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6.2 — Triangle Dilations
1. In class: A Solidify Understanding Task
Constructing a dilation from a given center
A dilation changes all distances from a fixed point by the same scale factor. The fixed point is the center of dilation. A figure before the dilation is the pre-image; the new figure is its image.
For a positive scale factor \(k\), a point \(P\) is sent to \(P'\) on the ray from the center \(O\) through \(P\), with
When \(k>1\), every point (other than the center) moves farther from \(O\); when \(0<k<1\), it moves closer. The center itself stays fixed.
To dilate a polygon: draw a ray from the center through each vertex, mark its image at the appropriate multiple of its original distance, and connect the image vertices in corresponding order.
Point \(A\) is 3 cm from center \(O\); point \(B\) is 5 cm away. Dilate a triangle with scale factor \(k=3\).
1. Extend the ray \(OA\) and place \(A'\) 9 cm from \(O\).
2. Extend \(OB\) and place \(B'\) 15 cm from \(O\).
3. Do the same for \(C\); connect \(A',B',C'\).
The new side lengths are three times their matching original lengths. In the packet's Problem 1, the center is \(M\), not the origin of a coordinate system; that does not change the method.
With \(k=\tfrac12\), mark each image vertex halfway along the segment from \(O\) to the original vertex. A point 8 cm from \(O\) goes to a point 4 cm from \(O\); a side 6 cm long becomes 3 cm.
In the packet's Problem 2, use the center \(N\). The original triangle has not changed: it is the center and the scale factor that determine where the new triangle appears.
Suppose a pentagon has two vertices \(P\) and \(Q\) that are 12 cm and 20 cm from center \(C\). For a reduction by \(\tfrac14\), place \(P'\) at 3 cm and \(Q'\) at 5 cm along rays \(CP\) and \(CQ\). Continue for all remaining vertices, then join them in order. No special rule is needed for pentagons.
Corresponding sides and proportional relationships
A dilation scales every side by the same factor. In particular, if \(\triangle ABC\) is dilated by \(k=3\) to \(\triangle A'B'C'\), then
\[\frac{A'B'}{AB}=\frac{B'C'}{BC}=\frac{A'C'}{AC}=3.\]All fractions compare corresponding sides in the same direction: enlarged length over original length. Turning every fraction upside down gives the equally valid common ratio \(\tfrac13\).
Suppose the original triangle has sides \(AB=4\), \(BC=5\), \(AC=3\). The enlarged copy uses \(k=3\); the reduced copy uses \(k=\tfrac12\).
| Side | Original | Enlarged \(k=3\) | Reduced \(k=\tfrac12\) |
|---|---|---|---|
| \(AB\) | 4 | 12 | 2 |
| \(BC\) | 5 | 15 | 2.5 |
| \(AC\) | 3 | 9 | 1.5 |
Enlarged to original: \(\tfrac{12}{4}=\tfrac{15}{5}=\tfrac93=3\).
Reduced to original: \(\tfrac24=\tfrac{2.5}{5}=\tfrac{1.5}{3}=\tfrac12\).
Enlarged to reduced: \(\tfrac{12}{2}=\tfrac{15}{2.5}=\tfrac9{1.5}=6\).
Side comparisons within the same triangle: \(\tfrac{AB}{AC}=\tfrac43=\tfrac{A'B'}{A'C'}=\tfrac{12}{9}\), and likewise \(\tfrac{AB}{BC}=\tfrac45=\tfrac{A'B'}{B'C'}=\tfrac{12}{15}\).
These equalities supply many valid proportionality statements, with different choices of corresponding side pairs and orientations. The packet asks for at least ten; rather than guessing ten unrelated equations, systematically use the three side pairs and the three figure-to-figure comparisons.
What happens to lines and angles?
A dilation with positive scale factor preserves angle measures and sends every line either to a parallel line or to the same line. The distinction depends on where the center lies.
• If a line does not pass through the center of dilation, its image is a different, parallel line.
• If the original line does pass through the center, its image lies on the very same line.
For a positive factor, the direction of a side is unchanged, and its length is multiplied by \(k\).
Place the center \(O\) at \((0,0)\) and the endpoints of a side at \(A=(2,0)\) and \(B=(4,0)\). If \(k=\tfrac32\), then \(A'=(3,0)\) and \(B'=(6,0)\). Both \(AB\) and \(A'B'\) lie on the x-axis.
The two supporting lines coincide. In school geometry, which usually requires parallel lines to be distinct, this is not a pair of parallel lines. A side not through \(O\), however, maps to a genuinely parallel line.
