Proof: AA (Angle-Angle) Similarity Theorem

Theorem Statement

If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar.

1. Given and Goal

Let there be two triangles, $\triangle ABC$ and $\triangle DEF$.
Assume $\triangle ABC$ is larger than $\triangle DEF$ without loss of generality.

A B C D E F Figure 1: Two triangles with congruent angles marked.

2. Construction (Superposition)

We will "superimpose" the smaller triangle $\triangle DEF$ onto the larger triangle $\triangle ABC$.

  1. Mark a point $P$ on side $AB$ such that $AP = DE$.
  2. Mark a point $Q$ on side $AC$ such that $AQ = DF$.
  3. Draw the line segment $PQ$.
A B C P Q Figure 2: Constructing points P and Q such that AP = DE and AQ = DF.

Check for Congruence: Consider $\triangle APQ$ and $\triangle DEF$.
1. $AP = DE$ (by construction).
2. $\angle A \cong \angle D$ (Given).
3. $AQ = DF$ (by construction).

By the SAS (Side-Angle-Side) Postulate, we have: $$ \triangle APQ \cong \triangle DEF $$

3. Establishing Parallel Lines

Since $\triangle APQ \cong \triangle DEF$, corresponding parts of congruent triangles are congruent (CPCTC). Therefore: $$ \angle APQ \cong \angle E $$

However, we were given that $\angle B \cong \angle E$. By the Transitive Property: $$ \angle APQ \cong \angle B $$

Notice that $\angle APQ$ and $\angle B$ are corresponding angles relative to the transversal line $AB$ crossing lines $PQ$ and $BC$. Since corresponding angles are equal, the lines must be parallel: $$ PQ \parallel BC $$

A B C P Q ∠APQ ∠B Figure 3: Since ∠APQ = ∠B, PQ is parallel to BC.

4. Basic Proportionality Theorem

Because $PQ \parallel BC$ inside $\triangle ABC$, we apply the Basic Proportionality Theorem (Thales's Theorem), which states that a line parallel to one side of a triangle divides the other two sides proportionally.

Thus: $$ \frac{AB}{AP} = \frac{AC}{AQ} = \frac{BC}{PQ} $$

Recall from our construction that $AP = DE$, $AQ = DF$, and since $\triangle APQ \cong \triangle DEF$, $PQ = EF$. Substituting these values into the equation:

$$ \frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF} $$

5. Conclusion

We have shown that the corresponding angles are congruent (Given) and the corresponding sides are in proportion (Proven).

$\therefore \triangle ABC \sim \triangle DEF$

Q.E.D.