Congruence & Similar Triangles | Math Learning Guide

📋 Prerequisites

Congruent triangles are identical copies; similar triangles are scaled copies. Similarity is the more powerful idea — it lets you find unknown sides using proportions, and it's behind most of the classic theorems in geometry. Master AA similarity and you can unlock almost any triangle problem.

📚 Key Concepts

Congruent Triangles

Two triangles are congruent if all corresponding sides and angles are equal (same shape and size).

Congruence criteria (enough to prove congruence):

SSS (Side-Side-Side): All three pairs of corresponding sides equal
SAS (Side-Angle-Side): Two sides and the included angle equal
ASA (Angle-Side-Angle): Two angles and the included side equal
AAS (Angle-Angle-Side): Two angles and a non-included side equal
HL (Hypotenuse-Leg): For right triangles only — hypotenuse and one leg equal

NOT enough: AAA (only proves similarity), SSA (ambiguous — can have two different triangles)

Similar Triangles

Two triangles are similar if corresponding angles are equal (same shape, different size).

Corresponding sides are in proportion (the scale factor).

If $\triangle ABC \sim \triangle DEF$ with scale factor $k$:

$ \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = k $
$ \angle A = \angle D $, $ \angle B = \angle E $, $ \angle C = \angle F $

Similarity Criteria

AA $(Angle-Angle): Two pairs of corresponding angles equal \\to the third pair is automatically equal.$ This is the most commonly used!
SAS~ (Side-Angle-Side similarity): Two sides in proportion, included angle equal
SSS~ (Side-Side-Side similarity): All three sides in the same proportion

Ratio of Areas

If the scale factor (side ratio) is $k$, then the ratio of areas is $k^2$:

\[ \frac{\text{Area}_1}{\text{Area}_2} = \left(\frac{\text{side}_1}{\text{side}_2}\right)^2 = k^2 \]

$Example: sides double (k=2), area quadruples (k^{2}=4). Sides triple, area is 9 \\times .$

Angle Bisector Theorem

If a bisector of an angle of a triangle divides the opposite side into segments of length $m$ and $n$:

\[ \frac{m}{n} = \frac{AB}{AC} \]

The angle bisector divides the opposite side in the ratio of the adjacent sides.

Parallel Lines & Similarity

If a line is drawn parallel to one side of a triangle, it creates a smaller triangle similar to the original.

This is the "basic proportionality theorem" (Thales' theorem):

$ DE \parallel BC \implies \frac{AD}{DB} = \frac{AE}{EC} $

Converse also works: if the sides are divided proportionally, the line is parallel.

How to Spot Similar Triangles

Look for parallel lines $(creates equal corresponding angles \\to AA)$
Look for shared angles $(common vertex angle + one more equal angle \\to AA)$
Look for right angles $(two right triangles sharing an acute angle \\to AA)$
Look for vertical angles + another pair of equal angles

⚠️ Common Mistake: Make sure you're matching the corresponding sides correctly. The side opposite angle A corresponds to the side opposite the equal angle in the other triangle. Label your triangles in order of correspondence ($\triangle ABC \sim \triangle DEF$ means A↔D, B↔E, C↔F). Also: area ratio is the square of the side ratio, not the same ratio!

💡 Key Insight: AA similarity is your workhorse. In competition problems, if you can find two pairs of equal angles, you have similar triangles and can set up proportions. The classic setup is a triangle with a line parallel to the base, or two right triangles sharing an acute angle. Once you establish similarity, set up the proportion carefully and solve for the unknown.

✏️ Example Problems

📝 Example 1 (Basic similarity proportion)

$\triangle ABC \sim \triangle DEF$. AB = 6, BC = 8, DE = 9.
What is the length of EF?

Step 1: Find the scale factor from $\triangle ABC$ to $\triangle DEF$.

Corresponding sides: AB ↔ DE, BC ↔ EF

Scale factor $ k = \frac{DE}{AB} = \frac{9}{6} = \frac{3}{2} = 1.5 $

Step 2: Apply the scale factor to BC to find EF.

$ EF = BC \times k = 8 \times \frac{3}{2} = 12 $

Alternative — set up proportion directly:

$ \frac{AB}{DE} = \frac{BC}{EF} \implies \frac{6}{9} = \frac{8}{EF} \implies 6 \cdot EF = 72 \implies EF = 12 $

Answer: 12

📝 Example 2 (Area ratio from similarity)

Two similar triangles have corresponding sides in ratio 2:5.
If the smaller triangle has area 12, what is the area of the larger triangle?

Step 1: Side ratio (small : large) = 2 : 5

Scale factor $ k = \frac{5}{2} $ (from small to large)

Step 2: Area ratio = $k^2$.

\[ \text{Area ratio} = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \]

Step 3: $Larger area = smaller area \\times area ratio.$

$ \text{Large area} = 12 \times \frac{25}{4} = 3 \times 25 = 75 $

Check: 75/12 = 25/4 = (5/2)$^{2}.$ ✓

Answer: 75

📝 Example 3 (Shadow problem — classic similar triangles)

A 6-foot tall person casts a 4-foot shadow. At the same time, a nearby tree casts
a 20-foot shadow. How tall is the tree (in feet)?

Setup: The sun's rays are parallel, so both the person and the tree form similar right triangles with their shadows.

Both triangles have: right angle (90°$) + same sun angle \\to AA similarity$

Step 1: Set up the proportion (height : shadow).

\[ \frac{\text{Person height}}{\text{Person shadow}} = \frac{\text{Tree height}}{\text{Tree shadow}} \]

Step 2: Plug in values. Let $h$ = tree height.

$ \frac{6}{4} = \frac{h}{20} $

Step 3: Solve for $h$.

$ 4h = 6 \times 20 = 120 $

$ h = 30 $

Check: 6/4 = 3/2, and 30/20 = 3/2. Same ratio. ✓

Answer: 30