# Notes on Congruence 1

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1 ongruence-1 Notes on ongruence 1 xiom 1 (-1). If and are distinct points and if is any point, then for each ray r emanating from there is a unique point on r such that =. xiom 2 (-2). If = and = F, then = F. Moreover, every segment is congruent to itself. xiom 3 (-3: Segment ddition). If,, =, and =, then =. xiom 4 (-4). Given any, and given and ray emanating from a point, then there is a unique ray on a given side of line such that =. xiom 5 (-5). If = and =, then =. Moreover, every angle is congruent to itself. xiom 6 (-6: SS). If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent. orollary 1 (orollary to SS). Given and segment =, there is a unique point F on a given side of such that = F. Proof. y -4 there is a unique ray G on the given side of such that = G. y -1, there is a unique point F on G such that = F. y SS, = F. Proposition 10. If in we have =, then =. Proof. onsider the correspondence of vertices,,. y hypothesis =, and by -5 =. Then by SS =. y the definition of congruent triangles =. Proposition 11 (Segment Subtraction). If, F, =, and = F, then = F. Proof. Suppose = F (R). y -1, there is a point G on F such that = G. y the R hypothesis F G. Since = by hypothesis, and = G we have = G by -3. y hypothesis, = F, so by -2, G = F. y the uniqueness in -1 G = F. This is a contradiction, so = F. Proposition 12. Given = F, then for any point between and, there is a unique point between and F such that =. Proof. y -1 there is a unique point on F such that =. Suppose is not between and F (R). y the definition of ray, either = F or F. Suppose = F. and are distinct points on by 1, but = and =. This contradicts -1, so F. Now suppose F. y -1 there is a unique point G on the ray opposite to such that G = F. y -3, G =. This contradicts the uniqueness part of -1, since =. Hence is between and F. efinition. < (or > ) means that there exists a point between and such that =. Proposition 13 (Segment Ordering). 1. xactly one of the following conditions holds (trichotomy): <, =, or >. 2. If < and = F, then < F. 3. If > and = F, then > F. 4. If < and < F, then < F (transitivity). Proof. 1 The statements of the propositions and many proofs are taken from the book uclidean and Non-uclidean Geometries by M. Greenberg.

2 ongruence-2 1. Suppose and are not congruent. y -1, there exists a point on such that =. y definition of ray, either or. If, then < by definition of <. Suppose. y Proposition 3.12, there is a unique point F on between and such that = F. Then > by definition of <. 2. Since <, there is a point P between and such that = P. Since = F, by Proposition 3.12, there is a unique point Q between and F such that P = Q. y -2, = Q, so < F. 3. Since >, there is a point P between and such that P =. Suppose = F. Then by -2, P = F, so > F. 4. Since <, there is a point P between and such that = P. Since < F, there is a point Q between and F such that = Q. y Proposition 3, there is a point R between and Q such that P = R. y -2, = R. Since R Q and Q F, we know by Proposition 3.3 that R F. Hence, < F. Proposition 14. Supplements of congruent angles are congruent. Proof. Suppose = F. Let P be the ray opposite to and let Q be the ray opposite to. We want to show P = FQ. F P Since the points,, and P are given arbitrarily on the sides of and P, by -1 we can choose the points, F, and Q on F and FQ such that =, = F, and P = Q. Then = F by SS. y the definition of congruent triangles, = F, and =. y -3, P = Q. Then again by SS, P = FQ. y the definition of congruent triangles P = FQ and P = Q. nd again by SS, P = FQ, so P = FQ. Proposition Vertical angles are congruent to each other. 2. n angle congruent to a right angle is a right angle. Proof. 1. y definition two angles are vertical if they allow labeling and where and are opposite, and and are opposite. Q

3 ongruence-3 isthesupplementto and isthesupplementto. y-5 =, so by Proposition 5 =. 2. Suppose is a right angle and F =. Suppose P is a point on the ray opposite to and Q is a point on the ray opposite F. We need to show F = Q. Since = F, by Proposition 5, their supplements are congruent, i.e. P = Q. y the definition of right angle = P, so by -5 = Q. gain by -5 F = Q. Proposition 16. For every line l and every point P there exists a line through P perpendicular to l. Proof. ither P lies on l or it does not. ssume first that P does not lie on l, and let and be any two points on l by I-2. y -4 there is a ray X such that X is on the opposite side of l as P and X = P. y -1 there is a point P on X such that P = P. Since P and P are on opposite sides of l, PP intersects l at a point Q between P and P. If Q =, P and P are supplementary. Since these angles are congruent, they are right angles, so PP l. If Q, then Q = Q by -1, so PQ = P Q by SS. y the definition of congruent triangles PQ = P Q. Hence PP l. Now suppose P lies on l. y Proposition 2.3 there is a point not on l. y the above argument we can construct a line perpendicular to l through this point, thereby obtaining a right angle. y -4, there is a unique ray on a particular side of l emanating from P such that P with one side contained in l is congruent to a right angle. y Proposition 3.15, P is a right angle. The side of this angle not contained in l is contained in a line perpendicular to l through P. Proposition 17 (S). Given and F with =, = F, and = F. Then = F. Proof. y -1, there is a unique point X on such that = X. y hypothesis = and = F, so = XF by SS. y the definition of congruent triangles = FX. y hypothesis = F, so by the uniqueness of -4 F = FX. y Proposition 2.3, = X. Hence = F. Proposition 18. If in we have =, then = and is isosceles. Proof. onsider the correspondence of vertices,, and. y -1 =. y hypothesis =, so by Proposition 3.17, =. y the definition of congruent triangles =, so by the definition of isosceles triangle is isosceles. Proposition 19 (ngle ddition). Given G between and, H between and F, G = FH, and G = H. Then = F. Proof. y the crossbar theorem we may choose G so that G. y -1 we may choose, F, and H so that =, = F, and G = H. y hypothesis G = H, so by SS G = H. Similarly, by hypothesis G = FH, so by SS G = FH. y the definition of congruent triangles G = H and G = HF. We next need to show, H, and F are collinear. y the definition of congruent triangles G = H and G = FH. Since, G, and are collinear G and G are supplementary. y Proposition 5 G is congruent to the supplement of H. enote this supplement by HX. y -4 HX is unique, so HX = HF. Then H is supplementary to FH, so F, H, and are collinear. Since H is between and F, H is in the interior of F, so H F by Proposition 3.7. Since G = H and G = HF, by -3 = F. y SS = F, so by the definition of congruent triangles = F. Proposition 20 (ngle Subtraction). Given G between and, H between and F, G = FH, and = F. Then G = H. Proof. We proceed as in the proof of Proposition Suppose G = H (R). y -4 there is a unique ray X on the same side of H such that G = HX. y the R hypothesis X. y hypothesis FH = G, and by R hypothesis XH = G, so by Proposition 10 = XF. y hypothesis = F, so by -5 F = XF. y the uniqueness part of -4 X =, but this is a contradiction, so G = H.