“Are corresponding segments parallel?” has a convention-sensitive edge case. A dilation sends the line containing each side to either a distinct parallel line or itself. If your class uses “parallel” only for distinct coplanar lines, the answer to “never, sometimes, always” is sometimes: the exception is a side whose line passes through the center. If coincident lines are also treated as parallel in the classroom's terminology, that answer changes. State the convention rather than presenting an unconditional rule.
A proof using dilation: the triangle midsegment theorem
The midpoint of a segment divides it into two equal parts. A midsegment of a triangle joins the midpoints of two sides.
The segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long.
Start with \(\triangle ABC\) and let \(D\) be the midpoint of \(AB\) and \(E\) the midpoint of \(AC\).
Step 1. Dilate triangle \(ABC\) about vertex \(A\), using \(k=\tfrac12\). Since \(AD=\tfrac12 AB\) and \(AE=\tfrac12 AC\), the new triangle's vertices are precisely \(A,D,E\).
Step 2. A dilation sends side \(BC\), whose line does not pass through \(A\), to side \(DE\) on a distinct parallel line. Hence \(DE\parallel BC\).
Step 3. The side-length scale factor is \(\tfrac12\), so \(DE=\tfrac12 BC\). Both claims are proved.
The packet uses the equivalent reverse construction: expand a triangle about \(A\) by \(k=2\). The original two outer vertices then become the midpoints of the sides of the larger triangle.
If \(D\) and \(E\) are midpoints of two sides of a triangle and the third side is 18 cm, then \(DE=\tfrac12(18)=9\) cm. If a midsegment is 6.5 cm, the third side is \(2(6.5)=13\) cm.
2. Ready review — Recognize basic angle relationships
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
When two lines intersect, vertical angles lie opposite one another and have equal measures. A linear pair consists of adjacent angles whose outside rays form a straight line; its measures total \(180^\circ\).
When a line (a transversal) crosses two other lines, angles between the two lines are interior. A pair of interior angles on opposite sides of the transversal is alternate interior; a pair on the same side is same-side interior. When the two crossed lines are parallel, alternate interior angles are equal and same-side interior angles are supplementary.
| Relationship | How to recognize it | What follows |
|---|---|---|
| Vertical angles | Opposite one another where two lines cross | Equal measures |
| Linear pair | Adjacent angles that together make a straight angle | Sum to \(180^\circ\) |
| Complementary | Two angles that form a right angle or otherwise add to \(90^\circ\) | Sum to \(90^\circ\) |
| Alternate interior | Between two lines, opposite sides of their transversal | Equal if the two lines are parallel |
| Same-side interior | Between two lines, same side of their transversal | Sum to \(180^\circ\) if parallel |
| Triangle sum | Three interior angles of any triangle | Sum to \(180^\circ\) |
Two opposite angles at an intersection are labeled \(2x+10\) degrees and \(4x-30\) degrees. They are vertical angles, so
\[2x+10=4x-30\quad\Rightarrow\quad 40=2x\quad\Rightarrow\quad x=20.\]Each angle is \(50^\circ\). The equation works because they are vertical, not merely because they look equal in the picture.
Angles \(a\) and \(b\) split a right angle. If \(a=37^\circ\), then \(b=90^\circ-37^\circ=53^\circ\). In a separate triangle, if two angles are \(48^\circ\) and \(61^\circ\), the third is \(180^\circ-48^\circ-61^\circ=71^\circ\).
The diagram appears to intend a perpendicular horizontal and vertical line, with the slanted ray splitting a right angle. That is why the intended name is complementary angles. The drawing does not explicitly mark the right angle, however; angle sum \(90^\circ\) requires that assumption.
Check your work — Ready 1–6 answer key
- a — Alternate interior angles.
- b — Vertical angles.
- e — Linear pair.
- f — Same side interior angles.
- c — Complementary angles (the intended vertical and horizontal rays are perpendicular).
- d — Triangle sum theorem.
3. Set — Find the center and scale factor from two figures
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
For the shared method, revisit constructing a dilation in the In class section.
When the original figure and its image are both supplied, work backward. Connect a vertex and its corresponding image vertex, such as \(P\) to \(P'\). Do the same for a second pair. The full lines usually meet at the dilation center. Then divide a matching image-side length by the pre-image-side length.
Use the same ratio for a second vertex pair and a second side. For the positive scale factors used here, each original vertex and its image must lie on the same ray from the center.
Two triangles have corresponding sides 4 cm and 6 cm. The line through \(A,A'\) intersects the line through \(B,B'\) at \(O\). Measurements give \(OA=5\) cm and \(OA'=7.5\) cm.
First, the side-length ratio is \(k=6/4=3/2\). Second, \(OA'/OA=7.5/5=3/2\), so this pair is consistent. Check that \(B'\) and \(C'\) also lie at 1.5 times the center distances. The dilation center is \(O\), and the image is an enlargement by \(\tfrac32\).
