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3 Similarity theorems
3.1. The AA rule
Theorem (AA). Two triangles are similar if two angles of one equal two angles of the other.
Proof. We can label the triangles 4 ABC, 4 A′B′C′^ so that ∠A = ∠A′^ and ∠B = ∠B′^. Suppose that c′^ = |A′B′| = k|AB| = kc with k > 0. If we knew that |B′C′| were equal to k|BC| then the two triangles would be similar by B5. In any case, by B1, we know
that there exists a point C′′^ on
B′C′^ for which |B′C′′| = k|BC| :
A B
C
A' B'
C'
C''
Triangles 4 ABC , 4 A′B′C′′^ now share proportional lengths adjacent to a common angle ∠B′^ = ∠B. They are therefore similar by B5 and as a consequence,
∠B′A′C′′^ = ∠A.
By hypothesis, this angle also equals ∠A′. The rays
A′C′,
AC′′^ must now coincide
since (by B3) they are determined by their angles with
A′B′. Therefore
C′^ = C′′
(
AC′^ can’t meet
B′C′^ in more than one point), and 4 ABC is similar to 4 A′B′C′.
3.2. Isosceles triangles
Recall that two triangles 4 ABC , 4 A′B′C′^ are said to be similar if corresponding an- gles are equal, and corresponding sides are proportional. The notion “corresponding” involves the bijection A 7 → A′, B 7 → B′, C 7 → C′,
assuming the vertices are ordered consistently.
Theorem. If two sides of a triangle are equal then the angles opposite these sides are equal.
Proof. Suppose that b = a. The idea is to re-label the vertices according to the scheme
A′^ = B, B′^ = A, C′^ = C,
so that α′^ = β and β′^ = α. We can now legitimately apply B5 with k = 1 to 4 ABC on the one hand, and 4 A′B′C′^ = 4 BAC, on the other, considered as two different triangles. They have a common angle ∠C = ∠C′, whose adjacent sides are propor- tional with k = 1 since a = b = a′^ and b = a = b′^. The SAS rule allows us to conclude that 4 ABC and 4 BCA are similar (indeed, congruent), and in particular ∠A = ∠A′ and ∠B = ∠B′. Thus, ∠A = ∠B or α = β.
A triangle satisfying the hypothesis of Theorem 1 is called isosceles.
Theorem. If two angles of a triangle are equal then the opposite sides are equal in length.
Proof. This result is the converse of Theorem 1, and is proved by applying AA in place of SAS.
3.3. The SSS rule
This section is devoted to establishing another known criterion for similarity. The proof is longer and is based on isosceles triangles.
Theorem (SSS). Two triangles 4 ABC , 4 A′B′C′^ are similar if their respective sides are proportional.
In symbols, this means
a′^ = ka, b′^ = kb, c′^ = kc for some k > 0 ⇒ α = α′, β = β′, γ = γ′.
Proof. We begin by constructing a triangle 4 A′B′C′′^ such that ∠A′B′C′′^ = ∠ABC and |B′C′′| = |B′C′| = k|BC|. The diagram assumes that β = ∠ABC < π/ 2 , but the following proof works without this assumption.
A B
C
A' B'
C'
C''
By construction, 4 ABC and 4 A′B′C′′^ share a common angle, and the sides adjacent to this angle are proportional. If follows, by the SAS rule, that they similar, and this tells us that ∠A′C′′B′^ = ∠ACB and |A′C′′| = k|AC|.
The middle triangle has all its lengths half as big as those of 4 ABC. So applying SSS (with k = 12 ) we deduce that 4 LM N ∼ 4ABC. The vertices are listed in the correct order, so it is ∠LN M that corresponds to ∠C.
We know (by B3) that the straight angle ∠AN B is the sum α+β +γ of the three angles shown, which (by B4) must equal π radians.
If a triangle has three equal sides it is called equilateral. By SSS, all its angles are equal.
Corollary. An equilateral triangle has all its angles equal to π/ 3 , and any two are similar.
4 Pythagoras’ theorem
In a triangle ABC with a right angle at C, the (length c of the) side opposite the right angle is called the hypotenuse. The following fundamental theorem lies at the heart of algebra, geometry and calculus.
Theorem (Pythagoras). If ∠BCA = π/ 2 then a^2 + b^2 = c^2.
Here is a “binomial proof” that relies on the concept of area (which we shall not how- ever develop in this course). Four congruent right-angled triangles are arranged cycli- cally around the inside of a square of length a + b :
a
b
b
c
(a + b)^2 = a^2 + 2ab + b^2
(a + b)^2 = c^2 + 4 · 12 ab
4.1. Two more proofs
Proof using B5. Suppose that 4 ABC has lengths a, b, c, and a right angle at C. Extend (or diminish) the segments CA, CB to CA′, CB′^ by a factor b so as to form the green triangle. Then scale by a factor a so as to construct the orange triangle 4 CB′′B′^ with
the ray
CB′^ bisecting the straight angle ∠A′CB′′. By B5 (SAS), the green and orange triangles have hypotenuses of length bc and ca respectively.
ab
b^2 a^2
bc ac
A' C
B'
B''
The big triangle now has angles ∠A′^ , ∠B′′^ in common with each of the smaller ones, and is therefore similar to 4 ABC by the AA rule. From the sides adjacent to B′^ , the factor of proportionality must be k = c, and it follows that
b^2 + a^2 = c · c,
by the definition of similarity.
Here is a variant of this argument in which we construct a similar diagram but work- ing within the given triangle ABC. It involves dropping a perpendicular from C to AB so that the two angles at P are π/ 2. It follows from the AA rule that the two smaller triangles are both similar to 4 ABC, and so
d a
a c
c − d b
b c
where d = |P B|. Then cd = a^2 , c(c − d) = b^2 ,
again giving c^2 = a^2 + b^2.
A C
B
P d
The problem with this proof is that as yet we have no direct way of constructing the perpendicular CP. We could fix P by requiring that |P B| = a^2 /c (as was done to get the computer to draw the picture!). In this case ∠CP B will be π/ 2 , but then the proof is very similar to the one we have already given above.
