Tuesday, December 16, 2014

Review for Final Exam, Continued (Day 81)

Today we continue our review for the final exam.

Most of the questions on this half, which cover Chapters 5 through 7 of the U of Chicago text, are mostly self-explanatory. Notice that in Question 47, we are given that ABCD is a trapezoid with one pair of opposite angles congruent and we are to prove that it is a parallelogram. In other words, we are using the inclusive definition where a parallelogram is a trapezoid. If teachers prefer the exclusive definition, they can change the Given section to: AB and CD are parallel and angles A and C are congruent, to prove that ABCD is a parallelogram.

For the bonus section today, let me include those last two proofs that I've been trying to squeeze in for a while, but couldn't until now. These are the proofs that the angle bisectors of a triangle are concurrent -- meeting at the incenter -- and that the medians are concurrent -- meeting at the centroid.

Just as with the circumcenter and orthocenter, we get these proofs from Dr. Hung-Hsi Wu. Let's do the incenter proof first. But before we can prove this, we must prove what Wu calls "Lemma 12," but what I will call the Angle Bisector Theorem:

Angle Bisector Theorem:
If a point is on the bisector of an angle, then it is equidistant from the sides of the angle.

Notice that this requires the concept of distance between a point and a line. This is usually defined to be the perpendicular distance from the point to the line. So we are going to begin with an angle O and a point P on the bisector of angle O. Since we want to consider the perpendicular distance from P to the sides of angle O, we choose points A and B on the sides of O such that OA and OB are perpendicular to PA and PB, respectively.

Here is Wu's proof, converted to two-column format:

Given: Ray OP bisects angle AOB, Ray OA perpendicular PA, Ray OB perpendicular PB
Prove: PA = PB

Statements                                                      Reasons
1. Ray OP bisects angle AOB, Ray OA ...      1. Given
2. Ray OB reflected over line OP is Ray OA 2. Side-Switching Theorem
3. PB' perpendicular to Ray OA                     3. Reflections preserve angle measure.
4. PB' is exactly PA                                        4. Uniqueness of Perpendiculars Theorem
5. PA = PB                                                      5. Reflections preserve distance.

I'm actually surprised that the U of Chicago text doesn't prove this theorem. It fits very nicely with the Perpendicular Bisector Theorem in Section 4-5.

We will also need the proof of the converse of this theorem:

Converse of the Angle Bisector Theorem:
If a point is equidistant from the sides of an angle, then it is on the bisector of the angle.

Given: PA = PB, Ray OA perpendicular PA, Ray OB perpendicular PB
Prove: Ray OP bisects angle AOB

Statements                                                      Reasons
1. PA = PB, Ray OA perpendicular PA, ...     1. Given
2. OP = OP                                                     2. Reflexive Property of Equality
3. Triangle POA congruent triangle POB       3. HL Congruence Theorem
4. Angle POA congruent angle POB              4. CPCTC
5. Ray OP bisects angle AOB                         5. Definition of angle bisector

Notice that this follows a common pattern that we've seen so far in the text -- a theorem is proved using reflections and symmetry, while its converse is proved using a Congruence Theorem (and in this case, HL). We could have proved the forward theorem in Section 4-5, but the converse had to wait until Section 7-5 when we proved HL. (The Perpendicular Bisector Theorem was proved in Section 4-5, while its converse could be proved using the Isosceles Triangle Theorem of Section 5-1, but the converse of that theorem in turn appears in Section 7-3, and its proof used AAS.)

Now we can prove the incenter theorem:

Angle Bisector Concurrency Theorem:
The three angle bisectors of a triangle meet at a point, called the incenter of the triangle. The incenter is the unique point equidistant from the three sides.

