Similar Triangles

Two triangles are similar if they have the same shape but potentially different sizes. Formally, $\triangle ABC \sim \triangle DEF$ implies two conditions: (1) all corresponding angles are equal, and (2) corresponding side lengths are proportional by a constant scale factor $k$, where $\frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA} = k$.

Why it works. Similarity is essentially a dilation. If you apply a scale factor $k$ to every side of a triangle, the angles remain invariant because the relative slopes of the lines do not change. If $k=1$, the triangles are congruent.

SAT use. The SAT often hides similarity by rotating or reflecting the triangles. Never assume sides are corresponding just because they are in the same visual position; always map them by their opposite angles.

Worked example. If $\triangle ABC$ has sides $3, 4, 5$ and $\triangle DEF$ has sides $6, 8, 10$, the scale factor $k=2$. The ratio of any corresponding sides is always $2$.

You do not need to check all three sides or all three angles to prove similarity. The Angle-Angle (AA) Criterion states that if two angles of one triangle are congruent to two angles of another, the triangles are similar.

Why it works. Because the sum of interior angles in any triangle is $180^\circ$, if two angles match, the third angle must match ($180 - (a+b)$). Once all angles match, the sides are forced into a constant ratio.

SAT use. Look for vertical angles, alternate interior angles (parallel lines), or shared angles. These are the 'hidden' clues that force AA similarity.

Worked example. In $\triangle ABC$ and $\triangle XYZ$, if $\angle A = \angle X$ and $\angle B = \angle Y$, then $\triangle ABC \sim \triangle XYZ$ by AA. Any missing side can now be found via proportion.

The order of letters in $\triangle ABC \sim \triangle DEF$ is a mathematical contract. It dictates that $\angle A = \angle D$, $\angle B = \angle E$, and $\angle C = \angle F$.

The Trap. The SAT frequently draws triangles in different orientations. A student might see a 'base' of $10$ in one triangle and a 'base' of $5$ in another and assume they correspond. If the angles don't match those positions, the ratio is wrong.

SAT use. Always identify the angles first. If $\angle A$ is the smallest angle in $\triangle ABC$, it must correspond to the smallest angle in $\triangle DEF$.

Worked example. If $\triangle ABC$ has angles $30^\circ, 60^\circ, 90^\circ$ and $\triangle PQR$ has angles $90^\circ, 30^\circ, 60^\circ$, you must match the side opposite the $30^\circ$ in the first to the side opposite the $30^\circ$ in the second.

Once similarity is established, the ratio of corresponding sides is constant: $\frac{side_1}{side_2} = \frac{side_3}{side_4}$.

The Method. Set up the proportion carefully: $\frac{Large_{side}}{Small_{side}} = \frac{Large_{side}}{Small_{side}}$. Cross-multiply to solve for the variable.

SAT use. The SAT often uses variables like $x$ or $y$ in these sides. You must be comfortable with algebraic proportions: $\frac{x}{x+2} = \frac{4}{6}$.

Worked example. $\triangle ABC \sim \triangle DEF$. $AB=4, BC=6, DE=x, EF=9$. The ratio is $\frac{BC}{EF} = \frac{6}{9} = \frac{2}{3}$. Thus, $\frac{AB}{DE} = \frac{4}{x} = \frac{2}{3}$. Solving gives $2x = 12$, so $x=6$.

If a line is drawn parallel to one side of a triangle intersecting the other two sides, it creates a smaller triangle similar to the original. This is the Side-Splitter Theorem.

Why it works. The small triangle shares the top angle with the large triangle, and the corresponding angles are equal due to the parallel lines (AA similarity).

SAT use. This often appears as a triangle with a line segment inside. The ratio of the segments on the sides is proportional: $\frac{AD}{DB} = \frac{AE}{EC}$.

Worked example. In $\triangle ABC$, $DE \parallel BC$. If $AD=2, DB=3, AE=x, EC=6$, then $\frac{2}{3} = \frac{x}{6}$. Thus $3x = 12$, so $x=4$.

When a triangle is nested inside another, the most common trap is failing to recognize the shared angle at the vertex.

The Setup. If $\overline{DE} \parallel \overline{BC}$ inside $\triangle ABC$, then $\triangle ADE \sim \triangle ABC$. The shared angle is $\angle A$.

SAT use. The SAT will often give you the lengths of the segments of the sides (e.g., $AD=4, DB=2$). You must use the full length of the side ($AB=6$) for your similarity ratio, not just the segment ($DB=2$).

Worked example. If $AD=3, DB=1$, the ratio of the small triangle to the large triangle is $\frac{3}{3+1} = \frac{3}{4}$.

When an altitude is drawn from the right angle to the hypotenuse of a right triangle, it creates three similar triangles: the original large one and two smaller ones nested inside.

The Logic. All three triangles share angles. The altitude $h$ is the geometric mean of the two segments of the hypotenuse: $\frac{x}{h} = \frac{h}{y} \Rightarrow h^2 = xy$.

SAT use. Look for a right triangle with an altitude. You can solve for any missing side using the ratios of the similar triangles.

Worked example. If the altitude splits the hypotenuse into segments of $4$ and $9$, the altitude $h = \sqrt{4 \times 9} = 6$.

If two triangles are similar with a scale factor $k = \frac{a}{b}$:

• The ratio of their perimeters is $k$.
• The ratio of their areas is $k^2$.

Why it works. Perimeter is a 1D measure (units), so it scales linearly. Area is a 2D measure (units$^2$), so it scales by the square of the linear factor.

SAT use. The SAT will give you the ratio of sides and ask for the ratio of areas, or vice versa. Don't forget to square or take the square root.

Worked example. If the ratio of sides is $2:3$, the ratio of perimeters is $2:3$, but the ratio of areas is $2^2:3^2 = 4:9$.

Problems involving shadows are classic similarity applications. At any given time, the sun's rays are parallel, meaning the triangle formed by an object and its shadow is similar to the triangle formed by a nearby object and its shadow.

The Setup. $\frac{Height_{object}}{Shadow_{object}} = \frac{Height_{pole}}{Shadow_{pole}}$.

SAT use. These are often word problems. Draw the triangles immediately. The 'angle of elevation' of the sun is the same for both objects.

Worked example. A $6$-foot person casts a $4$-foot shadow. A nearby tree casts a $20$-foot shadow. $\frac{6}{4} = \frac{x}{20} \Rightarrow 4x = 120 \Rightarrow x = 30$ feet.