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**J. Rainey’s E-Portfolio**

Math 6301-College Geometry Dr. N. Patterson 12/7/2006

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**Table of Contents Letter to Reader Mathematical Autobiography**

Proof-book Project Skemp Article Reflection Collection of “Best Work”

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Dear Reader, I hope this E-Portfolio gives reader’s insight into my teaching philosophy and how I write proofs. I would like to think that I am always evolving as a teacher and as a proof writer. It’s my sincere hope that this growth is evident in my works. My goal is to convey proofs in a succinct, logical and rigorous manner that’s easy to grasp. I tend to believe that the two-column proof is the easiest format to follow. I have tried to choose proofs that are challenging enough for the college geometry student, but not too intimidating for the high school geometry student to try. It is also important to note that technology has played a major part in my analysis of different theorems. The use of Geometer’s Skethpad has changed not only the way I analyze theorems, but also the way I teach. Therefore, I hope this E-Portfolio inspires math teachers to learn more about Geometer’s Skethpad and incorporate this technology into their teaching. Sincerely, Jena M. Rainey

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**Proof-Book Project Proof 1 Proof 2 Proof 3 Proof 4 Proof 5 Proof 6**

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**Collection of “Best Work”**

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**Review Question for Test one**

Question Two: O is the center of the circle. P, A, and B are points on the circle. AT and BT are tangents. Prove that AT = OB. SOLUTION: If two lines are perpendicular to the same line, then they are parallel to one another. Since m∠P = ½ m∠BOA, we know that m∠BOA = 90 because m∠P = 45 (Given). Hence OB⊥OA and AT⊥OA because a line tangent to a circle is ⊥to the radius and thus OB IIAT. Also note, TB⊥OB , because TB is tangent to the circle. Moreover, OA⊥OB because m∠BOA = 90. Thus, TB II OA. Hence OATB is a parallelogram and AT = OB by Th.6.3 which states that if a quadrilateral is a parallelogram, then its opposite sides are congruent.

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**Review Question for Test One**

Question Three: ABC is a tangent to the circle at B. DB = DE. Prove that DB bisects ∠EBC SOLUTION: My strategy is to show that m∠EBD=m∠DBC. Since ED=BD (Given) ∆EBD is an isosceles triangle and by the Base Angle Th., m∠DEB = m∠EBD. By the Inscribed Angle Th., m∠DEB = ½ m(arcBD). We also know that if a tangent and a chord intersect at a point on a circle, the measure of each angle formed is one-half the measure of its intercepted arc. Hence, m∠DBC = ½ m(arcBD). Thus, we can deduce m∠DEB = m∠DBC by substitution/transitive property. By further substitution, m∠EBD = m∠DBC.

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**Review Question for Test One**

Question Four: AB // DE, and DE = CE. Prove that AB = BC. SOLUTION: My strategy is to show that ∆BAC and ∆ DEC are similar equiangular triangles using properties of parallel lines. ∠B ≌ ∠EDC and ∠A ≌ ∠DEC by the Corresponding Angles Postulate. Hence, ∆BAC ∽∆DEC by AA Similarity. Therefore, AB/DE =BC/DC = AC/CE. Since DE=CE (Given), we know the AB=AC and thus ∠ABC = ∠DEC. But also note, ∠ACB ≌ ∠DEC because ∆DEC is also isosceles. Hence, ∆DEC is an equilateral triangle because if a triangle is equiangular, it’s equilateral. Therefore we can also deduce that ∆ABC is equiangular by definition of similar triangles and thus equilateral. Hence, AB≌BC.

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**m∠BCD = m∠CAB = m∠EAF = m∠EFA by transitive property/substitution.**

Review Question for Test One Question Six: A, B, and C are points on a circle. CD is a tangent to the circle at point C. FAB and EAC are line segments. EF = EA. Prove that ∠EFA = ∠BCD SOLUTION: ∆FEA is an isosceles triangle since EF≌EA. By the Base Angle Th., ∠EFA ≌ ∠ EAF. By the Vertical Angle Th., ∠EAF ≌ ∠CAB. By the Inscribed Angle Th., m∠CAB = ½ m(arc CB). We know that if a tangent and a chord intersect at a point on a circle, the measure of each angle formed is one-half the measure of its intercepted arc. Hence, m∠BCD = ½ m(arcCB). Therefore, m∠BCD = m∠CAB = m∠EAF = m∠EFA by transitive property/substitution.

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PAGE 51, #41 Prove that the Cevians joining the vertices of a triangle with the points of tangency of the incircle are concurrent. This is the Gergonne point of the triangle. SOLUTION: We know that two tangents from a point to a circle are congruent, therefore we can deduce CE=CD, BE=BF, and AF=AD Also note the following: CE/EB=BF/FA=AD/DC = 1 Thus, by the converse of Ceva’s Theorem the three segments are concurrent.

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Best Work #1 (HW# 1) I chose this problem because it was my first opportunity to use some of the new GSP features I had learned and I enjoyed doing this proof.

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Best Work #2 HW #2 I choose this problem because I recall this one being on the Praxis when I took it. Also, this concept is often tested on the Geometry EOCT exams.

