Overview

Target Contests: AMC 8, AMC 10

In competition geometry, you are rarely given every side length and angle. Instead, you must deduce missing information by finding relationships between different parts of a figure. The two most powerful tools for bridging these gaps are Congruence (finding exact identical copies of a triangle) and Similarity (finding scaled-up or scaled-down versions of a triangle).

1. Triangle Congruence (The "Equals" of Geometry)

Two triangles are congruent ($\cong$) if they are exactly the same size and shape. All corresponding angles are equal, and all corresponding side lengths are equal. If you prove two triangles are congruent, you instantly know that all their matching parts are identical (often abbreviated as CPCTC: Corresponding Parts of Congruent Triangles are Congruent).

You do not need to check all six parts (3 sides, 3 angles) to prove congruence. You only need three specific pieces of information:

• SSS (Side-Side-Side): If all three sides of one triangle equal all three sides of another, they are congruent.

• SAS (Side-Angle-Side): Two sides and the angle between them are equal.

• ASA (Angle-Side-Angle): Two angles and the side between them are equal.

• AAS (Angle-Angle-Side): Two angles and a non-included side are equal.

• HL (Hypotenuse-Leg): For right triangles only. If the hypotenuse and one leg match, they are congruent.

Why SSA Fails: The "Swinging Hinge" Proof

A common trap is assuming that two sides and a non-included angle (Side-Side-Angle) guarantee congruence. They do not. This is known as the ambiguous case. Let's prove geometrically why SSA fails to lock a triangle into a single, unique shape.

The Proof: Suppose we are given an angle $\angle A$, a fixed adjacent side of length $c$ (forming segment $AB$), and a fixed opposite side of length $a$. We want to construct $\triangle ABC$.

1. Draw the angle $\angle A$. The bottom ray extends infinitely (this will contain our unknown side $b$).
2. From vertex $A$, measure out length $c$ along the top ray to establish vertex $B$.
3. Now, we must attach the final side of length $a$ to vertex $B$ and drop it down to intersect the bottom ray at vertex $C$.
4. Imagine side $a$ is a pendulum hanging from point $B$. Let the shortest distance (the altitude) from $B$ to the bottom ray be $h$.
5. If $h < a < c$, the pendulum of length $a$ will swing and intersect the bottom ray at two distinct points, let's call them $C_1$ and $C_2$.
6. This creates two completely different triangles: $\triangle ABC_1$ (an obtuse triangle) and $\triangle ABC_2$ (an acute triangle).
7. Both triangles share the exact same Side ($c$), Side ($a$), and Angle ($\angle A$), but they are obviously not congruent. Therefore, SSA cannot be used to prove congruence.

2. Triangle Similarity (The "Scaled" of Geometry)

Two triangles are similar ($\sim$) if they have the exact same shape, but not necessarily the same size. Think of it as zooming in or out on a screen. In similar triangles:

1. All corresponding angles are exactly equal.
2. All corresponding sides are strictly proportional (they share a constant scaling factor, $k$).

Similarity Criteria

Proving similarity requires even less information than congruence:

• AA (Angle-Angle): This is the king of AMC geometry. Because the angles of a triangle always sum to $180^\circ$, if you know two angles match, the third must match. If you see parallel lines, look for AA similarity immediately.

• SAS Similarity: Two sides are proportional, and the included angle is exactly equal.

• SSS Similarity: All three pairs of sides share the exact same ratio.

3. The Two Most Important AMC Similarity Setups

A. The Area Scaling Law (The $k^2$ Rule)

If two triangles are similar with a side length ratio of $1 : k$, their perimeters also have a ratio of $1 : k$. However, their areas have a ratio of $1 : k^2$. If you draw a segment connecting the midpoints of two sides of a triangle (a midline), the small top triangle is similar to the whole triangle with a side ratio of $1 : 2$. Therefore, the small triangle holds exactly $1/4$ of the total area, leaving $3/4$ of the area for the trapezoid at the bottom.

B. The Right Triangle Altitude Theorem

If you draw an altitude from the right angle of a right triangle to its hypotenuse, you do not just get one right triangle—you get three, and they are all similar to each other by AA similarity.

