Exercise 1.1 · Q12: Similar triangles

Basics you need

Solution

1. Say exactly what "similar" means

   For triangles $ABC$ and $DEF$ (vertices matched $A \leftrightarrow D$⁠, $B \leftrightarrow E$⁠, $C \leftrightarrow F$⁠),

   $$\begin{gathered} ABC \sim DEF \\ \iff \text{matching angles are equal and}\\ \frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA} = k \text{ for some } k > 0. \end{gathered}$$

   So $DEF$ is $ABC$ with every length multiplied by $k$⁠. The number $k$ is the scale factor from $ABC$ to $DEF$⁠.

   The words "is similar to" have to be turned into something we can calculate with before we can prove anything. The scale factor $k$ does this: each of the three properties becomes a short statement about $k$⁠.
   Similarity means "same shape, possibly different size". Equal angles fix the shape; the common ratio $k$ records the change of size.

2. Reflexive: is every triangle similar to itself?

   Take any triangle $T = ABC$ and match each vertex with itself. The matching angles are equal (each angle equals itself), and every ratio is

   $$\frac{AB}{AB} = \frac{BC}{BC} = \frac{CA}{CA} = 1 > 0.$$

   So $T \sim T$ with scale factor $k = 1$⁠, that is $(T,T)\in R$⁠. Since $T$ was arbitrary, $R$ is reflexive.

   Reflexivity must hold for every triangle, so we take an arbitrary one. Matching each vertex to itself is always possible, and dividing a length by itself always gives 1, whatever the triangle.
   A scale factor of 1 means "no change of size". Lengths are positive, so $\frac{AB}{AB}$ is defined and equals 1.

3. Symmetric: if $T_1 \sim T_2$⁠, is $T_2 \sim T_1$⁠?

   Suppose $T_1 = ABC \sim T_2 = DEF$ with scale factor $k > 0$⁠: matching angles equal and

   $$\frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA} = k.$$

   Use the same matching in reverse ($D \leftrightarrow A$⁠, $E \leftrightarrow B$⁠, $F \leftrightarrow C$⁠). The matching angles are still equal, and taking reciprocals of the ratios gives

   $$\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} = \frac1k > 0.$$

   So $T_2 \sim T_1$ with scale factor $\frac1k$⁠, that is $(T_2,T_1)\in R$⁠. Hence $R$ is symmetric.

   If $T_2$ is $T_1$ enlarged by $k$⁠, then $T_1$ is $T_2$ shrunk by the same amount, that is scaled by $\frac1k$⁠. Taking reciprocals is allowed because no side has length 0 and $k \ne 0$⁠. Equal angles stay equal whichever triangle is named first.
   If $\frac{p}{q} = k$ with $p, q, k > 0$⁠, then $\frac{q}{p} = \frac1k$⁠, and $\frac1k > 0$ (Basic 3).

4. Transitive: does $T_1 \sim T_2$ and $T_2 \sim T_3$ give $T_1 \sim T_3$⁠?

   Suppose $T_1 = ABC \sim T_2 = DEF$ with scale factor $k$⁠, and $T_2 = DEF \sim T_3 = GHI$ with scale factor $m$⁠, under the matchings $A\leftrightarrow D\leftrightarrow G$⁠, $B\leftrightarrow E\leftrightarrow H$⁠, $C\leftrightarrow F\leftrightarrow I$⁠. Then

   $$\begin{aligned} DE &= k\,AB, & EF &= k\,BC, & FD &= k\,CA,\\ GH &= m\,DE, & HI &= m\,EF, & IG &= m\,FD. \end{aligned}$$

   Substituting the first row into the second row:

   $$GH = mk\,AB, \qquad HI = mk\,BC, \qquad IG = mk\,CA.$$

   So $\dfrac{GH}{AB} = \dfrac{HI}{BC} = \dfrac{IG}{CA} = km > 0$⁠. The angles also match: $\angle A = \angle D = \angle G$⁠, and likewise for the other two pairs. Hence $T_1 \sim T_3$ with scale factor $km$⁠, that is $(T_1,T_3)\in R$⁠, and $R$ is transitive.