Given: Ray AE bisects angle BAC, Ray BD bisects angle ABC, Rays AE and BD intersect at I
Prove: Ray CI bisects angle ACB, I equidistant from all three sides

Statements                                                      Reasons
1. Ray AE bisects angle BAC, ...                    1. Given
2. I equidistant from AC and AB                    2. Angle Bisector Theorem
3. I equidistant from BA and BC                    3. Angle Bisector Theorem
4. I equidistant from CA and CB                    4. Transitive Property of Equality
5. Ray CI bisects angle ACB                          5. Converse of Angle Bisector Theorem

Wu also proves uniqueness -- that is, that I is the only point equidistant from all three sides. He does this by using the Converse of the Angle Bisector Theorem to show that another point I' that is equidistant from any two of the sides must lie on the bisector of the angle those sides form. (That is, Wu proves the Converse of this Angle Bisector Concurrency Theorem!)

Now finally, we get to the Median Concurrency Theorem -- the only concurrency that is specifically mentioned in the Common Core Standards. This one is even more complicated -- it depends on the properties and sufficient conditions for a parallelogram in Sections 7-6 and 7-7. It also depends on what Wu calls Theorem 18 and the U of Chicago calls the Midpoint Connector Theorem:

Midpoint Connector Theorem:
The segment connecting the midpoints of two sides of a triangle is parallel to and half the length of the third side.

The U of Chicago proves this in Section 11-5, using coordinates. But Wu proves this without using coordinates at all, as follows:

Given: In triangle ABC, D and E are midpoints of AB and AC, respectively
Prove: DE | | BC, BC = 2DE

Statements                                                      Reasons
1. bla, bla, bla                                                1. Given
2. Let F be on ray DE such that DF = 2DE   2. Ruler Postulate (Point-Line)
3. E midpoint of DF                                       3. Definition of midpoint
4. ADCF is a parallelogram                           4. Sufficient Conditions (pgram test), part (c)
5. CF = AD                                                    5. Properties (pgram consequence), part (b)
6.  CF = BD                                                    6. Transitive Property of Equality
7. CF | | AD (same as CF | | BD)                    7. Definition of parallelogram
8. DBCF is a parallelogram                           8. Sufficient Conditions (pgram test), part (d)
9. DF = BC                                                    9. Properties (pgram consequence), part (b)
10. BC = 2DE                                                10. Transitive Property of Equality
11. DF | | BC (same as DE | | BC)                 11. Definition of parallelogram

In analyzing this proof, notice how similar steps 4-7 are to steps 8-11.

Now we can finally prove the Median Concurrency Theorem, Wu's Theorem 19:

Median Concurrency Theorem:
The three medians of a triangle meet at a point G, called the centroid of the triangle. On each median, the distance of G to the vertex is twice the distance of G to the midpoint of the opposite side.

Given: In triangle ABC, C', B', and A' are midpoints of AB, AC, and BC respectively.
(This way, the medians are AA', BB', and CC' as per Wu's notation.)
Prove: CC' meets BB' at a point G such that BG = 2GB'.
(Without loss of generality, the same proof will work if we change B or C to A.)

Statements                                                      Reasons
1. bla, bla, bla                                                1. Given
2. C'B' | | BC, C'B' = 1/2 BC                          2. Midpoint Connector Theorem (applied to tri. ABC)
3. Let M, N be midpoints of BG, CG            3. Ruler Postulate (Point-Line)
4. MN | | BC, MN = 1/2 BC                          4. Midpoint Connector Theorem (applied to tri. GBC)
5. C'B' | | MN                                                 5. Transitivity of Parallelism Theorem
6. C'B' = MN                                                 6. Transitive Property of Equality
7. MNB'C' is a parallelogram                        7. Sufficient Conditions (pgram test), part (d)
8. MB' and NC' bisect each other                  8. Properties (pgram consequence), part (c)
9. BM = MG, MG = GB'                               9. Definition of midpoint
10. BM = GB'                                               10. Transitive Property of Equality
11. BM + MG = GB' + GB'                          11. Addition Property of Equality
12. BG = 2GB                                               12. Segment Addition (Betweenness Theorem)

As I mentioned before, many texts use coordinates to prove theorems such as these. But we don't know that coordinates work unless we use similarity, and we don't know that similarity works, according to Wu, unless we first use theorems such as this one. It is at this point where Dr. David Joyce and Common Core are in the most agreement. In order to avoid circularity and ensure that no result is used before it is proved, we must avoid coordinates here.

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