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6.4 — Cut by a Transversal
1. In class: A Solidify Understanding Task
Parallel lines make similar triangles
A transversal is a line that crosses other lines. Start with two transversals meeting at a point called \(A\), and draw horizontal lines across them. Each horizontal line makes a triangle with vertex \(A\).
Take one of these triangles, \(\triangle ABC\). A second horizontal segment \(DE\) is parallel to its base \(BC\), and points \(D\) and \(E\) lie on the same two rays from \(A\). The triangles share angle \(A\). Because \(BC\parallel DE\), the corresponding angles where the transversals meet these bases are equal. Thus two pairs of angles match, so the triangles are similar by AA similarity (angle-angle).
You can also compare the middle and largest triangles: \(\frac{AD}{AF}=\frac{AE}{AG}=\frac{DE}{FG}\). There are many correct proportions, but each one must pair corresponding segments in the same order.
For a separate numerical example (not measurements of the diagram above), suppose \(BC\parallel DE\), \(AB=6\), \(AD=9\), \(AC=8\), and \(BC=10\). Find \(AE\) and \(DE\).
First find the scale factor from the smaller to the larger triangle: \(\frac{AD}{AB}=\frac96=\frac32\).
\[AE=AC\cdot\frac32=8\cdot\frac32=12.\] \[DE=BC\cdot\frac32=10\cdot\frac32=15.\]Check: \(6/9=8/12=10/15=2/3\). All three corresponding side ratios agree.
Segments between the parallel lines are proportional, too
Similarity initially gives ratios of whole triangle sides. We can use those ratios to compare just the portions of the transversals between two parallel lines.
Consider a separate original set of possible lengths (not measured from the figure): \(AB=10\), \(AC=8\), \(AD=15\), and \(AE=12\).
Subtract to find the pieces between the horizontal lines:
\[BD=AD-AB=15-10=5,\qquad CE=AE-AC=12-8=4.\]Compare each piece to the portion above it:
\[\frac{BD}{AB}=\frac5{10}=\frac12,\qquad \frac{CE}{AC}=\frac48=\frac12.\]The equality works for these numbers. But a single numerical check is not a proof for every possible diagram.
Why this relationship always works
The small and large triangles are similar, so \(\frac{AD}{AB}=\frac{AE}{AC}\). Since \(AD=AB+BD\) and \(AE=AC+CE\), substitute:
We subtracted 1 from both sides in the final step. The proof does not need measured lengths; it relies only on triangle similarity and segment addition.
Apply the same reasoning to the triangles \(\triangle ADE\) and \(\triangle AFG\). Because all three are similar, the ratio of left-side to right-side whole lengths is a constant \(k=AB/AC=AD/AE=AF/AG\). Subtract corresponding whole lengths:
\[BD=AD-AB=k(AE-AC)=k\,CE,\]\[DF=AF-AD=k(AG-AE)=k\,EG.\]Dividing these equations gives \(\frac{BD}{DF}=\frac{CE}{EG}\), provided the denominators are nonzero. This is why any two adjacent strips cut from parallel lines share the same ratio.
To reproduce the picture with ordinary nested triangles, choose parallel lines on the same side of \(A\), and do not use a line through \(A\) as a triangle base. Measurements from a hand-drawn sketch may differ slightly from exact ratios; the equalities follow from the parallel-line assumption, not measurement precision. In problem 4, Tristan writes \(\frac{BD}{AB}=\frac{CE}{AC}\): the numerator segments need not themselves be full sides of a triangle.
2. Ready review — Pythagorean theorem and similar right triangles
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.
In a right triangle, the hypotenuse is the side opposite the right angle and is always the longest. If the legs are \(a,b\) and the hypotenuse is \(c\), then
\[a^2+b^2=c^2.\]If the unknown is the hypotenuse, add the squares. If the unknown is a leg, subtract the square of the other leg from the hypotenuse’s square.
