Overview of Section 4.1 **Ratio, Proportion, Similar Polygons**

Informally, we can observe that two objects have the 'same shape.' The mathematical concept for this is

similarity.

Definition: Two figures are similar if and only of there exists a one-to-one correspondence between the figures such that the ratio between any two distances within one figure is the same as the ratio between the corresponding pair of distances in the other figure.Ratio -- a quotient of two real numbers. If r does not equal 0, the number s/r is the ratio of s to r, sometimes written s:r.

Proportion -- an equation stating that two or more ratios are equal.

Cross product property

Reciprocal Property

Denominator Addition Property

Ratio of Sum of Numerators to Sum of Denominators Property

Similarity of Polygons

Definition of Similar Polygons: Two polygons are similar if and only if there is a one-to-one correspondence between the vertices of one polygon and the vertices of the other polygon such that the corresponding angles are congruent and ratio of the lengths of corresponding sides is a constant.

Scale factor-- the ratio of corresponding lengths. Also called (elsewhere) thecoefficient of similitude,similarity coefficient, ordilution factor.

Now Solve This 4.11. Describe and sketch two quadrilaterals and two convex pentagons in which the corresponding sides are proportional but the polygons are not similar.

A square and a rhombus, for example would have sides proportional but they are not similar unless the rhombus is a square.

Others?

2. If two polygons P1 and P2 are similar ante the side of P! are proportional to the sides of P2 with scale factor r, with what scale factor are the sides of P2 proportional to the sides of P1?

1/r? Why or why not?

Theorem 4.1

Parallel projection preserves ratios of lengths of segments.That is,RECALL THAT THEOREM 1.29 PROVED THAT PARALLEL PROJECTION PRESERVES CONGRUENCE OF SEGMENTS.

Proof: The proof presented in the text assumes that

ABandBCcan each be parsed into some number of segments of the same length. If each of those segments onABandBCare of lengthxwithABhavingiof them andBChavingjof them, then we can writeAB = ixandBC = jx. That is, AC is covered with i + j segments each of length x.Parallel projection preserves

congruenceof the segments, so each of these i + j segments is mapped onto i + j segments on A'C'. The segments on A'C' will all be congruent, say of length y, so A'B' = iy and B'C' = jy. SoThe ratios of the segment lengths have each been shown to be equal to the ration of

numberof congruent segments of some common measure making up each segment. The argument assumes that the segments arecommensurable(that there exists a common measure.The familiar proof that the is irrational is an example that segments of length 1 and are not commensurable.

Two segments where each has a rational length are commensurable. Simply write each rationale length with a common denominator and use the unit fraction with that common denominator as the common measure.

Theorem 4.2 Side-Splitting Theorem

A line parallel to a side of a triangle that intersects the other two sides in distinct points splits the sides into proportional segments. That is

Plan: We can try to find triangles with a common altitude to the segments we want to compare as bases.

Open GSP FileTriangle AQP and Triangle APQ are two names for the same triangle. Hence the alternative expressions for the area are equal.

Taking PQ as the common base for Triangle PQB and Triangle QPC, the altitude to that base is the distance between the parallel lines. Hence the areas of these two triangles are equal.

QED

Theorem 4.3 The Converse of the Side Splitting Theorem

If a line divides two sides of a triangle proportionally (the ratio of the segments on one side equals the ration of the corresponding segments on the other side, then the line is parallel to the third side.Open GSP File

Corollary to the Side-Splitting Theorem (Open GSP File)

Now Solve This 4.21. State and Prove the

converseof the corollary to the side-splitting theorem2. Use the side splitting theorem to prove that parallel projection preserves the ratio of segments. The following figure is suggested:

Plan: A'BCD, E'BCF, and A'E'FD are parallelograms. Corresponding sides are congruent.By side-splitting theorem, BE:AE = BE':A'E'

Since BE' is congruent to CF and E'A' is congruent to FD, then BE:EA = CF:FD

Using the side-splitting theorem circumvents the previous issues about countability of the measure units in the component segments of the ratio.

Theorem 4.4: The AA similarity condition for triangles

If two angles on one triangle are congruent to two angles of another triangle, then the triangles are similar.PLAN: We are give two triangles with two angles congruent. If we can show that one triangle can be embedded in the other by matching one of the given angles, then we can use the

side-splitting theorem.Open GSP file.

Now Solve This 4.3Prove a proportion embedded in the proof of Theorem 4.4 . . .

Theorem 4.5: the SSS Similarity Condition (Converse of the AA similarity condition)

Theorem 4.6: The SAS Similarity ConditionGiven two triangles, if two pairs of corresponding sides are proportional and the included angles are congruent, then the triangles are similar.

Now Solve This 4.4Prove Theorem 4.6

Example 4.2The Harmonic MeanThe formula for the length of EG -- the line segment parallel to the bases of the trapezoid through the intersection of the diagonals -- is the harmonic mean of a and b. When we considered M and N as the midpoints of the sides of the trapezoid, the length of MN was the arithmetic mean of a and b.

Check this out.The animation suggests the theorem

Theorem: If a ≥ b > 0, then

AM ≥ HM with equality iff a = bProve this algebraically. That is

Open GSP file for Example 4.3.The proof above uses the similarity of triangles ADE and ABC. What if you had used the similarity of triangles ABC and EFB?EXTRA: Given a triangle ABC, CONSTRUCT the rhombus CDEF such that the rhombus shares the vertex at C with the triangle, D is on AC, E is on AB, and F is on BC.

If you want it, a construction is shown

HERE