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G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com 1 Proof of geometric relationships continues in this objective. We establish some very important relationships dealing with triangles and their angles. Prove that the interior angles of a triangle sum to 180. Informal Proof - Triangle Dissection An informal proof that is often used is the process of having our students create a triangle on a piece of paper, naming the three angles A, B, and C and then cutting out the triangle. When the triangle is cut out, the student should rip off the three angles, placing them together, vertex to vertex. They will see that the three angles form a straight line. Therefore, the sum of the three interior angles of a triangle is 180. Classic Proof – Use of Parallel Line Angle Relationships A formal two column proof of this theorem is done using the angle relationships found with parallel lines and a transversal. Since these angle relationships were established in objective G.CO.9, we are now able to apply them. Given: ABC B 2 Prove: m1 + m2 + m3 = 180 Construct an auxiliary line parallel to AC through B. STATEMENT AC || BD m4 + m2 + m5 = 180 m1 = m4 m3 = m5 m1 + m2 + m3 = 180 REASON Given (Auxiliary Line) Angles of a Straight Angle If , Alternate Interior ’s If , Alternate Interior ’s Substitution Property (Twice) 1 3 A D C G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com 2 Transformational Proof Given: ABC B 2 Prove: m1 + m2 + m3 = 180 A Translate ABC by vector AB to form straight ABB’ along the vector. The isometric properties of translation preserve angles, thus m1 = mB’BC’. Since ABB’ is a straight angle, we know that m2 + mCBC’ + mB’BC’ = 180Translations also preserve parallelism, therefore BC ' . Since AC BC ' m3 = ensuring that AC mCBC’ because alternate interior angles are congruent. By making two substitutions into the straight angle relationship of m2 + mCBC’ + mB’BC’ = 180 we arrive at the proof that m1 + m2 + m3 = 180 Determine mA. 1 C 3 B' 1 B =A' A C' 3 2 1 C 3 A triangle has 180 as its angle sum…. So… B 84° 180 - 84 - 35 = 61 35° mA = 61 C A NYTS Determine the measure of the angle. a) mC = ________ b) mB = ________ c) mA = ________ B B d) mA = ________ B B o 96° C 32° 44° 61° A A 60° A C C o A 180 – 96 - 44 = 40 40 = mC 180 – 61 - 60 = 59 59 = mB 180 – 90 - 32 = 58 58 = mA 50° C 180 – 50 = 130 130 / 2 = 65 = mA G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com 3 Prove that Base Angles of an Isosceles are Equal (Isosceles Triangle Theorem) Informal Proof - Paper Folding Prove that the base angles of an isosceles triangle are congruent. Create an isosceles triangle by using your compass to construct a circle. Then draw two radii (all radii of the same circle are congruent) and connect the endpoints with a segment. Label the radii, AB and CB . Then fold the paper until point A maps to point C. Crease the paper. Notice that when you do this A C. Therefore, the base angles of an isosceles triangle are congruent. B C D A Classic Proof – Triangle Congruence Given: ABC is an isosceles triangle, with base AC . Prove: A C Construct an auxiliary line that is a perpendicular bisector of AC . (B is B on the perpendicular bisector of AC because AB BC ) STATEMENT ABC is an Isosceles. BD is the bisector of B AD CD ADB& CDB are right ’s ADB CDB BD BD ABD CBD A C REASON Given Given (Auxiliary Line) Definition of a Bisector (D is the midpoint of AC ) Definition of a Bisector ( BD is to AC ) All Right Angles are Reflexive Property (Common Side) A D SAS CPCTC (Corresponding Parts of Congruent Triangles are Congruent) Alternative methods to prove congruent triangles: HL ( AB CB, Definition of Isosceles Triangle) or SSS ( AB CB, Definition of Isosceles Triangle and BD BD , Reflexive Property) C G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com 4 Transformational Proof Given: ABC is an isosceles triangle, with base AC . B Prove: A C Construct an auxiliary line that is a perpendicular bisector of AC . By the definition of isosceles triangle, AB CB. Because point B is equidistant to points A and C, B is on the perpendicular bisector of AC . A reflection over the perpendicular bisector would map A onto C, B onto B, and D onto D. Thus the isometric properties of a reflection then give us ABD CBD. Therefore, A C by corresponding parts of congruent triangles are congruent. C D A Symmetry Proof Given: ABC is an isosceles triangle, with base AC . B Prove: A C Construct an auxiliary line; BD , such that BD is the perpendicular bisector of AC. Established in G.CO.3, an isosceles triangle has reflectional symmetry about the perpendicular bisector of its base. Thus A C because A reflects onto C. Determine mB. D A C A triangle has 180 as its angle