The legs of a right triangle are 9 and 12. Find the hypotenuse.
\[\begin{aligned}c^2&=9^2+12^2=81+144=225\\c&=\sqrt{225}=15.\end{aligned}\]We take the positive square root because length cannot be negative.
A right triangle has hypotenuse \(\sqrt{37}\) and one leg 6. Find the other leg, \(x\).
\[\begin{aligned}x^2+6^2&=(\sqrt{37})^2\\x^2+36&=37\\x^2&=1\\x&=1.\end{aligned}\]Unlike a leg, the hypotenuse cannot be shorter than the other sides. Before solving, identify the side opposite the square corner in the packet diagram.
When two right triangles are known to be similar, corresponding legs and hypotenuses have the same scale factor. Match sides by their positions relative to the right angle, not by whether they appear horizontal or vertical.
A small right triangle has a leg of 5 and a hypotenuse of 13. A similar larger right triangle has the corresponding leg of 15. Find its hypotenuse, \(x\).
\[\frac{5}{15}=\frac{13}{x}\quad\Longrightarrow\quad 5x=15(13)\quad\Longrightarrow\quad x=39.\]Check using the scale factor: \(15/5=3\), so the hypotenuse also becomes \(13\cdot3=39\).
Check your work — Ready 1–6 answer key
- \(13\).
- \(1\).
- \(\sqrt{15}\).
- \(3\).
- \(\frac c{\sqrt{29}}=\frac{15}{5}\), so \(c=3\sqrt{29}\).
- \(\frac{x}{12}=\frac{26}{13}\), so the missing leg is \(24\).
3. Set — Write and solve proportions for parallel transversals
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.
First mark each pair of corresponding segments—the stretches lying between the same two parallel lines. Ratios only remain valid when the top, middle, and bottom points are paired consistently.
For the nested triangles in the In class section, three valid whole-side ratios are
\[\boxed{\frac{AB}{AD}=\frac{AC}{AE}=\frac{BC}{DE}}.\]With three parallel levels and two transversals, you can also compare the two consecutive gaps: \(BD/DF=CE/EG\). This remains true if the transversals intersect outside the region shown: the intercept theorem depends on the parallel lines, not on the point of intersection being visible.
Two useful arrangements of three equal ratios
In Set problem 7, the upper transversal has consecutive segments named \(a\) and \(b\), and the lower has matching consecutive segments \(d\) and \(c\). One useful set of three equal ratios is
\[\boxed{\frac{a}{d}=\frac{b}{c}=\frac{a+b}{d+c}}.\]The third fraction compares the combined lengths, not an additional small segment. For instance, if \(a=4,\ b=6,\ d=3,\ c=4.5\), all three fractions equal \(\tfrac43\).
In Set problem 8, the two transversals meet at \(B\) below three parallel horizontal lines. Here the three equal ratios may instead compare the full distances from \(B\) to the three parallel levels:
\[\boxed{\frac{BA}{BC}=\frac{BF}{BD}=\frac{BG}{BE}}.\]Other equivalent proportions are possible. For both arrangements, confirm which points lie on each transversal and pair corresponding positions carefully.
Suppose \(AB=4\), \(AD=10\), \(AC=5\), and \(BC=6\), where \(BC\parallel DE\). The larger triangle is \(10/4=2.5\) times as large.
Consequently \(AE=5(2.5)=12.5\) and \(DE=6(2.5)=15\). Three equal ratios are
\[\frac{AB}{AD}=\frac{AC}{AE}=\frac{BC}{DE}\quad\Longrightarrow\quad\frac4{10}=\frac5{12.5}=\frac6{15}.\]For missing lengths, decide whether the labels represent full lengths measured from the apex or the short segments lying between two parallel lines. Do not plug a complete side into a proportion intended for a short segment.
In the figure above, \(AB=4\), \(AC=5\), \(BD=6\), and \(CE=x\). Use the ratio of two corresponding parts:
\[\frac{BD}{AB}=\frac{CE}{AC}\quad\Longrightarrow\quad\frac64=\frac{x}{5}.\]Cross-multiply: \(4x=30\), so \(\boxed{x=7.5}\).