If a triangle and its reduced image share vertex \(A\), and the other image vertices lie halfway along rays \(AB\) and \(AC\), the center is \(A\). In particular, \(A'=A\) and \(k=\tfrac12\). This is a useful pattern for the nested triangles in Set Problem 14.
If traced corresponding-point lines do not meet consistently, or different sides give different ratios, the two drawings do not establish a single exact dilation. Printed diagrams may be schematic; confirm the intended geometry with the labeled correspondence rather than trusting approximate measurements alone.
Check your work — Set 7–14 answer key
- Plot/construction not shown. For every vertex \(P\), put \(P\prime\) on ray \(CP\) with \(CP\prime=2CP\), then join the image vertices.
- Plot/construction not shown. Each image vertex is the midpoint of the segment from \(C\) to its original vertex: \(CP\prime=\frac12CP\).
- Plot/construction not shown. Each image vertex lies on ray \(CP\) with \(CP\prime=3CP\).
- Plot/construction not shown. Each image vertex lies on ray \(CP\) with \(CP\prime=\frac14CP\).
- Center: the common intersection of the extended lines \(CC\prime,DD\prime,EE\prime\), to the upper left of the two triangles. The drawing suggests \(k\approx\frac12\). No lengths or scale are labeled, so that numerical factor is an estimate, not an exact given.
- Center: the common intersection of the lines joining corresponding vertices, above and to the left of the houses. The drawing suggests \(k\approx3\); exact lengths are not given.
- Center \(G\); the drawing suggests \(k\approx\frac14\). The ratios \(\frac{GH\prime}{GH},\frac{GI\prime}{GI},\frac{GJ\prime}{GJ}\) should agree; an exact numerical factor is not labeled.
- Center \(A\); the drawing suggests \(k\approx\frac13\). The ratios \(\frac{AB\prime}{AB}\) and \(\frac{AC\prime}{AC}\) should agree; an exact numerical factor is not labeled.
4. Go review — Classify and justify a transformation
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Earlier transformation work distinguishes four actions. A translation slides every point by one vector; a reflection flips a figure across a line; a rotation turns a figure through a fixed angle about a fixed point; a dilation scales distances from a fixed point.
| Transformation | What to verify, not just notice |
|---|---|
| Translation | Every corresponding point moves the same distance in the same direction. |
| Reflection | The same mirror line is the perpendicular bisector of each segment connecting corresponding points (except fixed points on the mirror). |
| Rotation | All corresponding points have equal distances to one center, and their turns around it have the same directed angle. |
| Dilation | All corresponding points lie on center-to-point rays and satisfy one common positive distance ratio; sides are scaled consistently. |
A triangle has vertices \(P=(0,0)\), \(Q=(3,0)\), \(R=(0,2)\). Its image has \(P'=(4,-1)\), \(Q'=(7,-1)\), \(R'=(4,1)\). Identify and prove the transformation.
Solution. Subtract corresponding coordinates:
\[P'-P=(4,-1),\quad Q'-Q=(4,-1),\quad R'-R=(4,-1).\]Every vertex moved 4 units right and 1 unit down. Thus the mapping is the translation \((x,y)\mapsto(x+4,y-1)\); the equal displacement of all three vertices establishes the claim.
Triangle \(PQR\) has \(P=(1,0)\), \(Q=(3,0)\), \(R=(1,2)\). Its image has \(P'=(0,1)\), \(Q'=(0,3)\), \(R'=(-2,1)\). Find and justify the transformation.
Solution. Each point follows the same quarter-turn about the origin:
\[(1,0)\mapsto(0,1),\quad(3,0)\mapsto(0,3),\quad(1,2)\mapsto(-2,1).\]The distances to the origin stay unchanged: \(1\) to \(1\), \(3\) to \(3\), and \(\sqrt5\) to \(\sqrt5\). Each point turns \(90^\circ\) counterclockwise. Therefore it is a \(90^\circ\) counterclockwise rotation about \((0,0)\).
Reflect \(P=(1,1)\), \(Q=(0,3)\), \(R=(1,4)\) across the line \(x=2\). The image vertices are \(P'=(3,1)\), \(Q'=(4,3)\), \(R'=(3,4)\). Explain why this is a reflection.
Solution. Each original and image point has the same \(y\)-coordinate. The midpoint of each horizontal joining segment has \(x=2\): for example, midpoint \(PP'=(2,1)\), midpoint \(QQ'=(2,3)\). A vertical line is perpendicular to every horizontal joining segment, so \(x=2\) is the perpendicular bisector of each. That establishes the reflection.
