- Focus on the angle markings rather than the apparent side lengths. - How many corresponding angle pairs are needed for the angle-angle criterion?

The marked angles show two pairs of corresponding congruent angles. Therefore, the triangles are similar by the angle-angle similarity criterion.

Yes. The triangles are similar by AA.

- One pair of congruent angles is already marked. - Look for an angle that belongs to both triangles.

1. The diagram gives \(\angle C\cong\angle E\). 2. The triangles also share the angle at \(A\), because \(B\) and \(D\) lie on the same ray from \(A\), and \(C\) and \(E\) lie on the other ray. 3. Therefore, \(\triangle ABC\sim\triangle ADE\) by AA similarity.

AA similarity.

- Try writing the vertices of one triangle in a row, then put the matching vertices beneath them. - When you write a similarity statement, which names must occupy corresponding positions?

1. The single-mark angles at \(A\) and \(D\) match. 2. The double-mark angles at \(B\) and \(E\) match. 3. The remaining vertices \(C\) and \(F\) match, so \(\triangle ABC\sim\triangle DEF\).

\(A\leftrightarrow D\), \(B\leftrightarrow E\), \(C\leftrightarrow F\); \(\triangle ABC\sim\triangle DEF\).

- Try pairing each angle in one triangle with an equal angle in the other. - What is the smallest number of matching angle pairs that proves triangle similarity?

1. Both triangles have an angle of \(42^\circ\) and an angle of \(78^\circ\). 2. Two matching angles prove the triangles are similar by AA, even though the angles are listed in different orders.

Yes. They have two matching angles, so they are similar by AA.

- Use the stated parallel segments rather than estimating angles from the drawing. - Treat \(AB\) and \(AC\) as transversals of the parallel segments. - Identify one corresponding-angle pair on each side of the triangle.

Because \(DE\parallel BC\), corresponding angles give \(\angle ADE\cong\angle ABC\) and \(\angle AED\cong\angle ACB\). Therefore, triangle \(ADE\sim\triangle ABC\) by AA.

\(\angle ADE\cong\angle ABC\) and \(\angle AED\cong\angle ACB\), so \(\triangle ADE\sim\triangle ABC\) by AA.

- Use the angle markings to match corresponding vertices before using side lengths. - Find the scale factor from a pair of corresponding sides whose lengths are known. - Apply the same scale factor to the side corresponding to \(8\).

The angle markings match \(A\) with \(D\) and \(B\) with \(E\), so \(AB\) corresponds to \(DE\) and \(AC\) corresponds to \(DF\). The scale factor from triangle \(ABC\) to triangle \(DEF\) is \(\frac{9}{6}=\frac{3}{2}\). Therefore, \(x=8\cdot\frac{3}{2}=12\).

\(x=12\)

- Match vertices by their angle markings before comparing side lengths. - Keep the vertex order consistent in the similarity statement. - Use one known pair of corresponding sides for the scale factor.

1. The one-arc angles give \(A\leftrightarrow Q\), and the two-arc angles give \(B\leftrightarrow R\). Therefore, \(C\leftrightarrow P\). 2. The correct statement is \(\triangle ABC\sim\triangle QRP\). 3. Corresponding sides \(AB=4\) and \(QR=6\), so the scale factor is \(\frac{6}{4}=\frac{3}{2}\).

\(\triangle ABC\sim\triangle QRP\), with scale factor \(\frac{3}{2}\).

- What must the three interior angles of triangle \(ABC\) add to? - After finding the missing angle, compare it with the labeled angles in the second triangle. - Match vertices by equal angle measures before writing the similarity statement.

