Theorem (NIB), The "The Adjacent Supplementary Angles" Theorem (Converse of Postulate 14) :

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1 More on Neutral Geometry I (Including Section 3.3) ( "NI" means "NOT IN OOK" ) Theorem (NI), The "The djacent Supplementary ngles" Theorem (onverse of ostulate 14) : If two adjacent angles are supplementary, then they form a linear pair. roof: Suppose that ray ) are supplementary. and are two adjacent angles (with common (We need to show that they form a linear pair.) y ostulate 4 (The Ruler lacement ostulate), there is a point on line such that. Thus, since and are both in the half-plane side of opposite the half-plane containing, and are on the same side of. Now, and form a linear pair by definition of linear pair. y ostulate 14, and are supplementary. y the supposition above, and are also supplementary. Therefore,./ m( ) = m( ). We see then that ray and are both rays in the same half-plane side of line such that each ray combines with ray to produce angle with the same measure r = 180 m( ). y ostulate 12 (The ngle onstruction ostulate), there is one and only one ray in that same half-plane side of which unions with to produce the angle with that particular angle measure, r = m( ) = m( ). y the uniqueness of this ray, we can conclude that = and so, =. Therefore, form a linear pair. Q

2 Section 3.3: Triangle ongruence onditions 2 Theorem (ngle-side-ngle ongruence ondition): If, in two triangles, the vertices of one triangle can be put into one-to-one correspondence with the vertices of the other triangle such that: Two angles and the included side of one triangle are congruent to the corresponding angles and included side of the other triangle, Then, the correspondence of triangle vertices is a congruence and the two triangles are congruent. roof: Let and be two triangles and consider the correspondence of vertices. Suppose that, and, and. [We need to show. ] [ We do this by showing that.], and then we apply ostulate 15 (SS) to conclude that Suppose, by way of contradiction, that and are not congruent.. Without loss of generality, we assume that >. [ Note: If <, then we rename the points,, as,,, and we rename the points,, as,,, respectively. fter these renamings, every statement made to this point is true. In particular, >. ] Since >, there exists a point T such that T and T =, by the Ruler lacement ostulate (ostulate 4). Note that T since T =. T Since m( ) = m( ), we conclude by ostulate 13 that m( T ) < m( ). In particular, m( T ) m( ). Since T =, T. Recall that T and. Therefore, T by SS, and T by and m( T ) = m( ), which contradicts that fact that m( T ) < m( ).. by SS. The S Triangle ongruence ondition is valid. Q

3 Theorem 3.3.2, The "ongruent ase ngles" Theorem (onverse of the "Isosceles Triangle Theorem"): If two angles in a triangle are congruent, then the sides opposite these congruent angles are congruent segments and the triangle is an Isosceles triangle. roof: The proof is a simple application of the -S- triangle congruence condition and is left as an exercise. 3 Theorem (ngle-ngle-side ongruence ondition): If, in two triangles, the vertices of one triangle can be put into one-to-one correspondence with the vertices of the other triangle such that: Two angles and the side opposite one of them in one triangle are congruent to the corresponding angles and corresponding opposite side of the other triangle, Then, the correspondence of triangle vertices is a congruence and the two triangles are congruent. roof: Let and be two triangles and consider the correspondence of vertices. Suppose that, and and. [ We show that using a proof-by-contradiction. ] Suppose, by way of contradiction, that and are not congruent.. Without loss of generality, we assume that >. Since >, there exists a point T such that T and T =, by the Ruler lacement ostulate (ostulate 4). Thus, T. T Now, T is an exterior angle of T and T is an interior angle of that triangle which is not adjacent to T. y Theorem (The xterior ngle Theorem), m( T ) > m( T ).

4 Note that T since T =, and recall that and T. T by ostulate 15 (SS). T by and, therefore, m( T ) = m( ). lso, m( ) = m( ) = m( T ) since T =. m( T ) = m( T ), which contradicts the fact that m( T ) > m( T ).. Recall that and. by SS. The S Triangle ongruence ondition is valid. Q 4 Theorem 3.3.4: SSS ongruence for Quadrilaterals Theorem 3.3.5, The "Larger Side opposite Larger ngle" Theorem: or two sides of a triangle and the angles opposite them, the angle opposite the larger side is the larger of the twos angles. Theorem The Triangle Inequality: The sum of the lengths of any two sides > the length of the third side. or proving the Side-Side-Side ongruence ondition, we will need the Hinge Theorem which is not proven here, but a proof of it can be found in the textbook. Theorem The Hinge Theorem: Given that two sides of one triangle are congruent to two corresponding sides of another triangle, then, the greater of the two included angles has the longer of the two corresponding third sides opposite these included angles. The Shorter Side is Opposite of the Smaller ngle. The Longer Side is Opposite of the Larger ngle.

5 Theorem (Side-Side-Side ongruence ondition): If, in two triangles, the vertices of one triangle can be put into one-to-one correspondence with the vertices of the other triangle such that: The three sides of one triangle are congruent to the corresponding sides of the other triangle, Then, the correspondence of triangle vertices is a congruence and the two triangles are congruent. roof: Let and be two triangles and consider the correspondence of vertices. Suppose that and and. 5 [ We show that using a proof-by-contradiction. Then, we conclude that by SS. ] Suppose, by way of contradiction, and are not congruent. m( ) m( ). Without loss of generality, we assume that m( ) > m( ). lso, recall that and. y Theorem (The Hinge Theorem), >, which contradicts the fact that.. by ostulate 15 (SS). The SSS Triangle ongruence ondition is valid. Q Theorem (NI), The rop a erpendicular Theorem. Given any line l and any point, there exists a unique line through which is perpendicular to line l. roof: onsider line l and a point. We first assume that point is not on line l. (See the figure.) l We must show that there exists a line through which is perpendicular to line l.

6 Let be any point on line l. If line is perpendicular to line l, then such a perpendicular line exists. 6 Suppose then that line is not perpendicular to line l. Let be a point on l so that is an acute angle. y the ngle onstruction ostulate, there is a unique ray in the half-plane on the side of line l which is opposite of the side containing such that H l m( ) = m( ). y the Ruler ostulate, there is a point on ray such that =. oints and are on different half-plane sides of l ; therefore, line l intersects the segment at some point with. y construction, =. y construction,. lso,. Therefore, by SS. by. and are right angles by the ongruent Supplementary ngles Theorem. Therefore, line is perpendicular to line l. Therefore, there exists at least one line through point which is perpendicular to l. To prove that there is only one such line through perpendicular to line l, suppose (by way of contradiction) that there were a second line through and perpendicular to line l intersecting line l at point. Without loss of generality, we can assume that. (See the figure.) Then, is an exterior angle of. y the xterior ngle Theorem, m( ) > m( ) since is a nonadjacent interior angle of. ut they are both right angles: m( ) = m( ) = 90, a contradiction. Thus, when point is not on line l, there is a unique line through which is perpendicular to line l. Line l We next assume that point is on line l. Since the point is on line l, then, by the ngle onstruction ostulate, in each half plane bounded by line l, there is one and only one ray from point forming an angle of measure 90. y the "djacent Supplementary ngles" Theorem (onverse of SMSG ostulate 14), the rays in the two half-planes are opposite rays that form the unique line through perpendicular to line l. Q l