Overview of Section 5.2 **Congruence and Euclidean Constructions**

Definition of Congruence:Two figures are congruent if there exists an isometry that maps one onto the other.Points:

Based on and consistent with Euclid's concept of congruence by superimposing.

Transformations as functions allow a more general and more rigorous definition of congruence.

Extends our concepts of congruence from polygons to more general plane figures.

If we have sequence of isometries, the composition is an isometry.

If an isometry maps S onto S' then its inverse maps S' onto S.

Now Solve This 5.7

The Ellipse

c(red) can be mapped to Ellipsec'' (black) by a composition of two isometries. For example first mapcwith the centerKto its image c' (blue) with its centerK' by a translation with vector LM. Then rotate 90 degrees aboutK'.Note: Try this constructing and transformation using

GeoGebra. It can not be done gracefully with GSP.GeoGebra is available for download free.

Application of Isometries to Construction Problems

Example 5.3.

Find the minimum distance from A to a point Y on the line and then to a point B on the same side of the line.

See

Problem 5.1.5inProblem Set 5.1.See

Minimal Path Problemsin EMAT 6600

Solution:

Reflect B in line

l.Construct AB'. The intersection with linelat X will determine the minimal path.First, the reflection of B in B' preserves the angle measure. Because AB' is a straight line, we have vertical angle and therefore the angle of incidence equals the angle of reflection WHEN we have constructed the minimal path.

Use triangle inequality with triangle AYB' to complete the proof.

Now Solve This 5.8

The Converse: If the angle of incidence equals the angle of reflection at a point X onl, then A-X-B is the minimal path.

Example 5.4.Given an acute angle A and an arbitrary point P in the interior of the angle, construct a triangle with one vertex at P and the other two vertices on each side of the angle so that the perimeter of triangle is a minimum.

Solution: Reflect P in each of the two sides to locate image point R and and image point S. Construct RS. Let the intersections of RS with the two sides of the triangle be X and Y. Then by the reflections RX = PX and YS = PY. The perimeter of the triangle PXY is PX + XY + YP but this is the same as RX + XY + YS = RS. Since RS is a straight line, this is the minimal path.

Example 5.5. The Shortest Highway Problem

A highway connecting two cities A and B as in the figure needs to be built so that part of the highway is on a bridge perpendicular to the parallel banks b1 and b2 of a river. Where should the bridge be built so that the path is as short as possible?Hint: Consider a translation of B (and the bank b2) by a vector determined by the length of the bridge. This essentially makes the river of width 0.

Example 5.6

Given three parallel lines, a, b, and c. and a point P on one of the lines, construct an equilateral triangle PQS with one vertex on each of the lines.

If we do a 60 degree rotation of line c about P, all of the points on C will be rotated 60 degrees and, in particular, some point of the image of c will intersect with line a. That will be point S.

Now Solve This 5.9

1. Consider constructions when P is on line c or line a. (See GSP file)

2. Find another triangle satisfying the conditions of the problem.

3. Similar construction when given three curves such as three circles or three lines that are not parallel.

4. Given four parallel lines, construct a square with a vertex on each line. When is it possible?

Example 5.7

In the figure, Triangle ABC is an equilateral triangle inscribed in a circle and P is any point on the minor arc A. Prove that PA + PC = PB.

Plan --

Although we have solved this previously by other methods, the intent here is to find a solution using transformations.

If we can move AP so that it is collinear with CP then we would have a segment that is the same length as AP + PC.

Then we could show that segment is the same length as BP by an appropriate isometry.