- 2.1 Basic measurements
- 2.2 Historical note
- 2.3 More on measurements
- 2.4 Congruence
- 2.5 Important remark about notation
- 2.6 The axiom for congruence
- 2.7 Exercises
- 2.8 Monotonicity of lengths and angles
- 2.9 Isosceles triangles
- 2.10 Congruence via
*SSS* - 2.11 Inequalities for general triangles
- 2.12 Congruence via
*AAS* - 2.13 Perpendicularity and orthogonality

The triangle with *vertices* is denoted by
, where is the point of intersection of the lines and ;
is the point of intersection of the lines and ; and is
the point of intersection of the lines and . These points of
intersection divide each of the lines into two unbounded half-lines
and one bounded line segment, called a *side* of the triangle.

The triangle, , defines numbers, the angle measures (also called the angles) at the vertices , and , and the lengths of the sides, which are the line segments , and .

The angle measure at for example the vertex is denoted by , or .

We denote the length of the side , for example, by . Until modern times, the side and its length were denoted by the same symbol, and the reader had to figure out which is which from the context. As with angles, when convenient, we will also use the same notation for a line segment and its length.

The pair of lines, and , for example, determines two angles; the question of which of these angles is determined by the triangle can be stated in words with difficulty; we will leave this as visually obvious.

For physical triangles, two triangles are congruent if they exactly match if you put one on top of the other. Another way of saying this, for ideal triangles, is that there is an isometry of the plane (a composition of rotation, translation and reflection) that maps one exactly onto the other.

It is common to refer to the above angle as ``Angle-Side-Angle'' or *ASA*.

For physical triangles this is essentially obvious. If you know the length of a side, and you know the two angles, then the lines on which the other sides lie are determined, so the third vertex is also determined.

**Remark:** One of these, *AAS*, is not obvious; in fact it is false
in spherical geometry.

In the following few exercises, when you are asked to prove something you
may assume that *AAS*, *ASA*, *SAS* and *SSS* are true. One other fact that
you may use is Thm. 3.4: the sum of the angles of a
triangle is . Note that *this is only for these exercises*; in
general we cannot assume things we have not proven or taken as an axiom,
because we may wind up applying circular reasoning (that is, giving a proof
that something is true which implicitly assumes it was true to begin with.)
But the main point of this exercise is to get you thinking about how
geometry works, so we can relax our restrictions a little.

A *quadrilateral* is a region bounded by four line segments;
that is, it is a four-sided figure. The quadrilateral with vertices,
, , and , in this order, is determined by the four line
segments connecting and , and , and , and connecting
and . For to form a quadrilateral, these segments must not
intersect except at the verticies.

A quadrilateral defines 8 pieces of information: the lengths of the four sides, and the measures of the four angles. Two quadrilaterals are congruent if these 8 pieces of information agree.

What is the minimal number of pieces of information one needs about two quadrilaterals to prove that they are congruent? (No response needed here, but you need the answer for the next question.)

So let us consider the case where they are different, and arrive at a contradiction. We may assume that (if not, just exchange the names on the triangles).

Apply the second part of Axiom 7 to find a line passing through the point and some point lying between and , so that .

By *ASA*,
. Therefore
. But we are given that
. Therefore,
. Since lies on the line determined by and , and lies
between them, this contradicts Axiom 6.

While an isosceles triangle is defined to be one with two sides of equal length, the next theorem tells us that is equivalent to having two angles of equal measure.

*Conversely, if
has
, then it is
isosceles, with base .*

We find a point and construct the line so that , and . (That this can be done follows from Axioms 6 and 7.) It is unclear where the point lies: it could lie inside triangle ; it could lie on the line between and ; or it could lie on the other side of the line . We need to take up these three cases separately.

For both of the remaining cases, we draw the lines and . We
observe that, by *SAS*,
. It
follows that
and that
. Hence
is isosceles, with base , and
is
isosceles with base . Since the base angles of an isosceles
triangle have equal measure,
and
.

- (i)
- .
- (ii)
- .

The next theorem says that in a triangle, if one angle is bigger than another, the side opposite the bigger angle must be longer than the one opposite the smaller angle. This is generalizes the fact that the base angles of isosceles triangles are equal (Thm. 2.2).

The converse of the previous theorem is also true: opposite a long side, there must be a big angle.

If , then, by the previous theorem, , which again contradicts our assumption.

The following theorem doesn't quite say that a straight line is the shortest distance between two points, but it says something along these lines. This result is used throughout much of mathematics, and is referred to as ``the triangle inequality''.

BASIC CONSTRUCTION.
Given a line , and any point , there is a line through
perpendicular to .

In any triangle, there are three special lines from each vertex. In
, the *altitude* from is perpendicular to
; the *median* from bisects (that is, it crosses
at a point so that ); and the *angle
bisector* bisects (that is, if is the point where the
angle bisector meets , then
).

By *AAS*,
. Hence . Since
is also the angle bisector, by *ASA*,
. Hence , from which it follows that .