Let right $\triangle ABC$ have its right angle at $C$, and let $CD$ be the altitude to hypotenuse $AB$. Because they share angles, $\triangle ACD \sim \triangle CBD \sim \triangle ABC$. Setting up the proportions yields two incredibly useful formulas:

1. $CD^2 = AD \cdot BD$ (The altitude is the geometric mean of the two hypotenuse segments).
2. $AC^2 = AD \cdot AB$ (A leg is the geometric mean of the adjacent segment and the whole hypotenuse).

Worked Examples

Example 1: Nested Triangles and Area

Problem: In $\triangle ABC$, point $D$ is on $AB$ and point $E$ is on $AC$ such that $DE \parallel BC$. If $AD = 3$ and $DB = 6$, and the area of $\triangle ADE$ is $10$, what is the area of the quadrilateral $DBCE$?

Solution: Because $DE \parallel BC$, the corresponding angles are equal ($\angle ADE = \angle ABC$ and $\angle AED = \angle ACB$). By AA similarity, $\triangle ADE \sim \triangle ABC$.

Next, find the ratio of the sides. The side of the small triangle is $AD = 3$. The corresponding side of the large triangle is the whole length $AB = AD + DB = 3 + 6 = 9$. The side ratio $k$ is $\frac{9}{3} = 3$.

Since the sides are scaled by a factor of $3$, the area is scaled by a factor of $k^2 = 3^2 = 9$. Area of $\triangle ABC = 9 \times (\text{Area of } \triangle ADE) = 9 \times 10 = 90$.

To find the area of quadrilateral $DBCE$, subtract the small triangle from the whole triangle: $\text{Area}(DBCE) = 90 - 10 = 80$.

Example 2: The Hourglass (Bowtie) Similarity

Problem: Trapezoid $ABCD$ has parallel bases $AB$ and $CD$. The diagonals $AC$ and $BD$ intersect at point $E$. If $AB = 4$, $CD = 10$, and the length of diagonal $AC$ is $21$, find the length of $AE$.

Solution: Because $AB \parallel CD$, the alternate interior angles are equal: $\angle EAB = \angle ECD$ and $\angle EBA = \angle EDC$. Also, the vertical angles at $E$ are equal. By AA similarity, $\triangle ABE \sim \triangle CDE$. (Be careful with the vertex order! Top-left corresponds to bottom-right).

The ratio of similarity between the top and bottom triangles is the ratio of their parallel bases: $\frac{AB}{CD} = \frac{4}{10} = \frac{2}{5}$. This means every segment in $\triangle ABE$ is $\frac{2}{5}$ the length of its corresponding segment in $\triangle CDE$. So, $\frac{AE}{CE} = \frac{2}{5}$, which means $CE = \frac{5}{2} AE$.

We know the whole diagonal $AC = AE + CE = 21$. Substitute for $CE$:

Example 3: Right Triangle Altitude

Problem: In right $\triangle PQR$ with right angle at $Q$, altitude $QS$ is drawn to hypotenuse $PR$. If $PS = 4$ and $SR = 9$, find the area of $\triangle PQR$.

Solution: Using the Right Triangle Altitude theorem, we know the three triangles are similar. The altitude squared equals the product of the two hypotenuse segments:

Now we have the height of the large triangle. The base of the large triangle is the entire hypotenuse $PR = PS + SR = 4 + 9 = 13$. The area of $\triangle PQR = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 13 \cdot 6 = 39$.

Common Pitfalls

• Incorrect Vertex Matching: When writing similarity statements (e.g., $\triangle ABC \sim \triangle DEF$), the order of the letters dictates which angles and sides match up. In the hourglass shape, $\triangle ABE$ is similar to $\triangle CDE$, not $\triangle DCE$. If you mess up the order, your side proportions will be inverted.
• Forgetting to square the ratio for Area: If a side ratio is $1:3$, the area ratio is $1:9$. Students constantly forget to square the ratio, or incorrectly apply the $k^2$ rule to perimeter (perimeter scales linearly, just like sides).
• Adding instead of Scaling: Similarity is strictly about multiplication/division. If a side goes from $5$ to $10$, the triangle is scaled by $\times 2$. It is not scaled by $+5$.

Practice Problems