Honors Geometry Unit 7 Review: Proportions and Similarity

Two polygons are similar when all corresponding angles are congruent and all corresponding sides are proportional. You need both. A 2-by-2 square and a 2-by-5 rectangle have matching 90° angles, but $\frac{2}{2} \neq \frac{2}{5}$, so they aren't similar.

The statement $\triangle ABC \sim \triangle DEF$ tells you the correspondence by letter order. $A$ matches $D$, $B$ matches $E$, and $C$ matches $F$, so $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$. That common ratio compares lengths in $ABC$ to corresponding lengths in $DEF$. The scale factor from $ABC$ to $DEF$ is its reciprocal.

Once you know $k$, perimeters scale by $k$ and areas scale by $k^2$. Area is two-dimensional, so the scale factor gets applied in two directions.

Worked example. $\triangle ABC \sim \triangle DEF$ with $AB = 6$, $DE = 9$, and $BC = 8$. Find $EF$. Set up $\frac{6}{9} = \frac{8}{EF}$ and cross multiply to get $6 \cdot EF = 72$, so $EF = 12$. The scale factor from $ABC$ to $DEF$ is $\frac{9}{6} = 1.5$. If $\triangle ABC$ has area 20, then $\triangle DEF$ has area $20 \cdot 1.5^2 = 45$.

Some families are always similar. Any two equilateral triangles, any two squares, and any two regular hexagons are similar, because their angles always match and their sides are all equal within each figure.

A ratio compares two quantities and can be written $a:b$, $\frac{a}{b}$, or "a to b." A proportion says two ratios are equal, as in $\frac{a}{b} = \frac{c}{d}$. By the Cross Products Property, if $\frac{a}{b} = \frac{c}{d}$ with $b, d \neq 0$, then $ad = bc$. That's how you solve for a missing term and how you check whether two ratios really are equal.

Watch whether a ratio is part-to-part or part-to-whole. If a bag has red and blue marbles in a $3:4$ ratio, the fraction that is red is $\frac{3}{7}$, not $\frac{3}{4}$.

When ratios compare lengths, use the same units on both sides. A map scale of $1:50{,}000$ with cities 4 cm apart gives $\frac{1}{50{,}000} = \frac{4}{d}$, so $d = 200{,}000$ cm, which is 2 km.

The geometric mean of positive numbers $a$ and $b$ is $\sqrt{ab}$, the value $x$ that makes $\frac{a}{x} = \frac{x}{b}$ true. For 4 and 9, it's $\sqrt{36} = 6$.

Worked example. Solve $\frac{x+1}{4} = \frac{9}{x+1}$, where $x+1$ is a length. Cross multiply to get $(x+1)^2 = 36$, so $x + 1 = 6$ or $x + 1 = -6$. A length can't be negative, so $x + 1 = 6$ and $x = 5$. Honors problems often hide a quadratic inside a proportion like this, so check each solution against the context.

In right $\triangle ABC$ with right angle at $C$, draw the altitude from $C$ to the hypotenuse $\overline{AB}$, meeting it at $D$. You now have three right triangles, $\triangle ACB$, $\triangle ADC$, and $\triangle CDB$. All three are similar by AA, because each smaller triangle has a right angle congruent to the original's right angle and shares an acute angle with the original.

Those similarities give two geometric mean rules. The altitude is the geometric mean of the two hypotenuse segments, so $CD^2 = AD \cdot DB$. Each leg is the geometric mean of the whole hypotenuse and the segment that touches that leg, so $AC^2 = AD \cdot AB$ and $BC^2 = DB \cdot AB$.

Worked example. The altitude splits the hypotenuse into $AD = 9$ and $DB = 16$, so $AB = 25$. The altitude is $CD = \sqrt{9 \cdot 16} = \sqrt{144} = 12$. Leg $AC$ touches segment $AD$, so $AC = \sqrt{9 \cdot 25} = \sqrt{225} = 15$. Leg $BC$ touches segment $DB$, so $BC = \sqrt{16 \cdot 25} = \sqrt{400} = 20$. The Pythagorean theorem confirms it, since $15^2 + 20^2 = 225 + 400 = 625 = 25^2$.

You can also run the rules backward. If $AC = 6$ and $AB = 9$, then $36 = AD \cdot 9$, so $AD = 4$ and $DB = 5$.

For triangles, you don't need every angle and side. AA needs two pairs of congruent angles. SAS needs two pairs of proportional sides with congruent included angles. SSS needs all three pairs of sides proportional. In proofs, AA is the usual route, and the angles typically come from parallel lines, vertical angles, or a shared angle at a common vertex.

The Triangle Proportionality Theorem (the side-splitter) says a line parallel to one side of a triangle cuts the other two sides proportionally. Its converse says that if a line cuts two sides proportionally, it's parallel to the third side. The Triangle Angle Bisector Theorem says an angle bisector divides the opposite side into segments proportional to the two adjacent sides.

Worked example. In $\triangle ABC$, $\overline{DE} \parallel \overline{BC}$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$. If $AD = 4$, $DB = 6$, and $AE = 5$, find $EC$. The side-splitter gives $\frac{4}{6} = \frac{5}{EC}$, so $4 \cdot EC = 30$ and $EC = 7.5$. To find $DE$ when $BC = 15$, switch to $\triangle ADE \sim \triangle ABC$. Since $\frac{DE}{15} = \frac{AD}{AB} = \frac{4}{10}$, $DE = 6$.

Indirect measurement uses the same logic. A 6 ft person casts a 4 ft shadow while a flagpole casts a 20 ft shadow. Assume the person and flagpole are vertical, the ground is level, and the sun's rays are parallel. The right triangles are similar by AA, so $\frac{6}{4} = \frac{h}{20}$ gives $h = 30$ ft.