Chapter 6: Similar Triangles
Learning Objectives
- Define similarity and identify corresponding parts using proper similarity notation
- Calculate the scale factor between similar figures and apply it to find missing sides
- Apply AA, SAS~, and SSS~ postulates/theorems to prove triangles are similar
- Use the Side-Splitter Theorem and Angle Bisector Theorem to solve proportion problems
- Set up and solve proportions from similarity statements to find unknown lengths
- Apply similar triangles to indirect measurement (shadow method, map scale)
6.1 Similar Polygons and Scale Factor
Two figures are similar (~) if they have the same shape but not necessarily the same size. One figure can be obtained from the other by a dilation (scaling), possibly combined with rigid motions. For polygons, similarity requires both congruent corresponding angles and proportional corresponding sides.
Definition: Similar Polygons
Polygon $ABCD\ldots \sim$ Polygon $EFGH\ldots$ if and only if:
- All corresponding angles are congruent: $\angle A \cong \angle E$, $\angle B \cong \angle F$, etc.
- All corresponding sides are proportional: $\dfrac{AB}{EF} = \dfrac{BC}{FG} = \dfrac{CD}{GH} = \cdots = k$
The constant ratio $k$ is called the scale factor from the first figure to the second.
Properties of Similar Polygons
- Corresponding angles are congruent.
- Corresponding sides are proportional with scale factor $k$.
- Ratio of perimeters $= k$.
- Ratio of areas $= k^2$.
For triangles, the similarity statement $\triangle ABC \sim \triangle DEF$ carries the following meaning:
- Angles: $\angle A \cong \angle D$, $\angle B \cong \angle E$, $\angle C \cong \angle F$
- Sides: $\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF} = k$
As with congruence, order matters — the first vertex of one triangle corresponds to the first vertex of the other.
Example 6.1 — Scale Factor and Corresponding Sides
$\triangle ABC \sim \triangle DEF$ with $AB = 6$, $BC = 9$, $CA = 12$, $DE = 4$. Find $EF$ and $FD$. Then find the scale factor and the ratio of areas.
Scale factor: $k = \dfrac{DE}{AB} = \dfrac{4}{6} = \dfrac{2}{3}$
Find $EF$: $\dfrac{EF}{BC} = k \Rightarrow EF = 9 \cdot \dfrac{2}{3} = \mathbf{6}$
Find $FD$: $\dfrac{FD}{CA} = k \Rightarrow FD = 12 \cdot \dfrac{2}{3} = \mathbf{8}$
Ratio of perimeters: $k = \dfrac{2}{3}$
Ratio of areas: $k^2 = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}$
Rectangle $ABCD \sim$ Rectangle $EFGH$ with $AB = 10$, $BC = 6$, $EF = 15$. Find $GH$ and the ratio of perimeters.
Show Answer
$GH$ corresponds to $BC$: $GH = 6 \cdot \dfrac{3}{2} = \mathbf{9}$.
Ratio of perimeters $= k = \dfrac{3}{2}$ (i.e., $EFGH$ is $\dfrac{3}{2}$ times the perimeter of $ABCD$).
Two similar triangles — $\triangle ABC$ (purple) and $\triangle DEF$ (blue) with scale factor $k = \frac{4}{3}$ from small to large.
Figure 6.1 — Similar Triangles with Scale Factor
6.2 AA, SAS, and SSS Similarity Postulates
Just as congruence has shortcuts (SSS, SAS, ASA, …), we do not need to verify all six pairs of corresponding parts to prove triangles similar. Three criteria are sufficient:
Triangle Similarity Postulates and Theorems
- AA (Angle-Angle): If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.
- SAS~ (Side-Angle-Side Similarity): If two sides of one triangle are proportional to two sides of another triangle, and the included angles are congruent, then the triangles are similar.
- SSS~ (Side-Side-Side Similarity): If all three pairs of corresponding sides of two triangles are proportional, then the triangles are similar.
Note: Similarity vs. Congruence Shortcuts
AA works for similarity because knowing two angles fixes the shape (the third angle is determined by the Triangle Angle Sum). There is no "AA" for congruence — you also need a side. Similarly, SSA does not work for either congruence or similarity in general.
Example 6.2 — AA Similarity
$\triangle PQR$ has $\angle P = 50°$ and $\angle Q = 70°$. $\triangle STU$ has $\angle S = 50°$ and $\angle T = 70°$. Are the triangles similar?
$\angle P = \angle S = 50°$ and $\angle Q = \angle T = 70°$. By AA, $\triangle PQR \sim \triangle STU$.
Note: $\angle R = \angle U = 180° - 50° - 70° = 60°$ (the third pair follows automatically).
Example 6.3 — SSS~ Similarity
$\triangle ABC$ has sides $6$, $8$, $10$. $\triangle DEF$ has sides $9$, $12$, $15$. Are the triangles similar?
