition, because he angles we want to
be congruent could be supplementary. However, if we did have a pair of supplementary angles we would
contradict something about the hypothesis.
15.
In some ways this is the reverse of the previous problem. Consider 4AM B and 4AM C we
have SSA congruence, which implies that either the triangles are congruent or else the angles ∠ABM
and ∠ACM are supplementary. Since we do not know anything else about the triangles, we cannot say
more than this.
16.
It should be clear how to apply the SSA theorem to this problem (which is okay to do, since
the SsA theorem is a slightly different result). This would imply that either the triangles are congruent
(which we want) or else ∠ABC and ∠DEF are supplementary. But you can prove that both angles
must be acute, which it makes it impossible for them to be supplementary.
−−→
17.
Using the fact that BD bisects ∠ABC, you can find a AAS congruence between 4ABC and
4ADC.
18. Using the scalene inequality we can determine that AO < XO, which implies that A is an interior
←→
point of the circle. Then, using the secant theorem, the line AX must be a secant line.
19.
By SSS congruence, we can establish that 4AOB ∼
= 4DOC (where O is the center of the
circle). Similarly, we have 4AOD ∼
4BOC.
You
can
then
find the congruence by using the fact that
=
m∠ABC = m∠ABO + m∠OBC and similarly, m∠CDA = m∠CDO + m∠ODA.
20.
First note that 4ABO is isosceles, so that ∠ABO ∼
= ∠BAO. From here, establish that we
cannot have both of these angle right or obtuse, because that would imply that the radius of the circle
was more than α/2. Then since both are acute, they are smaller than an obtuse angle.
21. If the points form a triangle, then the triangle inequality implies that 7 < AC < 13. However,
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the problem never stated that the points were on a triangle! If they are collinear, then if (ACB) is true
we have AC = 7 and if (ABC) is true we have AC = 13. So the correct answer is 7 ≤ AC ≤ 13 (with
the equals being important).
22.
We can use the exterior angle inequality to say that m∠ADC > m∠ABC, so it must be
larger than 60 degrees. On the other hand, we can say that m∠ADB > m∠ACB, which implies that
60 < m∠ADC < 120.
23. This is an isosceles triangle, which implies that the base angle ∠ABC and ∠ACB are congruent.
Using the exterior angle theorem we can see that their sum must be strictly less than 180 degrees, so
they must both be acute. Then we can use the scalene inequality to show the result.
24. There are two approaches to this problem. Either show that m∠ABC > m∠BAC by betweenness
results and use the scalene inequality, or use the fact that 4ADB is isosceles together with the triangle
inequality on 4BDC.
25.
We can immediately determine by the exterior angle inequality that m∠ABD > 50. On the
other hand, we have m∠ADC > 70, which implies that m∠ABD < 110.
26. We can immediately say by the Scalene inequality that CD < BC < BD and that AC < AB.
Using the fact that AB = BD we can say combine these to get AB = BD > BC > AC. Then using
the Scalene inequality on 4ABC, it should be clear how the measure of ∠BCA relates to the other
angles.
27.
Using the fact that AO, BO, CO and DO are all radial segments of the circle, and the fact
that ∠BOD is in a linear pair with both ∠AOB and ∠COD, you should quickly be able to prove
that 4ABO ∼
= 4DCO, which will immediately imply that m∠DCO = 15. (Do not worry about the
fact that this implies the angle sums in these triangles would be less than 180 degrees, since this could
happen in Hyperbolic geometry).
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