NOTES Properties of Similarity and Similarity Criteria G.SRT.3

G.SRT.3
Use the properties of similarity transformations to establish the AA criterion for two triangles to be similar. / This objective has been included in geometry for many years – developing the criteria for similarity. While it only specifically mentions AA, this is the appropriate place to also prove that SAS and SSS establish similarity. /
The criterion for similarity should be founded in similarity transformations.
(1) The student will be able to prove two triangles to be similar using the minimum requirements of AA, SAS and SSS.
(2) The student will be able to use the properties of similarity transformations to establish the AA, SAS and SSS criterion for two triangles to be similar. / The main area of focus here is to demonstrate how if two triangles have two angles known there is a sequence of similarity transformations that will map one triangle onto another. Once AA is done, SAS and SSS follow quite easily. / 1 – Use transformations to establish the three similarity criterion, AA, SSS and SAS.
2 – Link dilations and dilation properties to the similar triangles as often as you can. It will help to link the old concepts with the new ones.

NACS g.srt.3 PropertiesSimilarities Page 3 of 74/28/2014

NOTES Properties of Similarity and Similarity Criteria G.SRT.3

CONCEPT 1 – Use the properties of similarity transformations to establish the AA criterion for two triangles to be similar.

When studying congruent triangles in G.CO.8 some minimum requirements were established that would guarantee congruence through a single or sequence of isometric transformations. It was found that SSS, SAS, ASA, AAS and HL (and some special cases of ASS) were enough information to always establish congruence between two triangles. In a likewise manner, it is necessary to find the minimum requirements in two triangles to establish similarity. In order to do this, use a single or sequence of similarity transformations that would map one triangle onto the other.

Begin by seeing if knowing two congruent corresponding angles (AA) is a criterion for similarity.

Given: ÐA @ ÐD and ÐB @ ÐE
Prove: DABC ~ DDEF
To prove two triangles to be similar to find a sequence of similarity transformations that map DABC on to DDEF.
First dilate DABC by the scale factor, k = , to get DA’B’C’.
DABC ~ DA’B’C’ because dilation is a similarity transformation and using the properties of a dilation it is known that:
ÐA @ ÐA’ and ÐB @ ÐB’ and that
A’B’ = k · AB = (AB) = DE (A’B’ = DE) /
ÐA @ ÐA’ & ÐA @ ÐD and ÐB @ ÐB’ & ÐB’ @ ÐE so using the transitive property
ÐA’ @ ÐD and ÐB’ @ ÐE.
and A’B’ = DE due to the dilation.
Thus DA’B’C’ @ DDEF by ASA.
Therefore DABC ~ DDEF because DABC was mapped onto DDEF using only similarity transformations.
AA is a similarity criterion.

CONCEPT 2 – Use the properties of similarity transformations to establish the SAS criterion for two triangles to be similar.

Now investigate if knowing two corresponding proportional sides and the included corresponding congruent angle (SAS) is enough for establishing similarity. Just as it was important in congruence the angle must be the included angle – meaning that it is the one between the two sides.

Given: ÐA @ ÐD and
Prove: DABC ~ DDEF
To prove two triangles to be similar find a sequence of similarity transformations that map DABC on to DDEF. /
First dilate DABC by scale factor, k to get DA’B’C’.
A’B’ = AB · k = (AB) = DE (A’B’ = DE)
A’C’ = AC · k = (AC) = (AC) = DF (A’C’ = DF)
DABC ~ DA’B’C’ because of the similarity transformation, which means that:
ÐA @ ÐA’ /
ÐA @ ÐA’ & ÐA @ ÐD so using the transitive property, ÐA’ @ ÐD and A’B’ = DE and A’C’ = DF are equal due to the dilation.
Thus DA’B’C’ @ DDEF by SAS.
DABC ~ DDEF because DABC was mapped onto DDEF using only similarity transformations.
Thus SAS is a similarity criterion.

CONCEPT 3 – Use the properties of similarity transformations to establish the SSS criterion for two triangles to be similar.

Given:
Prove: DABC ~ DDEF
To prove two triangles to be similar find a sequence of similarity transformations that map DABC on to DDEF. /
First dilate DABC by scale factor, k to get DA’B’C’.
A’B’ = AB · k = (AB) = DE A’B’ = DE
A’C’ = AC · k = (AC) = (AC) = DF A’C’ = DF
B’C’ = BC · k = (BC) = (BC) = EF B’C’ = EF /
Thus DA’B’C’ @ DDEF by SSS.
DABC ~ DDEF because DABC was mapped onto DDEF using only similarity transformations.
SSS is a similarity criterion.

NACS g.srt.3 PropertiesSimilarities Page 3 of 74/28/2014

NOTES Properties of Similarity and Similarity Criteria G.SRT.3

1. Which of the following would be the criterion for establishing similarity in the two triangles given?
A) AA B) SAS
C) SSS D) Not enough info or not similar /
2. Which of the following would be the criterion for establishing similarity in the two triangles given?
A) AA B) SAS
C) SSS D) Not enough info or not similar /

3. If two sides are proportional, then the two triangles must be similar. T or F

4. Are the following two polygons ALWAYS, SOMETIMES, OR NEVER similar?

a) Two Equilateral Triangles ALWAYS SOMETIMES NEVER

b) Two Rectangles ALWAYS SOMETIMES NEVER

c) Two Isosceles Triangles ALWAYS SOMETIMES NEVER

5. Are the following pairs of triangles similar? If they are, then name their similarity criteria. (SSS, SAS, AA)

a) Yes / No ______/ b) Yes / No ______/ c) Yes / No ______
d) Yes / No ______/ e) Yes / No ______/ f) Yes / No ______

6. Prove that if two triangles have two congruent corresponding angles, then they must be similar.

Given: ÐB @ ÐK and ÐC @ ÐL
Prove: DABC ~ DHKL /

Answer:

1)  A

2)  B

3)  F

4)  a) Always

b) Sometimes

c) Sometimes

5)  a) Yes, SAS

b) No

c) Yes, SAS

d) Yes, SAS or AA

e) Yes, SAS

f) Yes, AA

6) 

Given: ÐB @ ÐK and ÐC @ ÐL
Prove: DABC ~ DHKL

To prove two triangles to be similar first find a sequence of similarity transformations that map DABC on to DHKL.
First dilate DABC by scale factor, k = to get DA’B’C’.
DABC ~ DA’B’C’ because dilation is a similarity transformation and using the properties of a dilation: ÐB @ ÐB’ and ÐC @ ÐC’ and B’C’ = k · BC = (BC) = KL (B’C’ = KL)
ÐB @ ÐB’ & ÐB @ ÐK and ÐC @ ÐC’ & ÐC @ ÐL so using the transitive property ÐB’ @ ÐK and ÐC’ @ ÐL

and B’C’ = KL due to the dilation.

Thus DA’B’C’ @ DHKL by ASA.

And DABC ~ DHKL because DABC was mapped onto DHKL using only similarity transformations.

AA is a similarity criterion.

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