
# 6.1: Similar triangles


Two triangles $$A'B'C'$$ and $$ABC$$ are called similar (briefly $$\triangle A'B'C' \sim \triangle ABC$$) if (1) their sides are proportional; that is,

$A'B' = k\cdot AB, B'C' = k \cdot BC \text{ and } C'A' = k \cdot CA$

for some $$k > 0$$, and (2) the corresponding angles are equal up to sign:

$\begin{array} {l} {\measuredangle A'B'C' = \pm \measuredangle ABC} \\ {\measuredangle B'C'A' = \pm \measuredangle BCA} \\ {\measuredangle C'A'B' = \pm \measuredangle CAB} \end{array}$

#### Remarks

• According to Theorem 3.3.1, in the above three equalities, the signs can be assumed to be the same.

• If $$\triangle A'B'C' \sim \triangle ABC$$ with $$k = 1$$ in 6.1.1, then $$\triangle A'B'C' \cong \triangle ABC$$.

• Note that "$$\sim$$" is an equivalence relation. That is,
(i) $$\triangle ABC \sim \triangle ABC$$ for any $$\triangle ABC$$.
(ii) if $$\triangle A'B'C' \sim \triangle ABC$$, then
$\triangle ABC \sim \triangle A'B'C'$
(iii) If $$\triangle A''B''C'' \sim \triangle A'B'C'$$ and $$\triangle A'B'C' \sim \triangle ABC$$, then
$\triangle A''B''C'' \sim \triangle ABC$

Using the new notation "$$\sim$$", we can reformulate Axiom V:

Theorem $$\PageIndex{1}$$ Reformulation of Axiom V.

If for the two triangles $$\triangle ABC$$, $$\triangle AB'C'$$, and $$k > 0$$ we have $$B' \in [AB)$$, $$C' \in [AC)$$, $$AB' = k \cdot AB$$ and $$AC' = k \cdot AC$$, then $$\triangle ABC \sim \triangle AB'C'$$.

In other words, the Axiom V provides a condition which guarantees that two triangles are similar. Let us formulate three more such similarity conditions.

Theorem $$\PageIndex{2}$$ Similarity conditions

Two triangles $$\triangle ABC$$ and $$\triangle A'B'C'$$ are similar if one of the following conditions holds:

(SAS) For some constant $$k > 0$$ we have

$$AB = k \cdot A'B', AC = k \cdot A'C'$$

and $$\measuredangle BAC = \pm \measuredangle B'A'C'.$$

(AA) The triangle $$A'B'C'$$ is nondegenerate and

$$\measuredangle ABC = \pm \measuredangle A'B'C', \measuredangle BAC = \pm \measuredangle B'A'C'.$$

(SSS) For some constant $$k > 0$$ we have

$$AB = k \cdot A'B', AC = k \cdot A'C', CB = k \cdot C'B'.$$

Each of these conditions is proved by applying Axiom V with the SAS, ASA, and SSS congruence conditions respectively (see Axiom IV and the Theorem 4.2.1, Theorem 4.4.1).

Proof

Set $$k = \dfrac{AB}{A'B'}$$. Choose points $$B'' \in [A'B')$$ and $$C'' \in [A'C')$$, so that $$A'B'' = k \cdot A'B'$$ and $$A'C'' = k \cdot A'C'$$. By Axiom V, $$\triangle A'B'C' \sim \triangle A'B''C''$$.

Applying the SAS, ASA or SSS congruence condition, depending on the case, we get that $$\triangle A'B''C'' \cong \triangle ABC$$. Hence the result.

A bijection $$X \leftrightarrow X'$$ from a plane to itself is called angle preserving transformation if

$$\measuredangle ABC = \measuredangle A'B'C'$$

for any triangle $$ABC$$ and its image $$\triangle A'B'C'$$.

Exercise $$\PageIndex{1}$$

Show that any angle-preserving transformation of the plane multiplies all the distance by a fixed constant.

Hint

By the AA similarity condition, the transformation multiplies the sides of any nondegenerate triangle by some number which may depend on the triangle.

Note that for any two nondegenerate triangles that share one side this number is the same. Applying this observation to a chain of triangles leads to a solution.