Overview of Section 1.2 Congruence of Triangles
Definition of Congruent Triangles
Axiom 1.1 SAS Congruence
Theorem 1.3 The Isosceles Triangle Theorem and its corollary
If two sides of an isosceles triangle are congruent, then the angles opposite these sides are congruent.
Corollary: An equilateral triangle is equiangular.
Theorem 1.4 ASA congruence condition
Now Solve This 1.1
Prove converse of Theorem 1.3. If two angles of a triangle are congruent, then the sides opposite those angles are congruent.
Prove an equilangular triangle is equilateral
Medians, Altitudes, Properties of Isosceles Triangles.
Now solve this 1.2
Relationship of median, altitude, angle bisector, and perpendicular bisector to base in isosceles triangles.
Theorem 1.5. Median of Isosceles triangle is perpendicular bisector of base as well as the angle bisector of the angle opposite the base.
Theorem 1.6 Every point on the perpendicular of a segment is equidistant from the endpoints of the segment.
Corollary 1.2 A point is equidistant from the endpoints of a segment if and only if it is on the perpendicular bisector of the segment.
Corollary 1.3 If two points are equidistant from the endpoints of a segment, then the line through the points if the perpendicular bisector of the segment.
Now Solve This 1.3
Construction of perpendicular bisectors, medians, altitudes, and angle bisectors of an acute scalene triangle.
Construction of perpendicular bisectors, medians, altitudes, and angle bisectors of an obtuse scalene triangle
Euclidean Constructions: Straightedge and compass constructions (Review rules, p. 22)
Equilateral triangle given a side AB
Perpendicular bisector of a segment
Perpendicular line through a point on the line
Properties of a kite
A kite is a quadrilateral that has two pairs of congruent adjacent sides; it is a quadrilateral created when two isosceles triangles share a common base. The common base is a diagonal of the kite.
Theorem 1.7. The diagonal of a kite connecting the vertices where the congruent sides intersect bisects the angles at those vertices and is the perpendicular bisector of the other diagonal.
SEE PROBLEM SET 1.2, Problem 3
Rhombus -- a quadrilateral in which all four sides are congruent.
Corollary 1.4: The diagonals of a rhombus are perpendicular bisectors of each other and each bisects a pair of opposite angles.
Now Solve This 1.4
1. Converse of Corollary 1.4: A quadrilateral in which the diagonals are perpendicular bisectors of each other is a rhombus.
2. State and prove a converse of Theorem 7?
A quadrilateral in which the diagonals are perpendicular to each other is a kite.
3. Classify kite and rhombus by angle bisectors
A quadrilateral in which the angle bisectors are perpendicular to each other is a kite.
A quadrilateral in which the angle bisectors are perpendicular bisectors to each other is a rhombus.
Construction: Bisect a given angle.
Using constructed kite:
Using constructed isosceles triangle:
Theorem 1.8: Side, Side, Side (SSS) Congruency Condition
SEE PROBLEM SET 1.2, Problem 4.
Theorem 1.9: Hypotenuse-Leg Congruence Condition
SEE PROBLEM SET 1.2, Problem 5.
Theorem 1.10: Exterior Angle Theorem -- The exterior angle of a triangle is greater than either of the remote interior angles.
Corollary 1.5: Through a point not on an line, there is a unique (one and only one) perpendicular to the line.
Now Solve This 1.5
Construction of a perpendicular to a line from a point P not on the line.
Theorem 1.11: Hypotenuse-Acute angle congruency condition. If the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and an acute angle of another right triangle, then the triangles are congruent.
Distance to a line -- the distance from a point P to a line l is the length of the segment connecting P with the foot of the perpendicular to l through P.
Theorem 1.12: A point is on the angle bisector of an angle if and only if it is equidistant from the sides of the angle.
Theorem 1.13: Given two non-congruent sides in a triangle, the angle opposite the longer side is greater than the angle opposite the shorter side.
Theorem 1.14: (Converse of Theorem 1.13) Given two non-congruent angles in a triangle, the side opposite the greater angle is longer than the side opposite the smaller angle.
Theorem 1.15: The triangle inequality. The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
Proof: Given triangle ABC with sides of length a, b, and c.
Prove a + b > c,
b + c > a,
and
a + c > b
Now Solve This 1.6
1. Show that given three segments a, b, and c such that a + b > c, it is not always possible to whose sides are a, b, and c.
Discussion: Consider a = 10, b = 3, and c = 6.
2. Construct three segments a, b, and c so that a triangle with these segments as sides will exist. Is the existence and therefore the construction of the triangle assured by Theorem 1.15? Justify.
Hiker's Path Problem
Reflection in a line -- a reflection in a line l assigns to each point P not on the line, a point P', the image of P, in such a way that l is the perpendicular bisector of PP'. If P is on l then P' = P.
Properties of Parallel Lines
Theorem 1.16: If two lines in the same plane are each perpendicular to a third line in that plane, then they are parallel.
Prove by indirect proof: Assume the two lines are not parallel. Therefore they meet at a point P. Therefore there are two lines from P to the third line that are perpendicular. This contradicts corollary 1.5.
Transversal
Interior Angles
Corresponding Angles
Alternate Interior Angles
Alternate Exterior Angles
Theorem 1.17: If two lines are cut by a transversal and a pair of its corresponding angles is congruent (or a pair of Alternate Interior Angles are congruent), then the two lines are parallel.
Now Solve This 1.7
1. Prove Theorem 1.17 by contradiction. Assume lines are not parallel and therefore meet.
Now triangle PAB has an external angle congruent to an internal angle, contradicting the Exterior Angle Theorem.
2. Construct a line through P and intersecting l at Q. Locate R on l and construct m through P by duplicating angle PQR with vertex at P and side PQ in common. This gives a pair of corresponding angles congruent and therefore l is parallel to m by Theorem 1.17.
3. Prove that the opposite sides of a rhombus are parallel.
Proof. A rhombus is a quadrilateral with four congruent sides. Take a pair of opposite sides. By Corollary 1.4 the diagonal bisects the opposite angles. Therefore congruent alternate interior angles exist and by Theorem 1.17 the lines are parallel.
4. Use a rhombus to solve problem 2.
Now Solve This 1.8.
Parallelogram -- a quadrilateral in which each pair of opposite sides is parallel. It is not possible to prove that the diagonals of a parallelogram bisect each other without assuming the parallel postulate.
Problem set 1.2 has 22 problems. The instruction in the text indicates the problems are to be worked without use of the parallel postulate. All of the material in this section has been structured to have a neutral geometry; all of the theorems and corollaries hold independent of the parallel postulate. So the 22 problems in Section 1.2 should only make use of undefined terms, definitions, axioms from the appendix and sections 1.1 and 1.2, and the theorems, corollaries, and Now Solve This items in sections 1.1 and 1.2.
The PDF version gives instructions to work 19 problems but only provides 12 and only 11 of those 12 are in the published version.
I have provided the text for all of the non-PDF problems in this link.