4 ongruence-4 Lemma 1. Given = F, then for any ray G between and, there is a unique ray H between and F such that G = FH. G H F Proof. y the rossbar Theorem we can choose G so that G. y -1 we can choose points and F such that = and = F. y SS = F, and by the definition of congruent triangles = F and = F. Then by Proposition 3.12 there is a unique point H on F such that G = FH. gain by SS, G = FH, so by the definition of congruent triangles G = FH. We only need to show that H is between and F. Since H is on F, H is between and F, so by Proposition 3.7 H is between and F. efinition. < F means there exists a ray G between and F such that = GF. Proposition 21 (Ordering of ngles). 1. xactly one of the following conditions holds (trichotomy): P < Q, P = Q, or P > Q. 2. If P < Q and Q = R, then P < R. 3. If P > Q and Q = R, then P > R. 4. If P < Q and Q < R, then P < R. (transitivity). Proof. This proofis verysimilarto the proofofproposition3.13. Forlabeling purposeswe say P =, Q = F, and R = GHI. 1. Suppose = F. y -4 there exists a unique ray X on the same side of F as such that = XF. X either is between F and or X is not between F and. If X is between F and, then < F. Suppose X is not between F and. Since X is on the same side of F as and X is not between F and, we know that X and F are on opposite sides of. y Lemma 2 and orollary 1 every point except on F is on the opposite side of as every point of X, so segment F does not intersect X. y a similar argument with the ray opposite X and F, we can show that segment F does not meet X. Hence is on the same side of X as F, so is interior to XF. Then is between X and F. y Lemma 4 there exists a unique ray Y between and such that F = Y. Hence > F. 2. Since < F, there exists a ray X between and F such that = XF. y Lemma 4 there is a unique ray HY between HG and HI such that YHI = XF. y -5 = YHI, so < GHI. 3. Since > F, there exists a ray X between and such that X = F. y -5 X = GHI, so > GHI. 4. Suppose < F and F < GHI. Since < F, there is a unique ray X between and F such that = XF. Since F < GHI, there is a unique ray HY between HG and HI such that F = YHI. y Lemma 4, there is a unique ray HZ between HY and HI such that XF = ZHI. y -5 = ZHI. Since HI HZ HY and HI HY HZ, by Lemma 3 HI HZ HZ. Then by definition < GHI.

5 ongruence-5 Proposition 22 (SSS). Given and F. If =, = F, and = F, then = F. Proof. y orollary 4, since = F we can pick a point G uniquely on the opposite side of as such that F = G. y the definition of congruent triangles = G and F = G. Then by -2 = G and = G. We will show that = G. Since and G are on opposite sides of, segment G intersects at X. y -3, X, = X, or X. The circumstances X = and X are equivalent to X = and X, respectively. Suppose X. onsider G. Since = G, G, by Proposition 3.10 G = G. Now consider G. Since = G, by Proposition 3.10 G = G. Then by Proposition 3.20 (angle subtraction) = G. y SS = G. Suppose = X. Since G = in G, by Proposition 10 = G. Then by SS = G. Suppose X. onsider G. Since = G, G, by Proposition 3.10 G = G. Now consider G. Since = G, by Proposition 3.10 G = G. Then by Proposition 3.19 (angle addition) = G. y SS = G. In all three cases = G. y the definition of congruent triangles = G. Since F = G, = G. y -5 =, so by SS = F. Proposition 23. ll right angles are congruent to each other. Proof. Suppose = and HF = HG are two pairs of right angles. ssume = HF (R). y Proposition 3.21 (a), either > HF or < HF. Without loss of generality suppose > HF. Then there is X between and such that X = HF. y Proposition 3.14 X = HG. Since = by hypothesis, and > HF by by R hypothesis, we have > HF by Proposition 3.21 (c). Since HF = HG by hypothesis, we have again by Proposition 3.21 (c) > HG. From above since X = HG we have > X by Proposition 3.21 (c). y Proposition 3.8 (c), since X is between and, we know that is between X and, since is the ray opposite to. Then < X, but this contradicts Proposition 3.21 (a). Hence = HF.

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