T5 Supplements of congruent angles are congruent.

Idea of proof. Use the definitions of congruent angles, supplementary angles and some algebra.

T6 Complements of congruent angles are congruent.

Idea of proof .

Definition . Two angles form a linear pair if they have a side in common and their remaining sides are opposite rays.

T6.5 Two angles which form a linear pair are supplementary.

Proof . Let angle(ABD) and angle(CBD) form a linear pair with m( angle(ABD) ) = and m(angle(CBD)) = .

By the trichotomy law of numbers, + < 180, or + = 180 or + > 180. We will show the first and last cases lead to contradictions and hence the middle case must hold true, that is, the angles are supplementary.

Case 1: Suppose + < 180. Then by PrA there is a point P on the D-side of line(AC) so that m(angle(PBC)) = + . The point P lies in the interior of angle(ABD) (why?). Hence D lies in the interior of angle(PBC) and so by AAA,

+ = m(angle(PBC) = m(angle(PBD)) + m(angle(DBC)) = m(angle(PBD)) + .

It follows by subtraction that = m(angle(PBD)). But m(angle(PBD)) < , since P is in the interior of angle(ABD). This is a contradiction and case 1 cannot occur.

Case 2 : Suppose + > 180. Then > 180 - > 0 and so by the protractor axiom there is a point P in the interor of angle(ABD) so that m(angle(DBP)) = 180 - . But now D is in the interior of angle(PBC) and so by AAA,

m(angle(PBC)) = m(angle(PBD)) + m(angle(BDC)) = (180 - ) + = 180. This contradicts the protractor axiom and we conclude that the case + = 180 must hold true. qed

Definition . Two angle are vertical angles if they have the same vertex and each side of one angle is the opposite ray of a side of the other angle.

T7 Vertical angles are congruent.

Idea of proof: Let angle(ABC) and angle(DBE) be vertical angles with ray(BA) and ray(BE) opposite rays. Then angle(ABC) and angle(ABD) form a linear pair (why?) and so are supplementary (why?). Also angle(ABD) and angle(DBE) form a linear pair and so are supplementary. Now use T5 to finish off the argument. qed.

Definition . Two triangles ABC and DEF are congruent under the correspondence A to D, B to E, and C to F if each pair of corresponding sides are congruent (that is, AB = DE, BC = EF and CA = FD) and each pair of corresponding angles are congruent (that is, m(angle(A)) = m(angle(D)), m(angle(B)) = m(angle(E)), m(angle(C)) = m(angle(F))).

SAS If two sides and the included angle of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.

T8 If two sides of a triangle are congruent, the angles opposite these sides are congruent.

Proof : Suppose ABC is a triangle with AB = AC. Then ABC is congruent with ACB by SAS since AB = AC,

angle(A) is congruent with angle(A) and AC = AB. So the remaining parts of ABC are congruent with their corresponding parts of ACB. In particular the angle opposite seg(AB), angle(C), is congruent with the angle opposite seg(AC), angle(B). qed

T 8.5 (ASA) If two angles and the included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.

Proof. Suppose ABC and DEF have AB = DE, m(angle(A)) = m(angle(D)), m(angle(B)) = m(angle(E)),

We will show that AC = DF and hence ABC and DEF are congruent by SAS. If AC <> DF, then one is less than the other. We can suppose AC < DF. Then use RA to construct F' on seg(AC) so that AF' = DF. Now by SAS, ABF' and DEF are congruent, and hence m(angle(ABF')) = m(angle(DEF)). Note that since F' is between A and C, m(angle(ABF')) < m(angle(ABF')) + m(angle(F'BC)) = m(angle(ABC)). But m(angle(ABC)) = m(angle(DEF)) (why?). Hence m(angle(DEF)) < m(angle(DEF)), a contradiction. qed

T9 If two angles of a triangle are congruent, then the sides opposite them are congruent.

Idea of Proof. Use ASA.

T10 A triangle is equilateral if and only if it is equiangular.

Idea of Proof. Use T9 to show the if part and T8 to show the only if part.

T11 An exterior angle of a triangle is greater in measure than either remote interior angle.

Restatement: Given triangle ABC. Show angle(ACB) < angle(CBD) and angle(CAB) < angle(CBD).

Idea of proof: Construct midpoint M of seg(CB). Extend seg(AM) to seg(AE) so that MB = ME. Now angle(AMC) and angle(BME) are congruent since they are vertical angles (T7) and so triangle(AMC) is congruent with triangle(EMB) by SAS. So angle(C) is congruent to its corresponding angle(MBE) which in turn is smaller in measure than angle(CBD) by AAA. The second conclusion is handled the same way, using instead of angle(CBD) its vertical partner. qed

T12 There is exactly one perpendicular to a line that contains a point not on the line.

Restatement: Given a line l and a point P not on l. Show there is one and only one line m thru P perpendicular to l.

Proof : To show there is one, pick points A and B on l and let = m(angle(PAB)).

Use the protractor and ruler axioms to construct a point Q on the non-P side of line l so that angle(QAB) has measure and QA = PA. Since Q is not on the P-side of l, the seg(PQ) intersects line l in a point, call it M. Now since AM = AM, we know by SAS that AMP is congruent with AMQ. Hence the corresponding angles angle(PMA) and angle(QMA) are congruent. They also form a linear pair and so the sum of their measures is 180 (T6.5). But then each angle must have measure 90 and so line(PQ) is perpendicular to line l.

To show there is no more than one line thru P perpendicular to l, suppose there is another. Then it will intersect line l at a point N different from M. But now the triangle PMN has an exterior angle of measure 90 with a remote interior angle of measure 90. This contradicts the exterior angle theorem. (T11) and so there is exactly one line thru P perpendicular to l. qed

T12.5 (SSS) If three sides of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.

Idea of Proof. Let ABC and DEF be triangles with AB = DE, BC = EF, and CA = FD. Use ASA to construct a congruent copy ABC' of DEF with C' on the opposite side of line(AB). Let M be the intersection of segment(CC') with line(AB). (M exists by PS). Use T8 to show angle(ACM) is congruent with angle(AC'M) and also angle(BCM) is congruent with angle(BC'M). Then use AAA to show angle(ACB) is congruent with angle(AC'B). Now use SAS to show triangle(ABC) is congruent with triangle(ABC') and hence with triangle(DEF) qed

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T13 If two sides of a triangle are not congruent, the angles opposite them are not congruent and the larger angle is opposite the larger side.

Proof : Let ABC be a triangle with seg(AB) not congruent to seg(BC). Then AB < BC or AB > BC. We can suppose AB < BC. Use RA to construct point D between B and C so that BD = BA. Then by T8, m(angle(BAD)) = m(angle(BDA)). Now angle(BDA) is an exterior angle of CAD and so by T11 m(angle(BDA)) > m(angle(DCA)). Also m(angle(BAD)) < m(angle(BAC) (why?). From this it follows that m(angle(C)) < m(angle(BAC)) (how exactly?) qed

T14 If two angles of a triangle are not congruent, then the sides opposite them are not congruent and the larger angle is opposite the larger side.

Idea of Proof: Call the angles alpha and beta and the sides opposite a and b respectively. Assume alpha < beta.

Then a can't be equal to b by T8. Also b can't be less than a (use RA, T8, T11, AAA)

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