1. In \(\triangle ABC\), the angles sum to \(180^\circ\), so \(\angle C=180^\circ-48^\circ-67^\circ=65^\circ\). 2. The diagram gives \(\angle A=48^\circ\) and \(\angle D=48^\circ\), so those angles are congruent. 3. The diagram gives \(\angle F=65^\circ\), and step 1 shows \(\angle C=65^\circ\), so a second pair of angles is congruent. 4. Therefore, the triangles are similar by AA, with \(A\leftrightarrow D\), \(B\leftrightarrow E\), and \(C\leftrightarrow F\).

a) \(\angle C=65^\circ\) b) \(\angle A=\angle D=48^\circ\) and \(\angle C=\angle F=65^\circ\), so the triangles are similar by AA. c) \(\triangle ABC\sim\triangle DEF\)

- Is knowing two angles enough to work out the third angle of a triangle? - What would you need to match between the triangles to use AA?

1. The third angle in the first triangle is \(180^\circ-50^\circ-60^\circ=70^\circ\). 2. The third angle in the second is \(180^\circ-50^\circ-70^\circ=60^\circ\). 3. Both have angles \(50^\circ\), \(60^\circ\), and \(70^\circ\), so they are similar by AA, even though the listed angles are in a different order.

Yes. Both triangles have angles \(50^\circ\), \(60^\circ\), and \(70^\circ\), so two matching angles establish AA similarity.

- Recall exactly how many corresponding angle pairs the AA criterion requires. - Use the fact that the three angles in a triangle sum to \(180^\circ\). - Try constructing two different triangles that share one angle but not a second angle.

1. The statement is not correct. AA requires two pairs of corresponding congruent angles, not just one pair. 2. For example, one triangle could have angles \(60^\circ,50^\circ,70^\circ\), while another could have angles \(60^\circ,40^\circ,80^\circ\). Both contain a \(60^\circ\) angle, but the other angles do not match, so the triangles are not similar.

No. One equal angle is not enough for AA. For example, \((60^\circ,50^\circ,70^\circ)\) and \((60^\circ,40^\circ,80^\circ)\) are not similar triangles.

- Identify the right angle in each height-shadow triangle. - Parallel sun rays create another pair of equal angles. - Match height with height and shadow length with shadow length. - Set up one proportion using corresponding sides.

The person and flagpole are both vertical, so each height-shadow triangle has a right angle. Parallel sunlight creates the same acute angle with the ground in both triangles. Therefore, the triangles are similar by AA. Using corresponding height-to-shadow ratios, \(\frac{6}{4}=\frac{h}{18}\), so \(h=27\,\text{ft}\).

The flagpole is \(27\,\text{ft}\) tall.

- Identify the transversals crossing the parallel lines. - Find the full length \(AB\) from its two parts. - Match the smaller triangle's sides with those of the full triangle.

1. Corresponding angles on the parallel lines give \(\angle ADE=\angle ABC\) and \(\angle AED=\angle ACB\). 2. Thus \(\triangle ADE\sim\triangle ABC\) by AA. 3. Since \(AD=6\) and \(DB=3\), \(AB=9\) and the enlargement factor is \(9/6=3/2\). 4. Then \(AC=8\cdot\frac{3}{2}=12\), so \(EC=12-8=4\). 5. Also, \(BC=10\cdot\frac{3}{2}=15\).

a) \(\angle ADE=\angle ABC\) and \(\angle AED=\angle ACB\). b) \(EC=4\) and \(BC=15\).

- Look at the pair of angles formed at the intersection \(O\). - After establishing similarity, identify which side corresponds to \(AO\). - Use the same scale factor on the side corresponding to \(OB\).

1. The diagram gives \(\angle A\cong\angle C\). 2. Angles \(\angle AOB\) and \(\angle COD\) are vertical angles, so they are congruent. 3. Therefore, \(\triangle AOB\sim\triangle COD\) by AA. 4. The scale factor from \(\triangle AOB\) to \(\triangle COD\) is \(\frac{CO}{AO}=\frac{4}{6}=\frac{2}{3}\). 5. Since \(OB=9\) corresponds to \(OD=x\), \(x=\frac{2}{3}\cdot9=6\).

a) The triangles are similar by AA. b) \(x=6\)

- The exterior angle and the adjacent interior angle at \(B\) form a straight angle. - After finding the interior angle at \(B\), compare angle measures across the two triangles. - Two matching angle pairs are enough to establish AA similarity.