   The statement we want does not mention $T_2$⁠, so we eliminate it: the sides of $T_2$ appear in both sets of equations, and substituting removes them. Geometrically, enlarging by $k$ and then by $m$ is a single enlargement by $km$⁠. We use two different letters because the two scale factors need not be equal.
   Equality is transitive ($\angle A = \angle D$ and $\angle D = \angle G$ give $\angle A = \angle G$⁠), and the product of two positive numbers is positive (Basic 3).

5. Conclude the first part

   $R$ is reflexive (Step 2), symmetric (Step 3) and transitive (Step 4), so by definition $R$ is an equivalence relation in $A$.

   An equivalence relation is defined as a relation with these three properties, so nothing more is needed. The textbook's Example 2 does the same for congruence, which is the special case of similarity with scale factor $k = 1$ (see Example 2).

6. Why we may compare sides in increasing order

   To test two particular triangles, we need to know which side matches which. Suppose a triangle with sides $a \le b \le c$ is similar to a triangle with sides $a' \le b' \le c'$⁠, with scale factor $k > 0$⁠. Every side of the second triangle is $k$ times a side of the first, and multiplying by a positive $k$ keeps the order:

   $$a \le b \le c \quad\Longrightarrow\quad ka \le kb \le kc.$$

   So the shortest side must match the shortest, the middle the middle, and the longest the longest, giving $\dfrac{a'}{a} = \dfrac{b'}{b} = \dfrac{c'}{c}$⁠. In other words: two triangles are similar exactly when their sides, each listed in increasing order, have three equal ratios. If these three ratios are not all equal, no other matching can work either.

   A triangle has six possible ways of matching its sides with another's. This step shows only one of them, "in increasing order", can ever give equal ratios, so we only need to check that one. This turns "are they similar?" into one short calculation per pair.
   Multiplying both sides of an inequality by a positive number keeps the inequality the same way round (Basic 3).

7. Test the three pairs

   The sides are already listed in increasing order: $T_1 : 3, 4, 5$⁠; $T_2 : 5, 12, 13$⁠; $T_3 : 6, 8, 10$⁠. (All three are right triangles: $3^2+4^2 = 25 = 5^2$⁠, $5^2+12^2 = 169 = 13^2$⁠, $6^2+8^2 = 100 = 10^2$⁠.)

   Pair	Ratios of sides (in order)	All equal?	Related?
   $T_1, T_3$	$\frac63 = 2,\ \frac84 = 2,\ \frac{10}{5} = 2$	yes, $k = 2$	yes
   $T_1, T_2$	$\frac53,\ \frac{12}{4} = 3,\ \frac{13}{5}$	no	no
   $T_2, T_3$	$\frac65,\ \frac{8}{12} = \frac23,\ \frac{10}{13}$	no	no

   So $T_1 \sim T_3$ (scale factor 2, and by symmetry $T_3 \sim T_1$ with factor $\frac12$⁠), while $T_2$ is similar to neither. In full, restricted to these three triangles,

   $$\begin{gathered} R \text{ contains } (T_1,T_1),\ (T_2,T_2),\ (T_3,T_3),\\ (T_1,T_3),\ (T_3,T_1), \text{ and no other pairs.} \end{gathered}$$

   The angles confirm this. In a right triangle the smallest angle $\theta$ has $\sin\theta = \frac{\text{shortest side}}{\text{hypotenuse}}$⁠. For $T_1$ and $T_3$⁠, $\sin\theta = \frac35 = \frac{6}{10}$⁠, so $\theta \approx 36.87^\circ$ in both. For $T_2$⁠, $\sin\theta = \frac{5}{13}$⁠, so $\theta \approx 22.62^\circ$⁠. $T_2$ has a different shape.

   Each pair needs its own check because similarity is a statement about two specific triangles. For the unrelated pairs a single mismatch of ratios is enough, by Step 6. Being right-angled is not enough for similarity: all three triangles have a $90^\circ$ angle, but similarity needs all angles equal.
   Only the ordered matching can work (Step 6), so unequal ordered ratios prove the triangles are not similar. Basics: proving and disproving

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