# More on Neutral Geometry I - More on Neutral Geometry...

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More on Neutral Geometry I (Including Section 3.3) ( "NIB" means "NOT IN BOOK" ) Theorem (NIB), The "The Adjacent Supplementary Angles" Theorem (Converse of Postulate 14) : If two adjacent angles are supplementary, then they form a linear pair. Proof: Suppose that ABD and DBC are two adjacent angles (with common ray BD ----→ ) are supplementary. (We need to show that they form a linear pair.) By Postulate 4 (The Ruler Placement Postulate), there is a point on line AB ←---→ such that A – B – F . Thus, since F and C are both in the half-plane side of BD ←---→ opposite the half-plane containing A, F and C are on the same side of BD ←---→ . Now, ABD and DBF form a linear pair by definition of “linear pair”. By Postulate 14, ABD and DBF are supplementary. By the supposition above, ABD and DBC are also supplementary. Therefore, DBC DBF 2245 ∠ ./ ( ) ( ) m DBC m DBF = . We see then that ray BC ----→ and BF ----→ are both rays in the same half-plane side of line BD ←---→ such that each ray combines with ray BD ----→ to produce angle with the same measure r = 180 ° ( ) m ABD . By Postulate 12 (The Angle Construction Postulate), there is one and only one ray in that same half-plane side of BD ←---→ which unions with BD ----→ to produce the angle with that particular angle measure, r = ( ) ( ) m DBC m DBF = . By the uniqueness of this ray, we can conclude that BC ----→ = BF ----→ and so, DBC DBF = . Therefore, ∠ g1827g1828g1830 g1853g1866g1856 ∠ g1830g1828g1829 form a linear pair. Q E D B A D C F
2 Section 3.3: Triangle Congruence Conditions Theorem 3.3.1 (Angle-Side-Angle Congruence Condition): 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. Proof: Let ABC and DEF be two triangles and consider the correspondence of vertices ABC barb2leftbarb2right DEF . Suppose that CAB FDE 2245 ∠ , and CBA FED 2245 ∠ , and AB DE 2245 . [We need to show ABC DEF 2245 ∆ . ] [ We do this by showing that AC DF 2245 , and then we apply Postulate 15 (SAS) to conclude that ABC DEF 2245 ∆ .] Suppose, by way of contradiction, that AC and DF are not congruent . AC DF .