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) the coefficient of similitude, similarity coefficient, or dilution factor.

Now Solve This 4.1

1. 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.


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,


Open GSP file.

Proof: The proof presented in the text assumes that AB and BC can each be parsed into some number of segments of the same length. If each of those segments on AB and BC are of length x with AB having i of them and BC having j of them, then we can write AB = ix and BC = jx. That is, AC is covered with i + j segments each of length x.

Parallel projection preserves congruence of 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. So

The ratios of the segment lengths have each been shown to be equal to the ration of number of congruent segments of some common measure making up each segment. The argument assumes that the segments are commensurable (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 File

Triangle 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 4.2

1. State and Prove the converse of the corollary to the side-splitting theorem

2. 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 4.3

Prove 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 Condition

Given two triangles, if two pairs of corresponding sides are proportional and the included angles are congruent, then the triangles are similar.


Now Solve This 4.4

Prove Theorem 4.6




Example 4.2 The Harmonic Mean

The 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 = b

Prove this algebraically. That is





GSP file?


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



PROBLEM SET 4.1 (19 Problems)