Congruence II Propositions

    This topic proves the well-known Alternate Interior Angle and Exterior Angle Propositions much as Euclid did for the Euclidean plane. However, Euclid had gaps in these proofs; in fact, his gaps may not be noticed until trying to follow his proofs on a sphere. This topic uses the Crossbar Proposition to close the gap in the Exterior Angle Proposition. Also the SAA Congruence Criterion and the Hypotenuse-Leg Congruence Criterion are detailed. Propositions concerning midpoints, bisectors, and relations between sides and angles of a triangle are proven.

Definition (Interior Angles) Let be a transversal to lines and with meeting at and at Choose points and on such that choose points and on such that and are on the same side of and Then the four angles: and are called interior angles.  The pairs ( and ) and ( and ) are called alternate interior angles.

Proposition (Alternate Interior Angle)
    (i) If two lines cut by a transversal have a pair of congruent alternate interior angles, then the two lines are parrallel.
    (ii) Two lines perpendicular to the same line are parallel.
    (iii) If is any line and is any point not on there exists at least one line through parallel to
    
        Proof. (i): Given Assume the contrary and meet at a point Say is on the same side of as and There is a point on such that by the Segment Shift Axiom. By the Segment Congruence Axiom is congruent to itself, so that by the SAS Axiom. By the Congruent Triangles Definition, Since is the supplement of must be the supplement of by the the Special Angles Proposition and the Angle Shift Axiom. This means that lies on and hence and have the two points and in common, which contradicts the Point Uniqueness Proposition. Therefore,
    (ii): If and are both perpendicular to the alternate interior angles are right angles by the Right Angle Definition; and hence are congruent by the Fourth Postulate of Euclid Proposition. By part (i), and are parrallel.




    (iii):  By the Point-Line Perpendicular Property, there is a line through perpendicular to and there is a unique line through perpendicular to Since and are both perdicular to part (ii) implies

Definition (Exterior Angle) An angle supplementary to an angle of a triangle called an exterior angle of the triangle and the two angles of the triangle not adjacent to this exterior angle are called the remote interior angles.

Proposition (Exterior Angle Property) An exterior angle of a triangle is greater than either remote interior angle.

    Proof. We will show that and Consider the remote interior angle If then is parallel to by the Alternate Interior Angle Proposition, which contradicts the hypothesis that these lines meet at Assume for a contradiction, Then there is a ray between and such that This ray intersects in a point by the Crossbar Proposition. By the Alternate Interior Angle Proposition, and are parallel which contradicts that lies on both of them. Thus, For remote angle use the same argument applied to which is congruent to by the Special Angles Proposition and by using the Angle Ordering Proposition.

Definition (Foot) If a perpendicular is dropped from a point not on a line to then the point at which the perendicular intersects is called its foot.

Proposition (SAA Congruence Criterion) Given and Then

    Proof. Assume side is not congruent to side Then or Case 1: If then there is a point such that and Then Hence, It follows that This contradicts the Proposition. Case 2: Similiarily, with a point between and Therefore, and so

Proposition (Hypotenuse-Leg Criterion) Two right triangles are congruent if the hypotneuse and a leg of one are congruent repsectively to the hypotenuse and a leg of the other.

Proposition (Midpoints) Every segment has a unique midpoint.

    Proof. Given any segment it will be shown that there exists a unique point such that and By the Points Not On Line Proposition, there exists a point not on By the Angle Shift Axiom, there exist a unique ray on a side of opposite to the side containing such that Since and are on opposite sides of segment must intersect at a point by the Opposite Sides Definition. By the Special Angles Proposition, By the Segment Shift Axiom, there exists a point on ray such that By the Linear Decomposition Proposition, Hence, and by the Angle Definition. Thus, by the SAA Proposition, and so by the Congruent Triangles Definition.  The Order Axioms states that exactly one of the following holds: or But both and imply or respectively. Both of these contradict by the Segment Ordering Proposition. Therefore, and so is a midpoint of by the Midpoint Definition. To prove that is unique suppose that is another midpoint of and without loss of generality, assume Then by the Segment Relation Definition. By the Segment Ordering Proposition, By the Betweenneess Property Proposition, and thus by the Segment Relation Proposition. By the Segment Ordering Proposition, Therefore, by the Midpoint Definition Therefore, there can be no other midpoint besides Every segment has a unqiue midpoint.

Proposition (Bisectors)
    (i) Every angle has a unqiue bisector
    (ii) Every segment has a unique perpendicular bisector.

Proposition (Larger Angle Larger Side) In a triangle the greater angle lies opposite the greater side and the greater side lies opposite the greater angle.

Proposition (Side Angle Comparison) Given and   if and then if and only if

    Proof. Assume that Then there exists a ray between and such that and assume that is given such that by the Segment Shift Axiom. By the Segment Congruence Axiom, and the assumption So by the SAS Proposition. Also, intersects at a point by the Crossbar Proposition. If then by the Segment Definition which implies that By the Congruent Triangles Definition and thus by the Segment Ordering Proposition. Assume By the Order Axiom, exactly one of the following holds: or The case can not hold, else the Interior Angle Propoerty Proposition would imply that is not in the interior of However, is in the interior of by the Ray Between Rays Definition. Hence there are two cases two consider, namely and





Assume that Then by the Pappus Proposition, By the Angle Relation Definition and By the Angle Ordering Proposition, and By the Larger Side Larger Angle Proposition, and then by the Segment Ordering Proposition



Assume that By the Pappus Proposition By the Angle Relation Definition, By the Exterior Angle Proposition, and By the Angle Ordering Proposition, and By the Larger Side Larger Angle Proposition, and then by the Segment Ordering Proposition
    Conversely, assume that By the Angle Ordering Proposition exactly one of the following hold: or Suppose then by the first part of this proposition which contradicts the assumption that by the Angle Ordering Proposition. Suppose that then by the SAS Proposition. Whence, by the Congruent Triangles Definition but this contradicts the Angle Ordering Proposition; so can not hold. Therefore,

Proposition (Segment Comparison)
    (i) If and then
    (ii) Given any triangle and point such that then or
    
        Proof. (i): By theExtension Axiom there exists a point on such that By the Right Angle Definition is a right angle.



By the Exterior Angle Proposition Since by the Larger Angle Larger Side Proposition. By the Exterior Angle Proposition and Since By the Larger Angle Larger Side Proposition, Therefore,



    (ii): By the Point-Line Perpendicular Property, there exists apoint on such that is a right angle. By the Segment Ordering Proposition, exactly one of the following must hold, or Suppose then by the Segment Comparison Proposition, Suppose then by the Segment Comparison Proposition, Suppose then by the Segment Comparison Proposition, Suppose then by the Segment Comparison Proposition,

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Cite this as:
Congruence Ii Propositions
Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
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