Question

the dilation $d_{o,3} (x, y) \to (3x, 3y)$ is performed on the pre-image $\triangle abc$ to make a similar triangle. which statements are true? check all that apply. $\square$ $\angle a$ corresponds to $\angle a$. $\square$ $\angle aac$ corresponds to $\angle b$. $\square$ $\overline{cb}$ corresponds to $\overline{ca}$. $\square$ segment $aa$ is parallel to segment $cc$. $\square$ $\triangle abc \sim \triangle aac$.

Explanation:

Step1: Recall Dilation Properties

Dilation preserves angle measures and creates similar figures, so corresponding angles are equal, and corresponding vertices/angles/segments are related by the dilation center (O here, scale factor 3).

Step2: Analyze Each Statement

- **∠A corresponds to ∠A'**: In similar triangles (from dilation), corresponding angles (at corresponding vertices) are equal. A and A' are corresponding vertices, so ∠A ↔ ∠A'. True.
- **∠A'AC' corresponds to ∠B**: ∠A'AC' is not a corresponding angle to ∠B. Corresponding angles should be at corresponding vertices. False.
- **$\overline{CB}$ corresponds to $\overline{C'A}$**: $\overline{CB}$ (from C to B) should correspond to $\overline{C'B'}$ (or $\overline{C'A'}$? Wait, dilation: C→C', B→B', so $\overline{CB}$ ↔ $\overline{C'B'}$, not $\overline{C'A}$. False.
- **Segment A'A is parallel to segment C'C**: A'A and C'C are both vertical? Wait, A: let's find coordinates. A: (-2,3), A': (-9,9); C: (-3,1), C': (-9,3). So A'A: from (-2,3) to (-9,9) – slope? Wait, no: A'A is from A (-2,3) to A' (-9,9)? Wait, no, dilation center O (0,0). Wait, coordinates: A: let's check grid. A is at x=-2, y=3? Wait, the grid: x=-2, y=3 (A), C at x=-3, y=1, B at x=-1, y=1? Wait, A' is at x=-9, y=9, C' at x=-9, y=3. So A'A: from A (-2,3) to A' (-9,9)? No, wait dilation is $D_{O,3}$, so (x,y)→(3x,3y). So A: let's find A's original? Wait, maybe A is (x,y)→(3x,3y). So if A' is (-9,9), then A is (-3,3)? Wait, maybe I misread coordinates. Anyway, A'A: vector from A to A' is 3x - x = 2x? No, dilation: O is center, so OA' = 3 OA, so A, O, A' are colinear? Wait, no, A'A: wait, A and A' – if dilation center is O, then lines AA' and CC' should be parallel? Wait, A: let's take A as (-3,3) (since 3*(-3)=-9, 3*3=9 → A' (-9,9)), C as (-3,1) (3*(-3)=-9, 3*1=3 → C' (-9,3)). So A: (-3,3), A': (-9,9); C: (-3,1), C': (-9,3). Then A'A: from (-3,3) to (-9,9) – slope (9-3)/(-9+3)=6/(-6)=-1. C'C: from (-3,1) to (-9,3) – slope (3-1)/(-9+3)=2/(-6)=-1/3. Wait, no, maybe I messed up. Wait, A: (-2,3)? Wait, the grid: x=-2 (A), y=3; A' at x=-9, y=9 (3*(-3)=-9? Wait, maybe A is (-3,3)? Then 3*(-3)=-9, 3*3=9 (A'). C: (-3,1) → 3*(-3)=-9, 3*1=3 (C'). So A: (-3,3), A': (-9,9); C: (-3,1), C': (-9,3). Then A'A: from (-3,3) to (-9,9) – vector (-6,6), slope -1. C'C: from (-3,1) to (-9,3) – vector (-6,2), slope 2/(-6)=-1/3. Wait, no, maybe A and A' are on the same line from O? Wait, O is (0,0). A (-3,3): line OA is y = -x (wait, no, (-3,3): y = -x? No, (-3,3): slope 3/(-3)=-1, yes. A' (-9,9): slope 9/(-9)=-1, same line. C (-3,1): line OC: slope 1/(-3)=-1/3. C' (-9,3): slope 3/(-9)=-1/3, same line. So A'A is along OA, C'C along OC? No, A'A is from A to A' (both on OA), C'C from C to C' (both on OC). Wait, OA and OC are different lines (slope -1 vs -1/3), so A'A and C'C are not parallel? Wait, no, maybe I made a mistake. Wait, A: (-3,3), A': (-9,9) – so A'A is vertical? No, x from -3 to -9 (left 6), y from 3 to 9 (up 6) – so it's a line with slope (9-3)/(-9+3)=6/(-6)=-1. C: (-3,1), C': (-9,3) – x left 6, y up 2 – slope (3-1)/(-9+3)=2/(-6)=-1/3. Not parallel. Wait, but maybe the segments are both horizontal? Wait, no. Wait, maybe the problem's A'A and C'C: A is (-2,3), A' (-9,9); C (-3,1), C' (-9,3). So A'A: from (-2,3) to (-9,9) – slope (6)/(-7)≈-0.857. C'C: from (-3,1) to (-9,3) – slope (2)/(-6)=-1/3. Not parallel. Wait, maybe the statement is wrong? Wait, no, maybe I misread coordinates. Wait, A: let's check the red triangle (A'C'B'? No, A'C'A? Wait, the red triangle is A'C'A? No, A' (top), C' (middle left), A (bottom right? No, the grid: A' is at x=-9, y=9, C' at x=-9, y=3, A at x=-2, y=3. So A'A is from A (-2,3) to A' (-9,9)? No, A is at x=-2, y=3 (right of A'), C' at x=-9, y=3 (same x as A', y=3). So A'A: from A (-2,3) to A' (-9,9) – vertical? No, x from -2 to -9 (left), y from 3 to 9 (up). C'C: from C (-3,1) to C' (-9,3) – x left 6, y up 2. Wait, maybe the segments are both horizontal? No. Wait, maybe the problem has a typo, but according to dilation, lines from A to A' and C to C' are both passing through O? No, A'A is from A to A', not through O. Wait, no, dilation: the lines AA' and CC' should be concurrent at O? Wait, A (-2,3), A' (-6,9)? Wait, maybe I miscalculated the scale factor. Wait, the grid: A is at x=-2, y=3 (so 3x=-6, 3y=9), but A' is at x=-9, y=9. Oh! Wait, the scale factor is 3, but maybe the center is not O (0,0)? Wait, the problem says $D_{O,3}$, so O is (0,0). Then (x,y)→(3x,3y). So A (-2,3) should go to (-6,9), but in the graph, A' is at (-9,9). So maybe my coordinate reading is wrong. Let's check the graph: A' is at x=-9, y=9 (9 on y-axis), A is at x=-2, y=3? No, x=-2 is two units left of y-axis, but A' is nine units left. So maybe A is at x=-3, y=3: 3*(-3)=-9, 3*3=9 (matches A' at (-9,9)). Then C is at x=-3, y=1: 3*(-3)=-9, 3*1=3 (matches C' at (-9,3)). So A: (-3,3), C: (-3,1), B: (-1,1) (since B is at x=-1, y=1, 3*(-1)=-3, 3*1=3, so B' would be (-3,3)? No, B is at x=-1, y=1, so B' should be (-3,3). Wait, but in the graph, B' is not shown, but A' is (-9,9), C' (-9,3), A (-3,3), C (-3,1), B (-1,1). So A'A: from A (-3,3) to A' (-9,9) – vector (-6,6), slope -1. C'C: from C (-3,1) to C' (-9,3) – vector (-6,2), slope -1/3. Not parallel. Wait, but maybe the segments are both horizontal? No. Wait, maybe the statement is true? Wait, no, maybe I made a mistake. Wait, A'A: from A (-3,3) to A' (-9,9) – is this vertical? No, x decreases by 6, y increases by 6. C'C: from C (-3,1) to C' (-9,3) – x decreases by 6, y increases by 2. Not parallel. So this statement is false? Wait, but maybe the answer includes this? Wait, no, let's re-examine.
- **△ABC ~ △A'AC'**: Check similarity. △ABC: A (-3,3), B (-1,1), C (-3,1) – it's a right triangle with legs 2 (horizontal: from -3 to -1) and 2 (vertical: from 1 to 3). △A'AC': A' (-9,9), A (-3,3), C' (-9,3) – right triangle with legs 6 (horizontal: from -9 to -3) and 6 (vertical: from 3 to 9). So ratios: 6/2=3, 6/2=3, and included angles are right angles. So △ABC ~ △A'AC' by SAS similarity (right angle, legs in ratio 3). True.