Check ratios of corresponding sides (shortest to shortest, etc.):
$$\frac{6}{9} = \frac{2}{3}, \quad \frac{8}{12} = \frac{2}{3}, \quad \frac{10}{15} = \frac{2}{3}$$
All three ratios are equal, so by SSS~, $\triangle ABC \sim \triangle DEF$ with scale factor $k = \dfrac{2}{3}$.
Two triangles have two pairs of congruent angles. Without knowing any side lengths, can you conclude the triangles are similar? Which postulate applies?
Show Answer
AA Similarity — $\triangle PQR$ (purple) and $\triangle STU$ (blue) share two pairs of congruent angles, making them similar.
Figure 6.2 — AA Similarity: Two Triangles with Equal Angle Pairs
6.3 Using Similar Triangles: Proportional Sides
Once a similarity is established, we can set up proportions using corresponding sides and solve for unknown lengths by cross-multiplication. Two important theorems extend this idea inside a single triangle.
Side-Splitter Theorem
If a line is parallel to one side of a triangle and intersects the other two sides, then it divides those two sides proportionally.
In $\triangle ABC$, if $DE \parallel BC$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$, then:
$$\frac{AD}{DB} = \frac{AE}{EC}$$
Midsegment (special case): If $D$ and $E$ are midpoints, then $DE \parallel BC$ and $DE = \dfrac{1}{2}BC$.
Angle Bisector Theorem
An angle bisector of a triangle divides the opposite side in the ratio of the two adjacent sides.
If $\overline{AD}$ bisects $\angle A$ in $\triangle ABC$ and meets $\overline{BC}$ at $D$, then:
$$\frac{AB}{AC} = \frac{BD}{DC}$$
Example 6.4 — Side-Splitter Theorem
In $\triangle ABC$, $DE \parallel BC$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$. Given $AD = 4$, $DB = 6$, $AE = 5$, find $EC$.
By the Side-Splitter Theorem:
$$\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{4}{6} = \frac{5}{EC}$$
Cross-multiplying: $4 \cdot EC = 6 \cdot 5 = 30$, so $EC = \mathbf{7.5}$.
$\triangle ABC \sim \triangle DEF$ with $AB = 8$ and $DE = 12$. The perimeter of $\triangle ABC = 30$. Find the perimeter of $\triangle DEF$.
Show Answer
Ratio of perimeters $= k$, so: Perimeter of $\triangle DEF = 30 \cdot \dfrac{3}{2} = \mathbf{45}$.
AP Tip: On the AP exam, always write the similarity statement in the correct vertex order first, then set up ratios using the corresponding vertices. A ratio error from a mismatched similarity statement is one of the most common mistakes.
Side-Splitter Theorem — $DE$ (green, dashed) is parallel to $BC$ and divides sides $AB$ and $AC$ proportionally.
Figure 6.3 — Side-Splitter Theorem: Parallel Line Creates Proportional Segments
6.4 Indirect Measurement and Applications
Similar triangles provide a powerful method for measuring distances or heights that cannot be measured directly. Two common techniques are the shadow method and the mirror method.
Shadow Method
At the same time of day, the sun's rays are parallel. A person and a nearby object each cast shadows, forming two similar triangles (each with the same sun angle). Setting up the proportion:
$$\frac{\text{person's height}}{\text{person's shadow}} = \frac{\text{object's height}}{\text{object's shadow}}$$
Mirror Method
A mirror placed flat on the ground reflects the top of an object. The incident and reflected angles are equal, forming two similar triangles — one involving the observer and one involving the object.
Map Scale and Scale Drawings
A map scale states the ratio of map distance to actual distance. Setting up a proportion with the scale ratio allows conversion between map and real-world measurements.
Example 6.5 — Shadow Method
A person $5.5$ ft tall casts a $4$ ft shadow. At the same time, a tree casts a $32$ ft shadow. Find the height of the tree.
$$\frac{5.5}{4} = \frac{h}{32}$$
Cross-multiplying: $4h = 5.5 \times 32 = 176$, so $h = \mathbf{44}$ ft.
Example 6.6 — Map Scale
A map has scale $1$ inch $= 50$ miles. Two cities are $3.5$ inches apart on the map. Find the actual distance.
$$\frac{1}{50} = \frac{3.5}{d}$$
Cross-multiplying: $d = 3.5 \times 50 = \mathbf{175}$ miles.
A $6$-ft person casts a $9$-ft shadow. A nearby flagpole casts a $45$-ft shadow at the same time. How tall is the flagpole?
Show Answer
$9h = 6 \times 45 = 270$
$h = \mathbf{30}$ ft.
AP Tip: In geometry proofs, always state which similarity postulate or theorem (AA, SAS~, SSS~) justifies your conclusion. Simply saying "the triangles look similar" is not sufficient for full credit.