sum…. So… A 180 - 54 = 126 54° 126 / 2 = 63 mB = 63 C B Determine the measure of the requested angle. a) mC = ________ b) mA = ________ A A c) mC = ________ d) mD = ________ A A 98° C 78° B B 36° C C 65° B D B C G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com 180 – 98 = 82 130 / 2 = 41 = mC 180 – 78 -78 = 24 24 = mA 5 65 = mC 180 – 36 = 144 144 / 2 = 72 = mACB 180 – 72 = 108 180 – 108 = 72 72 / 2 = 36 = mD Prove that the external angle of a triangle is equal to the sum of the two internal remote angles. (Exterior Angle Theorem) A First of all we need to define these new terms. An external angle of a polygon is the angle formed between a side’s extension and its adjacent side. No part of this angle exists in the interior of the triangle. (ACD is an external angle.) D C B A The remote angles of a triangle are those angles that are not the linear pair with the external angle being referenced to. o (A and B are remote angles to external ACD.) x B D C Informal Proof – Angle Dissection A Prove that the external angle of a triangle is equal to the sum of the two internal remote angles. An informal proof that is often used is the process of having our students create a triangle on a piece of paper, naming the three angles A, B, and C. Cut out the triangle and then draw a line on a piece of paper and then lie the triangle down on the line so that one side is extended by it. The student should rip off the two remote angles and then place them together, vertex to vertex in the external angle. They will see that the two remote angles exactly fit into the external angle. Therefore, the external angle of a triangle is equal to the sum of the two internal remote angles. B C D C D Classic Proof – Triangle Congruence Given: ABC with external angle, ACD. A Prove: mACD = mB + mA B C D G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com STATEMENT mB + mA + mACB = 180 mACB + mACD = 180 mACB + mACD = mB + mA + mACB mACD = mB + mA 6 REASON sum 180 Linear Pairs are supplementary Substitution Property (or Transitive) +/- Property of Equality Transformational Proof A Given: ABC with external angle, ACD. Prove: mACD = mB + mA B C D A Translate ABC by vector BC placing ABC in the interior of external ACD. One property of a translation is that it produces parallel lines and so AB || AB '|| AC . Thus BAC A’CA because alternate interior angle are equal (or a rotation of 180 about the midpoint of AC ) A' B D C = B' C = B'' A A' B Determine mBCD. C = B' D The external BCD is equal to the sum of the two remotes, A and B. B 92° 46° C D mBCD = 92 + 46 mBCD = 138 A Determine the missing information. a) mBCD = ___________ C = B'' b) mB = _________ c) mA = _________ mBCA = _________ mBCA = _________ G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com B 7 B B 42° 73° 74° C A D 21° D C 65° C 160° A D A 73 + 74 = 147 = mBCD 65 - 21 = 44 = mB 180 - 65 = 115 = mBCA 160 - 42 = 118 = mA 180 - 160 = 20 = mBCA Prove the Midsegment Theorem (that the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length) Informal Proof B This isn’t necessarily proving the relationship but we are going to work backwards to show that it works. Translate ABC by vector AB resulting in BB’D. Translate ABC by vector AC resulting in CDE. 2 1 C B' These two transformations create AB’E, where B is the midpoint of AB ' and D is the midpoint of B ' E , D B thus BD is the midsegment. BD || AE Because of the translation by vector 1 AB and BD = AE because of the translation of AC 2 to BD and to CE. 3 A 2 1 3 A E C Transformational Proof - Translation Given: ABC, where D is the midpoint of AB and E is the B midpoint of BC . 1 Prove: DE || AC and DE = AC 2 D A E C G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com 8 B Translate DBE by vector DA . If the translation of E by vector DA lies on AC , then DE || AC by the properties of a translation. Using an indirect proof, we assume that L (the image of E) does not lie on AC. A contradiction occurs (provided below), so L the image of E and lies on AC. In addition, by the properties of E D A C L translations, DE || AC . Translate DBE by vector EC (vector EC is the same as vector DL because they are parallel and the same distance). We know that DE = AL by the previous translation, and we know the DE = LC by the current translation, therefore, AL = LC (transitive). Thus, AL + LC = AC (Segment Addition Postulate), DE + DE = AC (Substitution), and 2DE = AC (Simplify). Lastly, by the Multiplication / 1 Division Property, DE AC . 2 Determine x. The midsegment theorem tells us that the midsegment is half its parallel side. x ½ (24 cm) = x 24 cm 12 cm = x Determine the missing information. a) x = ____________ b) x = ____________ c) x = ____________ 68° 13.5 cm x x = 27 cm 3x x 42 cm 3x = 21 x=7 x = 68