Step3: Confirm True Statements

From above: ∠A corresponds to ∠A' (true), △ABC ~ △A'AC' (true), and wait, what about the parallel statement? Wait, maybe I messed up coordinates. Let's recalculate A'A and C'C slopes. A (-3,3), A' (-9,9): slope = (9-3)/(-9+3) = 6/(-6) = -1. C (-3,1), C' (-9,3): slope = (3-1)/(-9+3) = 2/(-6) = -1/3. Not parallel. So only first and fifth statements? Wait, no, the fifth is △ABC ~ △A'AC'. Wait, the options:

1. ∠A corresponds to ∠A': True.
2. ∠A'AC' corresponds to ∠B: False.
3. $\overline{CB}$ corresponds to $\overline{C'A}$: False.
4. A'A || C'C: False (slopes -1 and -1/3, not equal).
5. △ABC ~ △A'AC': True (SAS similarity: right angle, legs ratio 3).

Wait, but also, is A'A parallel to C'C? Wait, maybe the segments are both horizontal? No, A'A has slope -1, C'C slope -1/3. Wait, maybe the problem's A'A and C'C are vertical? Wait, A (-3,3) to A' (-9,9): x from -3 to -9 (left 6), y from 3 to 9 (up 6) – not vertical. C (-3,1) to C' (-9,3): x left 6, y up 2 – not vertical. So parallel? No. So the true statements are: ∠A corresponds to ∠A', and △ABC ~ △A'AC'. Wait, but let's check the first statement again: in similar triangles, corresponding angles (at corresponding vertices) are equal. A and A' are corresponding vertices (since dilation maps A to A'), so ∠A (at A) corresponds to ∠A' (at A'). True.

Answer:

The true statements are:

• $\boldsymbol{\angle A}$ corresponds to $\boldsymbol{\angle A'}$.
• $\boldsymbol{\triangle ABC \sim \triangle A'AC'}$.

(So the checked boxes should be the first and fifth options.)