Lesson · Similar triangles
Similar triangles
Congruence copies a triangle; similarity rescales it. Two similar triangles agree in all three angles, while every length of one is the same multiple \(k\) of the matching length of the other - so a single similarity is worth three angle equalities and a whole proportion of sides. Most of this chapter is learning to see where such pairs hide.
Two criteria
Definition
Triangles \(ABC\) and \(A'B'C'\) are similar, written \(ABC \sim A'B'C'\), when \(\angle A = \angle A'\), \(\angle B = \angle B'\), \(\angle C = \angle C'\) and
the ratio of similarity. The order of the letters records which vertex matches which.
Theorem
(AA) If two angles of one triangle equal two angles of another, the triangles are similar. (The third pair is then equal for free, since the angles of each triangle sum to \(180^\circ\).)
Theorem
(SAS-ratio) If \(\angle A = \angle A'\) and the sides enclosing these angles are proportional, \(A'B' : AB = A'C' : AC\), then \(ABC \sim A'B'C'\).
AA is the workhorse, because equal angles are cheap: vertical angles, base angles of an isosceles triangle, alternate angles at parallels, the two halves of a bisected angle. Two of them settle the shape completely. SAS-ratio takes over when the hypothesis already speaks in ratios.
Warning
Read every ratio off the correspondence of vertices, never off the picture. And remember that one proportion can be regrouped: \(DA : DB = DK : DC\) and \(DA : DK = DB : DC\) say the same thing, since both cross-multiply to \(DA \cdot DC = DB \cdot DK\). Swapping the means like this is often the one step between the proportion you were given and the pair of sides your criterion needs.
Where similar triangles hide
Theorem
(Intercept theorem) A line parallel to the side \(BC\) of a triangle \(ABC\), meeting \(AB\) at \(M\) and \(AC\) at \(N\), cuts the two sides in equal ratios: \(AM : AB = AN : AC = MN : BC\). Conversely, if \(AM : AB = AN : AC\), then \(MN \parallel BC\).
This is not a new fact but similarity in disguise: the parallel makes the corresponding angles at \(M\) and \(B\) and at \(N\) and \(C\) equal, so \(AMN \sim ABC\) by AA, and the equal ratios are the proportion of sides. The midline theorem of the previous course is the case \(k = \tfrac12\).
The second hiding place is the right triangle cut by the altitude from its right angle. The foot \(D\) splits the hypotenuse into \(p = AD\) and \(q = DB\), and all three triangles in the figure - the two small ones and the whole - carry the same pair of acute angles, so \(ADC \sim CDB \sim ACB\). Three similar triangles give three proportions, and each proportion says that one segment is a geometric mean: a length \(x\) with \(x^2 = uv\).
Theorem
In a right triangle with legs \(a\), \(b\), hypotenuse \(c\) and altitude \(h\) to the hypotenuse, whose foot cuts the hypotenuse into \(p\) (adjacent to \(b\)) and \(q\) (adjacent to \(a\)),
All three relations are one similarity each: \(ADC \sim CDB\) gives \(p : h = h : q\), and pairing each small triangle with the whole gives the other two. Adding the last two even recovers Pythagoras: \(a^2 + b^2 = (p + q)\,c = c^2\).
The square law
Principle
If two figures are similar with ratio \(k\), their areas are in the ratio \(k^2\). For triangles this is immediate: base and height both scale by \(k\), so the area scales by \(k \cdot k\).
Example
An equilateral triangle of side \(n\) is the unit equilateral triangle magnified \(n\) times, so its area is \(n^2\) unit triangles - visible directly in the triangular lattice, where side \(2\) covers \(4\) cells and side \(3\) covers \(9\). Whenever a picture is drawn on such a lattice, every ratio of areas becomes a ratio of integers before any surd can appear.
The square is what makes the law easy to misuse: doubling every length quadruples the area, and a similarity ratio read off carelessly enters the answer twice. Which is exactly the point of the warm-up.
Warm-up
Example
A line parallel to the side \(BC\) of a triangle \(ABC\) meets \(AB\) at \(M\) and \(AC\) at \(N\), so that \(AM : MB = 2 : 3\). The triangle \(AMN\) has area \(8\). Find the area of the triangle \(ABC\).
Which two triangles are similar here?
\(MN \parallel BC\) makes the corresponding angles at \(M\) and \(B\) equal, and the angle at \(A\) is shared, so \(AMN \sim ABC\) by AA.
The ratio is not \(2 : 3\).
The similarity compares \(AM\) with the whole side \(AB = AM + MB\), so \(k = \tfrac{AM}{AB} = \tfrac{2}{5}\). Taking the printed ratio \(2 : 3\) for the similarity ratio is the standard slip.
Now apply the square law.
Areas scale by \(k^2 = \tfrac{4}{25}\), so the area of \(ABC\) is \(8 \cdot \tfrac{25}{4} = 50\).