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Best Work #2 Cont’d First examine triangle OPR. Since we know that the sum of the measures of the interior angles of a triangle is 180 degrees, we can deduce m + m + r = 180 (Substitution Property) 2m+ r = 180 r = m. Next consider triangle PQR, we can extablish the following: (m + n) + (n + k) + (k + m) = 180 2k + 2m + 2n = 180 2k + 2n = m. Since r = m and 2k + 2n = m, we can now deduce r = 2k + 2n (Substitution or Transitive Property) 1/2 r = k + n 1/2 mAngle POR = mAngle OQR + mAngle OQP = mAngle PQR. (Substitution/Angle Addition Postulate) mAngle POR = 2 mAngle PQR (Multiplication Property) Therefore, the measure of the central angle (Angle POR) is equal to twice the measure of the inscribed angle (Angle PQR). SOLUTION Proof: Construct a circle with center O and let Angle PQR be an acute angle inscribed in the circle. Next, construct segments PR and OQ. Segments OP, OR, and OQ are radii of the circle and therefore congruent. Hence, triangles OPR, OQR, and OPQ are isosceles triangles. Since base angles of an isosceles triangle are congruent, we have mAngle ORQ = mAngle OQR mAngle ORP= mAngle OPR mAngle OPQ = mAngle OQP The angles can be labeled as follows: k = mAngle ORQ = mAngle OQR m = mAngle ORP = mAngle OPR n = mAngle OPQ = mAngle OQP and r = mAngle POR.

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Best Work #3 (p.271 #24) I chose this problem because Ptolemy’s Theorem seems to be a very useful theorem that I haven’t introduced my students to in the past. I know plan to do so in the future.

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Best Work #4 (p.271 #14) I chose this problem because I enjoy working problems involving special triangles and it’s proof is very succinct. In the past I have had a tendency to say too much.

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**SKEMP ARTICLE REFLECTION**

Relational Understanding and Instrumental Understanding Skemp’s article is brilliant. For years I have struggled with trying to balance my students’ desire to understand instrumentally with my desire to teach relationally. As a student, I have been taught in both ways and I can see the value in both approaches. However, I recognize that relational understanding is real understanding, and I am constantly looking for new teaching strategies that will help my students develop a relational understanding of concepts. I think that most teachers probably use the instrumental approach to understanding because it is easier to teach that way. In order to teach relational understanding, the teacher’s knowledge base must be really strong. Writing proofs is a great exercise in relational understanding because it requires you to justify and explain every step. There is no way to write a proof successfully without having a clear understanding of the problem being addressed. Also, proof writing is all about rigorously communicating mathematical ideas. Skemp’s article has reaffirmed many of my beliefs about teaching and I will continue to strive for relational understanding in my classroom. One of the reasons that I chose to take College Geometry is because I want to improve my proof writing skills. I strongly believe that developing stronger proof writing skills will help me become a better relational teacher, which is my ultimate goal.

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**Mathematical Autobiography**

I have always excelled in mathematics. I like math because it is not subjective like so many other subjects are. Also, math relates to everything. Therefore it seemed to make sense for me to major in mathematics at Clark Atlanta University. I had received a full Academic Scholarship to enter a 5-year BS/MS program in Mathematics. In college, I was able to work closely with some extraordinary mathematicians (Dr. J. Ernest Wilkins, Dr. Abdulalim Shabazz and Dr. Ronald Mickens). Dr. Wilkins was one of my professors and served as my mentor for my Master’s Thesis. Dr. Shabazz was also my professor and chaired the math department. I didn’t take a course from Dr. Mickens, but was given an opportunity to assist him with some research he was doing that involved mathematical modeling of infectious diseases. The life stories of the aforementioned three men are so extraordinary, that one can not help but be influenced by their work and their passion for mathematics and mathematical research. They certainly influenced me to be a life-long learner of mathematics and to share what I have learned with others. They also help me to see how math played such an important role in so many things. Dr. Shabazz would always say, “Math is life and life is math”. I often share this quote with my students. I come from a long line of educators (My grandmother and my mother were teachers). Therefore, when I moved to California to work as a Software Engineer (minored in Computer Science), everyone knew that I would eventually come back home and teach. . Since returning to Atlanta, I have completed further studies at Georgia State in Math Education and Educational Leadership. As a result of these studies, I earned a second Master’s in Math Education and a certification in Educational Leadership. Currently, I am pursuing an Ed.S in Math Education. If you would like to know about my work with students, please go to:

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Monge’s Theorem Monge's theorem says that for three disjoint circles of unequal radii, with no one contained in any other, the pairs of external tangents meet in three points that are collinear. SOLUTION My strategy for solving this problem is to consider 3 right cones on the three circles as bases. Next, if you observe their common tangent planes you notice that they intersect in three lines A', B' and C' corresponding to the three points A, B and C in the theorem. Now, A' and B' intersect at the apex of one of the cones. Similarly, B' and C' intersect and so do C' and A'. Thus A', B' and C' are coplanar, and so A, B and C are collinear.