1. The exterior angle at \(B\) is \(118^\circ\), so the adjacent interior angle is \(\angle ABC=180^\circ-118^\circ=62^\circ\). 2. The second triangle has \(\angle E=62^\circ\), so \(\angle ABC\cong\angle E\). 3. The diagrams also give \(\angle A=47^\circ\) and \(\angle D=47^\circ\). 4. Therefore, \(\triangle ABC\sim\triangle DEF\) by AA, with \(A\leftrightarrow D\), \(B\leftrightarrow E\), and \(C\leftrightarrow F\).

The interior angle at \(B\) is \(62^\circ\) because it forms a linear pair with the \(118^\circ\) exterior angle. Since \(\angle B=\angle E=62^\circ\) and \(\angle A=\angle D=47^\circ\), \(\triangle ABC\sim\triangle DEF\) by AA.

- Treat \(AB\) and \(AC\) as transversals crossing the parallel segments. - Identify which angles in the small triangle correspond to angles in the large triangle before using side lengths. - Combine the two labeled pieces of \(AB\) before forming a similarity ratio.

1. Because \(DE\parallel BC\), transversal \(AB\) gives \(\angle ADE\cong\angle ABC\), and transversal \(AC\) gives \(\angle AED\cong\angle ACB\). 2. Therefore, \(\triangle ADE\sim\triangle ABC\) by AA. 3. From the diagram, \(AD=4\) and \(DB=2\), so \(AB=6\). 4. The small-to-large scale factor is \(\frac{AD}{AB}=\frac{4}{6}=\frac{2}{3}\). 5. Since \(DE=6\) corresponds to \(BC\), \(\frac{6}{BC}=\frac{2}{3}\), so \(BC=9\).

a) \(\angle ADE\cong\angle ABC\) and \(\angle AED\cong\angle ACB\) because \(DE\parallel BC\), so the triangles are similar by AA. b) \(BC=9\)

- Which angle pairs are marked alike? - What can two matching angles tell you about triangles? - Once you know the triangles are similar, match sides in the same order.

1. The markings show \(\angle A=\angle D\) and \(\angle B=\angle E\). Two matching angles establish AA similarity. 2. Corresponding sides are \(AB\leftrightarrow DE\) and \(AC\leftrightarrow DF\). 3. The proportion is \(\frac{DF}{3}=\frac{6}{4}\), so \(DF=\frac{6}{4}\cdot3=4.5\,\text{cm}\).

The marked angles give AA similarity. \(\frac{DF}{3}=\frac{6}{4}\), so \(DF=4.5\,\text{cm}\).

- Each height is perpendicular to the level ground. - How do the sun's rays make another angle equal in both triangles? - Match height with height and shadow with shadow.

1. Both height-shadow triangles have a right angle and the same angle of sunlight, so they are similar by AA. 2. Let the stick's shadow be \(s\). The ratio gives \(\frac{12}{8}=\frac{3}{s}\). 3. Solve: \(s=\frac{3\cdot8}{12}=2\,\text{ft}\).

\(2\,\text{ft}\). Both right triangles share the same sunlight angle, so corresponding sides are proportional.

- One angle pair is already known to correspond. - Express each triangle's third angle using the \(180^\circ\) angle sum. - The second matching angle does not have to use the same named position in both triangles. - Check every possible correspondence of the two remaining angles.

Since \(\angle A=\angle D=50^\circ\), one angle pair already matches. Triangle \(ABC\) has \(\angle C=180^\circ-50^\circ-(2x+10)^\circ=(120-2x)^\circ\). Triangle \(DEF\) has \(\angle F=180^\circ-50^\circ-3x^\circ=(130-3x)^\circ\). One possibility is \(\angle B=\angle E\): \(2x+10=3x\), giving \(x=10\). The other possible correspondence is \(\angle B=\angle F\): \(2x+10=130-3x\), giving \(x=24\). Both values produce valid positive triangle angles, so both guarantee AA similarity, with different vertex correspondences.

\(x=10\) or \(x=24\)