Center \(O=(1,1)\). Original vertices \(A=(3,3)\), \(B=(7,3)\), \(C=(3,5)\) correspond to \(A'=(2,2)\), \(B'=(4,2)\), \(C'=(2,3)\). Identify and justify the transformation.
Solution. From \(O\), the vectors to original vertices are \((2,2)\), \((6,2)\), \((2,4)\). The image vectors are \((1,1)\), \((3,1)\), \((1,2)\), exactly half as large. Hence the center is \(O\) and \(k=\tfrac12\). As a second check, \(AB=4\) while \(A'B'=2\).
Check your work — Go 15–18 answer key
- Rotate \(90^\circ\) clockwise about the grid intersection one square right of \(C\). Relative to this center, \(D=(-7,0),B=(-3,-2),C=(-1,0)\) map by \((x,y)\mapsto(y,-x)\) to \((0,7),(-2,3),(0,1)\), the primed vertices. Plot/construction not shown.
- Reflect across the line of slope 1 through the grid intersection one square below \(A\). Relative to that intersection, the reflection is \((x,y)\mapsto(y,x)\): \(A=(0,1),B=(2,4),C=(0,9),D=(-2,4)\) map to \(A\prime=(1,0),B\prime=(4,2),C\prime=(9,0),D\prime=(4,-2)\). Plot/construction not shown.
- Translate 7 grid squares right and 1 square up. Each corresponding vertex has the same displacement, so the translation maps every segment as well. Plot/construction not shown.
- Dilate by factor \(\frac12\) about the center of the pentagons (intersection of the lines joining corresponding vertices). All five image vertices lie halfway from that center to their originals. Plot/construction not shown.
Additional Learning Resources
These are optional supplements targeted to the day's skills. The links require an internet connection, the lesson text and diagrams work offline, while mathematical typesetting requires an internet connection.
| Resource and direct link | What it teaches / when to use it | Scope / What to skip |
|---|---|---|
| Khan Academy — Dilating points | A 3-minute walkthrough of finding image points from a given center and factor; supports Main 1–2 and Set 7–10. | Close match. Focus on tracing the rays and measuring center-to-vertex distances. |
| Khan Academy — Dilations: center | Find the dilation center from corresponding points and check their relative distances; supports Set 11–14. | Close match. The full short video stays within this lesson's geometry. |
| Math Open Reference — Midsegment of a Triangle | Written explanation and an interactive figure showing why a triangle midsegment is parallel to and half the third side. | Close match. Use the main property and illustration; construction links are optional. |
| Khan Academy — Identifying transformations | Worked visual comparisons of translations, rotations, reflections and dilations; supports Go 15–18. | Close match. To meet this packet's proof requirement, also use the evidence checklist above. |
Teacher review — Substantive packet issues
1. Main task, Problems 4–5 (p. 7): “never, sometimes, always parallel.” A dilation maps each side's supporting line either to a distinct parallel line or onto itself. Under the usual school definition that parallel lines are distinct, some corresponding sides are coincident rather than parallel when the center lies on the supporting line. The task should explicitly distinguish “parallel or coincident” from “parallel.” A precise revision would ask students to determine when the supporting lines are parallel and distinct, and when they coincide.
2. Ready, Problem 5 (p. 9): missing right-angle marking. The intended answer is complementary angles, apparently using a horizontal and vertical ray that are perpendicular. A true right angle is not marked, so the sum of 90° relies on visual convention instead of an explicit given. Mark the right angle or say that the two rays are perpendicular.
Other construction and identification exercises rely on drawing and measurement rather than stated numerical data; students should check their sketches rather than infer exact ratios from apparent proportions alone.
Coverage map
| Original packet | Objective | Instruction here |
|---|---|---|
| Main 1–2 · p. 6 | Construct triangle dilations, scale factors 3 and 1/2, distinct centers | In class |
| Main 3 · p. 7 | Find proportional relationships among three triangles | In class |
| Main 4–5 · p. 7 | Parallel or coincident supporting lines under dilation | In class |
| Main 6 · p. 8 | Prove midpoint theorem from a dilation | In class |
| Ready 1–6 · p. 9 | Recognize six angle relationships | Section 2 |
| Set 7–10 · p. 10 | Dilate triangles, parallelogram, pentagon about a named center | In class |
| Set 11–14 · p. 11 | Recover center and scale factor from corresponding figures | Section 3 |
| Go 15–18 · p. 12 | Classify and justify translations, rotations, reflections, dilations | Section 4 |
The supplementary examples and figures are original illustrations, not reproductions of the packet's numbered drawings. The lesson deliberately stops before the general polygon-similarity conjectures introduced in Section 6.3.
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