Practice Problems
$\triangle RST \sim \triangle XYZ$ with $RS = 15$, $ST = 20$, $RT = 25$, $XY = 9$. Find $YZ$, $XZ$, the scale factor from $\triangle RST$ to $\triangle XYZ$, and the ratio of their areas.
Show Solution
$YZ = 20 \cdot \dfrac{3}{5} = \mathbf{12}$; $XZ = 25 \cdot \dfrac{3}{5} = \mathbf{15}$.
Ratio of areas $= k^2 = \left(\dfrac{3}{5}\right)^2 = \dfrac{9}{25}$.
$\triangle ABC$ has $\angle A = 40°$, $\angle B = 65°$. $\triangle DEF$ has $\angle D = 40°$, $\angle F = 75°$. Are the triangles similar? If so, write the correct similarity statement.
Show Solution
Matching pairs: $\angle A = \angle D = 40°$, $\angle B = \angle E = 65°$, $\angle C = \angle F = 75°$.
Yes, by AA: $\triangle ABC \sim \triangle DEF$.
Two triangles have sides $5$, $6$, $7$ and $10$, $12$, $14$. Are they similar? By which theorem?
Show Solution
All three ratios are equal, so yes, the triangles are similar by SSS~ with scale factor $k = \dfrac{1}{2}$.
In $\triangle PQR$, $ST \parallel QR$ with $S$ on $\overline{PQ}$ and $T$ on $\overline{PR}$. Given $PS = 6$, $SQ = 4$, $PT = 9$, find $TR$.
Show Solution
$\dfrac{6}{4} = \dfrac{9}{TR} \Rightarrow 6 \cdot TR = 36 \Rightarrow TR = \mathbf{6}$.
A $12$-ft ladder leans against a wall forming a triangle with the ground. A $6$-ft ladder makes a similar triangle. If the base of the $12$-ft ladder is $5$ ft from the wall, how far is the base of the $6$-ft ladder from the wall?
Show Solution
All corresponding lengths scale by $k$, so the base distance $= 5 \cdot \dfrac{1}{2} = \mathbf{2.5}$ ft.
$\triangle ABC \sim \triangle DEF$ with $AB = x + 2$, $DE = 9$, $BC = 8$, $EF = 6$. Find $x$.
Show Solution
$\dfrac{x+2}{9} = \dfrac{8}{6} = \dfrac{4}{3}$.
$3(x + 2) = 36 \Rightarrow x + 2 = 12 \Rightarrow x = \mathbf{10}$.
The angle bisector from $A$ in $\triangle ABC$ meets $\overline{BC}$ at $D$. Given $AB = 8$, $AC = 12$, $BC = 15$, find $BD$ and $DC$.
Show Solution
Let $BD = 2t$ and $DC = 3t$. Then $BD + DC = BC$: $2t + 3t = 15 \Rightarrow t = 3$.
$BD = \mathbf{6}$, $DC = \mathbf{9}$.
AP Problem: Prove that the midsegment of a triangle is parallel to the third side and half its length. Use coordinates: $A = (0, 2a)$, $B = (0, 0)$, $C = (2b, 0)$. Let $M$ be the midpoint of $\overline{AB}$ and $N$ be the midpoint of $\overline{AC}$. Show $MN \parallel BC$ and $MN = \tfrac{1}{2}BC$.
Show Solution
$M = \text{midpoint of } AB = \left(\dfrac{0+0}{2}, \dfrac{2a+0}{2}\right) = (0, a)$.
$N = \text{midpoint of } AC = \left(\dfrac{0+2b}{2}, \dfrac{2a+0}{2}\right) = (b, a)$.
Show $MN \parallel BC$:
Slope of $MN = \dfrac{a - a}{b - 0} = 0$. Slope of $BC = \dfrac{0 - 0}{2b - 0} = 0$. Both slopes are $0$, so $MN \parallel BC$. $\square$
Show $MN = \tfrac{1}{2}BC$:
$MN = \sqrt{(b-0)^2 + (a-a)^2} = b$.
$BC = \sqrt{(2b-0)^2 + 0^2} = 2b$.
Therefore $MN = b = \dfrac{1}{2}(2b) = \dfrac{1}{2}BC$. $\square$
📋 Chapter Summary
Similarity Criteria
If two angles of one triangle are congruent to two angles of another, the triangles are similar. (Third angle automatically follows.)
Two pairs of sides are proportional AND the included angles are congruent.
All three pairs of corresponding sides are proportional: $\dfrac{a}{a'} = \dfrac{b}{b'} = \dfrac{c}{c'}$.
The ratio of corresponding sides: $k = \dfrac{\text{image side}}{\text{original side}}$. Perimeters scale by $k$; areas scale by $k^2$.
Geometric Mean
In a right triangle, the altitude to the hypotenuse creates two smaller triangles similar to each other and to the original.
The geometric mean of $a$ and $b$ is $\sqrt{ab}$. Appears in right-triangle altitude relationships: $h^2 = pq$ where $p$, $q$ are the two segments of the hypotenuse.