Exercises: Establish AA Similarity Criterion

$58\degree$

$112\degree$

$68\degree$

$180\degree$

If two sides of one triangle are proportional to two sides of another triangle, the triangles are similar.

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

If all three angles of one triangle are congruent to all three angles of another triangle, the triangles are similar.

If two angles of one triangle are congruent to two angles of any polygon, the triangles are similar.

$\angle P' = 3 \times \angle P$

$\angle P' = \angle P$ and $P'Q' = 3 \cdot PQ$

$\angle P' = \angle P$ and $P'Q' = PQ$

$P'Q' = 3 \cdot PQ$ and $\angle P' = \frac{1}{3} \angle P$

$\triangle ABC \cong \triangle DEF$ because two pairs of angles are equal.

$\triangle ABC \sim \triangle DEF$ by AA, because $\angle A = \angle D$ and $\angle B = \angle E$.

The triangles cannot be similar because the third angle $\angle C$ is not given.

More information about the side lengths is needed to determine similarity.

Yes — $\angle P = \angle X$ and $\angle Q = \angle Z$, so $\triangle PQR \sim \triangle XYZ$ by AA.

Yes — $\angle P = \angle X$ and $\angle R = \angle Z$, so $\triangle PQR \sim \triangle XYZ$ by AA.

No — only one pair of angles is equal.

No — we need to check all three angle pairs first.

$\triangle ABE \sim \triangle DCE$ with correspondence $A \leftrightarrow D$, $B \leftrightarrow C$, $E \leftrightarrow E$.

$\triangle ABE \sim \triangle CDE$ with correspondence $A \leftrightarrow C$, $B \leftrightarrow D$, $E \leftrightarrow E$.

The triangles are not similar because no angle measures are given.

$\triangle ABE \sim \triangle DCE$ with correspondence $A \leftrightarrow C$, $B \leftrightarrow D$, $E \leftrightarrow E$.

Yes — $\triangle ABC \sim \triangle DEF$ by AA, because $\angle C = \angle F = 90\degree$ and $\angle A = \angle D = 28\degree$.

No — right triangles are only similar if their hypotenuses are proportional.

Yes — all right triangles are similar to each other.

Cannot determine — we need to know the side lengths.

$\triangle ABC \sim \triangle DEF$ by AA, and the triangles are also congruent because they share two equal angles.

$\triangle ABC \sim \triangle DEF$ by AA, with scale factor $k = \frac{3}{2}$ from $\triangle ABC$ to $\triangle DEF$.

The triangles are congruent by ASA because two angles and the included side are given.

The triangles are not similar because their sides are different lengths.

The triangles must be congruent, since all three angles are equal.

The triangles must be similar, since all three angles are equal — but they may not be congruent.

Nothing can be concluded without knowing at least two angle pairs; three angles matching is more than required.

The triangles must be similar, and their scale factor must equal 1.

$\angle A$ is shared, and $\angle ADE = \angle ABC$ because $DE \parallel BC$ (corresponding angles). Two pairs match, so $\triangle ADE \sim \triangle ABC$ by AA.

$\angle ADE = \angle ABC$ and $\angle AED = \angle ACB$, so $\triangle ADE \sim \triangle ABC$ — but $\angle A$ cannot be used because it appears in both triangles.

$\angle A = 38\degree$ and $\angle ADE = \angle ABC = 74\degree$, so AA applies with these two pairs, giving the same result as choice A.

We cannot determine similarity without knowing the lengths of $DE$ and $BC$.

Yes — all corresponding angles are equal, so the rectangles are similar by an angle criterion analogous to AA.

No — the student's reasoning would only work for triangles. Rectangles with all right angles may not be similar (for example, a $1 \times 2$ rectangle and a $1 \times 4$ rectangle both have four $90\degree$ angles but are not similar).

Yes — equal angles always guarantee similarity for any polygon.

No — you would also need all four angles to be equal, not just two.

Maya is correct — she does need to verify all three angle pairs before applying AA.

Maya has confused AA with ASA. She only needs two angle pairs, not three.

Maya should also check the side lengths before concluding similarity.

Maya only needs one angle pair to conclude similarity by AA.

Jordan is correct — if all three angles match, the triangles must be congruent.

Jordan confused similarity and congruence. Similarity (proved by AA) guarantees equal angles and proportional sides, but not necessarily equal sides. Congruence requires $k = 1$, which needs side information to verify.

Jordan is wrong because AA only proves two angles match, not all three.

Jordan is correct that the triangles are congruent, but the reasoning should reference ASA, not AA.