Flashcards in Chapter 3: Similarity Deck (18):
Chapter 3 notation:
-[ABC] = area of △ABC = bxh/2
-~ denotes Similarity
-△ABC ~ △DEF means ∠A==∠D, ∠B==∠E, ∠C==∠F and AB/DE = BC/EF = AC/DF = k (proportionality constant/magnification factor)
-k>1 => △ABC is larger than △DEF ; 0 △ABC is smaller than △DEF ; k=1 => △ABC == △DEF
Given parallel lines l and m and 2 △s with bases on m and opposite vertices on l, the ratio of areas of the triangles is the ratio of the lengths of their bases.
[ABC] / [DEF] = BC/EF
Thm 3.2.3 (Parallel Lines Preserve Ratios):
Suppose l, m, and n are parallel lines, met by transversals t, t' at A, B, C and A', B', C' resp. Then, AB/BC = A'B'/B'C'
Fact (Useful in proofs, like Lemma 3.3.6):
If P and Q are pts on AB, then
AP/PB = AQ/QB => P=Q
Lemma 3.3.6 (Partial Converse to Parallel Lines Preserve Ratios):
Let P and Q be pts on AB and AC, resp. If AP/PB = AQ/QC, then PQ||BC.
Parallel Lines Preserve Ratios is not necessarily reversible.
AB/CB = DE/EF does not mean BE||CF.
Thm 3.4.7 (Angle Bisector Thm):
Let D be a pt on side BC of △ABC
1. For D ∈ BC, AD is the internal bisector of ∠BAC iff AB/AC = DB/DC
2. For D ∉ BC, AD is the external bisector of ∠BAC iff AB/AC = DB/DC
Pf provided with aside.
Similar describes objects with the same shape, but possibly different sizes.
Thm 3.2.4 (Useful in proofs):
In △ABC, suppose DE||BC. If D and E are on lines AB and AC, resp, and D≠A, D≠B, D≠C, then △ABC ~ △ADE.
Both congruency and similarity are 'equivalence relations'. That is, they are both:
-reflexive: △ABC ~ (==) △ABC
-symmetric: △ABC ~ (==) △DEF iff △DEF ~ (==) △ABC
-transitive: If △ABC ~ (==) △DEF and △DEF ~ (==) △GHI then △ABC ~ (==) △GHI
-Congruency => Similarity
Thm 3.3.1 (AAA):
Two triangles are similar iff all 3 corresponding angles are congruent.
Note: Fails for quads.
Thm 3.3.5 (AA):
Two triangles are similar iff 2 corresponding angles are congruent (equivalent to Thm 3.3.1).
Thm 3.3.3 (sAs) (small s for side ratios as opposed to big S for congruency/same length):
If in △ABC and △DEF, AB/DE = AC/DF and ∠A = ∠D, then △ABC ~ △DEF.
Thm 3.3.4 (sss):
△ABC ~ △DEF iff AB/DE = BC/EF = AC/DF.
Thm 3.4.1 (Pythagoras Thm):
If the hypotenuse of a right triangle has length C, and the other 2 sides have length a and b, then c^2 = a^2 + b^2.
Thm 3.4.2 (Converse to Pythagoras):
In △ABC, if (AB)^2 = (BC)^2 + (AC)^2, then ∠C = 90.
Pf left as exercise.
A quad. ABCD is cyclic iff (AC)(BD) = (AB)(CD) + (AD)(BC).
Pf provided (long).