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Inscribed Angle Theorem The Inscribed Angle Theorem states that the measure of an angle inscribed in a circle is half the measure of the arc it intercepts. It follows that all inscribed angles that intercept a given arc have equal measure. (PROOF ON NEXT SLIDE) My strategy for solving this proof is consider the following three cases:

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**Inscribed Angle Th. (cont’d)**

SOLUTION: CASE 1: Center of circle is on QR An exterior angle of a triangle is the sum of the other two interior angles, so the measure of POR is equal to the sum of the measures of OPQ and OQP. OPQ is an isosceles triangle (because radii OQ and OP are equal) so the measures of angles OPQ and OQP are equal. So the measure of angle POR is twice the measure of angle PQR. CASE 2: Center of circle is not inside angle By constructing a diameter from Q through O meeting the circle at point a S: Observe that angle ROS is twice angle RQS. By the same reasoning, angle POS is twice angle PQS. The measure of angle POR is the difference of the measures of POS and ROS, and the measure so the measure of angle POR is twice the difference of the measures of PQS and RQS, and that difference is the measure of PQR, so the measure of POR is twice the measure PQR.

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**Inscribed Angle Th. (cont’d)**

Case 3: The center of the circle is inside the angle In case 3, if we construct a diameter QD, we observe that ∠ SOR is twice ∠ SQR, and ∠ POS is twice ∠PQS. So ∠ POR (the sum of ∠ POS and ∠ SOR) is twice ∠PQR (sum of ∠ PQS and ∠ SQR).

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Law of Sines Theorem (MORE ON NEXT SLIDE) The Law of Sines Theorem states: a/(sin A) = b/(sin B) = c/(sin C) = the diameter of the circumscribed circle. SOLUTION: In order to solve this problem, draw an altitude of the triangle (extending side b, if necessary): From the diagram, it is clear that h=c sin A, and that h= a sin C. (If side b needed to be extended, because angle A is larger than 90 degrees, then note that the sine of A is equal to the sine of the supplement of A, so even though h is to the right of A, it is still true that h=c sin A.) So c sin A = a sin C, so c/(sin C) = a/(sin A) By drawing a different altitude, the same can be shown of any two sides, proving the following: c/(sin C) = a/(sin A) = b/(sin B).

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**Law of Sines Theorem (cont’d)**

SOLUTION CONT’D: (Part 2) Now we must show that a/(sin A) is equal to the diameter of the circumscribed circle. Consider ∠ A, which is subtended by chord a of the circle: ∠A is the same as ∠ D in the diagram, below, because it is subtended by the same chord. (Or, perhaps ∠ D is supplementary to ∠ A, which happens if ∠ A is obtuse). Since triangle BCD is a right triangle, we know that a/(sin D) is the length of BD, which is the diameter of the circle. Since m∠A = m∠D or is supplementary to D, their sines are equal. Thus, we can deduce: a/(sin D) = a/(sin A) = the diameter of the circle. Hence, the second part is proved.

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**Page 22 (# 19) What would you need to know to prove the Crossbar Theorem?**

CROSSBAR THEOREM: Let P be a point interior to ∠XYZ. Then the ray AP crosses the segment XZ. SOLUTION: In order to prove this theorem you need to know Pasch’s Theorem which states that if A, B, and C are distinct, noncolinear points and l is a line that intersects segment AB, then l also intersects either segment AC or segment BC. Consider the diagram on the right: Choose a point M such that X *Y *M (Y is between X and M). We can then apply the Pasch theorem to ∆ZMX and conclude that YP crosses either XZ or MZ. Since M and X are on opposite sides of YZ, it follows that M and P are on opposite sides of YZ and hence all points of MZ are on the side of YZ opposite to all points on YP. Therefore, YP does not cross MZ. Next, we can eliminate the possibility that the opposite ray, YQ, crosses either DZ or XZ. We can easily observe that both of these segments are on the opposite side of MX from YQ. Hence, the only possibility not eliminated is that YP crosses XZ.

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**Review Question from Test 1 Question Sixteen: ABCD is a parallelogram**

Review Question from Test 1 Question Sixteen: ABCD is a parallelogram. AX ≌CW. Prove XD ≌ BW. Solution: My strategy for solving this problem is to show ∆DAX ≌ ∆BCW Since ABCD is a parallelogram AD ≌BC and ∠A ≌ ∠C. Since we are given that AX ≌ WC, we know that ∆DAX ≌ ∆BCW by SAS. Hence, XD ≌ BW by CPCTC (Corresponding Parts of Congruent Triangles are Congruent.)

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Review Question from Test 1 Question Twenty: AB is a diameter of the circle, center O AB ≌ BC. Prove that AD ≌ DC Solution: My strategy for solving this problem is to show ∆ADB ≌ ∆ADB. Since we are give that AB ≌ BC we know that ∆ABC is an isosceles triangle and therefore ∠A ≌ ∠C. Also, we know that if diameter is the hypotenuse of a triangle inscribed in a circle, then the triangle is a rt. triangle. Hence, we can further deduce that m ∠ADB = m ∠CDB = 90. Hence, ∆ADB ≌ ∆ADB by AAS. Therefore, AD ≌ DC by CPCTC.

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