MATH 5200/7200 Exam 2
Dr. McCrory, October 16, 2001

In the following proofs you may use only our lists of axioms and basic theorems. You
should prove any other facts you use, even if they have already been proved in
homework. (Exceptions to this requirement are stated in problems 1 and 2.)

Proof: Consider the parallelogram ABCD:

(1) First we show that AB = CD. Since AB and CD are parallel, and BD is a transversal,
the alternate interior angles DBA and BDC are equal. Since AD and BC are parallel and
BD is a transversal, the alternate interior angles ADB and CBD are equal. Also DB = BD.
Therefore, by side-angle-side congruence, the triangles DBA and BDC are (positively)
congruent. Thus, by definition of congruent triangles, AB = CD.

(2) Assume that the diagonals AC and BD intersect at a point E. (For a proof of this
fact, see the note on exam 2.) We show that triangles EBA and EDC are congruent.

Angle EBA equals angle DBA, and angle EDC equals angle BDC. By step (2), angles DBA
and BDC are equal. Thus angles EBA and EDC are equal. Also, angle AEB equals angle
CED. (To see this from the axioms, note that a(AEB) + a(BEC) = a(AEC) = 180, and
a(BEC) + a(CED) = a(BED) = 180.) Therefore, by angle-angle similarity, triangles EBA
and EDC are (positively) similar. But AB = CD by step (2), so triangles EBA and EDC are
congruent.

(3) By the definition of congruence, EB = ED and EA = EC. Thus E is the midpoint of BD
and E is the midpoint of AC, which is what we wanted.

2. Prove the following theorem: Suppose that two circles c1 and c2 intersect in two
points A and B. Suppose that C and D are points on c1, and E and F are points on c2, such
that A lies on the segment CE and B lies on the segment DF. Prove that the lines CD
and EF are parallel.

You should use the theorem that if a quadrilateral is inscribed in a circle, then its
opposite angles are supplementary. (You don't have to prove this theorem.)

There are really several cases, but you should consider only the case illustrated by the
diagram below. Hint: Draw segment AB.

Proof: By adding the line segment AB, we divide the quadrilateral CDEF into two
quadrilaterals CDBA and ABFE, with CDBA inscribed in circle c1 and ABFE inscribed in
circle c2. Thus the opposite angles DCA and ABD are supplementary in quadrilateral
CDBA, and the opposite angles FBA and AEF are supplementary in the quadrilateral
BFEA. Furthermore, angles ABD and FBA are supplementary, since their sum is angle
FBD, which is a straight angle. In summary, we have

a(DCA) + a(ABD) = 180,

a(ABD) + a(FBA) = 180,

a(FBA) + a(AEF) = 180.

We conclude by algebra that

a(DCA) = a(FBA),

a(ABD) = a(AEF),

a(DCA) + a(AEF) = 180.

By the same side interior angle theorem, applied to the transversal CE of the two lines
CD and EF, the lines CD and EF are parallel. QED

3. Give reasons for each step in the following proof outline. The reasons for each step
can involve axioms or basic theorems, as well as previous steps in the proof outline. You
do not have to write out a complete proof.

Theorem: The bisector of an angle of a triangle divides the opposite side into
segments that are proportional to the adjacent sides.

Proof outline: The theorem states that if ABC is a triangle, then the bisector of the
angle ABC intersects the side AC at a point S such that AS/SC = AB/BC.

1. The bisector of angle ABC intersects line AC at a point S between A and C.

Let the angle bisector be ray BX. It follows from the definition of angle
bisector that ray BX is between ray BA and ray BC. By the betweenness
axiom, the ray BX intersects line AC at a point S between A and C.

2. Angle ABS equals angle SBC.

This is the definition of angle bisector.

3. Let L be a line through A parallel to the bisector BS.

This follows from the existence of parallel theorem, applied to the line BS
and the point A.

4. Let R be the point of intersection of L and the line BC.

a(BCS) + a(CSB) + a(SBC) = 180 by the triangle angle sum theorem. So
|a'(BCS)| + |a'(CSB)| < 180. Let Y be a point on line L so that a'(CAY) and
a'(CSB) have the same sign. By the corresponding angles theorem a(CAY) =
a(CSB). So |a'(BCS)| + |a'(CAY)| < 180. By the parallel axiom, the rays CB
and AY intersect at a point R.

5. Angle ABS equals angle BAR.

This follows from the alternate interior angles theorem applied to the
parallel lines BS and AR (steps 3 and 5) and the transversal AB.

6. Angle ARB equals angle SBC.

This follows from the corresponding angles theorem applied to the parallel
lines BS and AR (steps 3 and 5) and the transversal RB.

7. Angle ARB equals angle BAR.

This follows from steps 6, 2, and 4.

8. RB = AB.

This follows from step 7 and the isosceles triangle theorem applied to the
triangle BAR.

9. Triangle ARC is similar to triangle SBC.

This follows from angle-angle similarity. a(ARC) = a(ARB) by the angle
axiom and a(ARB)= a(SBC) by step 6. a(RCA) = a(BCS) by the angle axiom.

10. AC/SC = RC/BC.

This follows from step 9 and the definition of similarity.

11. AC = AS + SC.

S is between A and C (by step 1).

12. RC = RB + BC.

B is between R and C (by the construction of R in step 5).

13. AS/SC = RB/BC.

By 10, 11, 12, (AS + SC)/SC = (RB + BC)/BC, so AS/SC + 1 = RB/BC + 1, so
AS/SC = RB/BC.

14. AS/SC = AB/BC.

This follows from steps 13 and 8.