Overview
Two triangles are similar when their corresponding angles are equal and their corresponding sides are proportional. The SAT primarily tests the AA (Angle-Angle) similarity criterion, which is sufficient to establish similarity because once two angles are equal the third must match as well (angle sum = 180°). Similar triangle problems appear in the geometry domain and frequently involve indirect measurement (shadow problems), parallel lines, or the midsegment theorem.
Key Points
1. AA Similarity Criterion
If two angles of one triangle are equal to two angles of another triangle, the triangles are similar.
∠A = ∠D and ∠B = ∠E
→ Triangle ABC ~ Triangle DEF
→ Third angles are automatically equal: ∠C = ∠F
Common sources of equal angles on the SAT:
- Shared angle (same vertex in both triangles)
- Vertical angles at an intersection point
- Corresponding angles when parallel lines are cut by a transversal
- Right angle (90°) in both triangles
2. Setting Up Proportions
Once similarity is established, corresponding sides are in the same ratio (the scale factor k):
Side of Triangle A / Corresponding side of Triangle B = k
Always match sides that are opposite the same angle (corresponding position).
Proportion template:
a₁/a₂ = b₁/b₂ = c₁/c₂ = k
where a, b, c are corresponding sides of the two triangles.
Example (shadow problem): A 6 ft person casts a 4 ft shadow. A nearby tree casts a 10 ft shadow. Height of tree?
Person height / Person shadow = Tree height / Tree shadow
6 / 4 = h / 10
h = 15 ft
3. Scale Factor Effects on Perimeter and Area
| Measurement | Scales by |
|---|---|
| Side lengths | k |
| Perimeter | k |
| Area | k² |
Example: If triangle A has sides 3, 4, 5 and similar triangle B has sides 6, 8, 10, then k = 2. Perimeter of B = 2 × perimeter of A. Area of B = 4 × area of A.
4. Parallel Lines Creating Similar Triangles
When two parallel lines are cut by two transversals meeting at a point, the two resulting triangles are similar by AA:
- The shared angle at the intersection point is common to both triangles
- Corresponding angles formed by the parallel lines with the transversals are equal
This configuration appears frequently in hard module problems where a point is inside or above two parallel lines.
5. Midsegment Theorem
A midsegment connects the midpoints of two sides of a triangle.
Midsegment length = ½ × (length of the third side)
Midsegment is parallel to the third side
The triangle formed by a midsegment is similar to the original triangle with scale factor k = ½, so its area is (½)² = ¼ of the original area.
Pitfalls and Common Mistakes
Pitfall 1: Inverting the scale factor If triangle A has side 6 and similar triangle B has the corresponding side 15, the ratio A:B = 6:15 = 2:5. Writing it as 5:2 (backwards) gives a wrong answer when cross-multiplying to find an unknown side. Fix: Label both triangles clearly (which is A, which is B) before writing the proportion. Always keep the same triangle in the numerator on both sides of the equation.
Pitfall 2: Using non-corresponding sides in the proportion Pairing sides that face different angles (e.g., shortest side of one triangle with longest side of the other) produces a meaningless ratio. Fix: Identify the angle each side is opposite to, then pair sides opposite equal angles across the two triangles.
Pitfall 3: Confusing angle equality with side equality In similar triangles, corresponding angles are EQUAL; corresponding sides are PROPORTIONAL, not necessarily equal. Students sometimes set sides equal instead of forming a ratio. Fix: Angles → write ”=” directly. Sides → write a proportion with a ratio.
Pitfall 4: Applying scale factor directly to area without squaring If the linear scale factor is k = 3, the area ratio is 9 (not 3). Forgetting to square the scale factor for area is a frequent calculation error. Fix: Memorize the rule: side scales by k, area scales by k².
Related Entries
- Right_Triangles_Pythagorean
- Properties_Circles
- Coordinate_Geometry
- Lines_Angles_Parallel
- Trigonometric_Functions
Quick Reference Card
| Concept | Rule |
|---|---|
| AA criterion | Two equal angle pairs → triangles similar |
| Corresponding sides | a₁/a₂ = b₁/b₂ = c₁/c₂ = k |
| Perimeter ratio | Same as side ratio (k) |
| Area ratio | k² |
| Midsegment | = ½ × third side; parallel to third side |
| Parallel lines + transversals | Create similar triangles (AA via corresponding/vertical angles) |
| Shadow problems | Set up person:shadow = tree:shadow proportion |
| Top trap | Inverting the ratio direction |