Alternatively, \(AD=4+6=10\), so similarity gives \(AE/AC=AD/AB=10/4\); \(AE=12.5\), and \(CE=AE-AC=12.5-5=7.5\).
Three parallel lines cross two transversals. Between lines 1 and 2 the left segment is 2 and the right is 3. Between lines 2 and 3 the left segment is \(x\) and the right is 9. Find \(x\).
\[\frac{2}{x}=\frac{3}{9}\quad\Longrightarrow\quad 3x=18\quad\Longrightarrow\quad\boxed{x=6}.\]Both fractions compare the upper strip to the lower strip. A common mistake is to flip one fraction but not the other.
Check your work — Set 7–14 answer key
- Assuming the three nearly vertical lines are parallel, \(\frac ad=\frac bc=\frac{a+b}{c+d}\). Parallelism is suggested but not marked; these equalities need that given.
- \(\frac{BA}{BC}=\frac{BF}{BD}=\frac{BG}{BE}\), since the three cross-segments are marked parallel.
- Assuming the three upward-sloping lines are parallel, \(\frac25=\frac n{6.5}\), so \(n=2.6\). The diagram does not mark this parallelism.
- \(\frac23=\frac k8\), so \(k=\frac{16}{3}\).
- Intended parallel lines: \(BC\parallel ED\); then \(\triangle ABC\sim\triangle AED\) by AA. This parallelism is not given by markings, so the drawing alone does not prove it.
- Intended parallel lines: \(AB\parallel FG\parallel DE\); then \(\triangle CBA\sim\triangle CFG\sim\triangle CDE\) by AA. The photograph does not establish exact parallelism or collinearity; these must be assumed for the model.
- Intended parallel lines: \(GF\parallel IH\); then \(\triangle JFG\sim\triangle JHI\) by AA. Parallelism is not marked.
- Intended parallel lines: \(KI\parallel FJ\) and \(LH\parallel GJ\). With those assumptions, \(\triangle GKI\sim\triangle GFJ\sim\triangle LFH\) by AA. No extra similarity involving \(HI\) follows without an additional parallel-line given.
4. Set — Identify parallel segments and similar triangles
For each diagram, work backward from matching angles: a transversal crossing parallel segments forms equal corresponding angles. Two equal angle pairs prove AA similarity. Name the triangles in a consistent order.
In a triangle \(\triangle ABC\), points \(D\) and \(E\) lie on \(AB\) and \(AC\), respectively, and \(DE\parallel BC\).
Angle \(A\) is shared, angle \(D\) corresponds to angle \(B\), and angle \(E\) corresponds to angle \(C\). Thus
\[\boxed{\triangle ADE\sim\triangle ABC}.\]Check the vertex order. Writing \(\triangle ADE\sim\triangle ACB\) would incorrectly pair \(D\) with \(C\).
Some packet drawings contain more than two similar triangles. Identify one parallel pair at a time, and begin each similarity statement with the same apex, where practical. For the photo in problem 12, use the overlaid lettered white lines as the intended geometric construction; visual parallelism in a perspective photo by itself is not a rigorous given.
Several figures use strokes that look parallel but do not explicitly mark or state every needed parallel relationship. A diagram’s appearance does not prove that lines are exactly parallel, so the indicated proportions and triangle similarities are not logically guaranteed by an unmarked drawing alone. These questions are evidently intended to assume the relevant near-parallel segments really are parallel. Treat that intended assumption as part of the problem, especially for the photographed example in problem 12.
5. Go review — Similar slope triangles always give the same slope
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.
A line’s slope measures vertical change divided by horizontal change:
\[m=\frac{\text{rise}}{\text{run}}=\frac{y_2-y_1}{x_2-x_1}.\]A slope triangle consists of a horizontal run, a vertical rise (or fall), and the segment of the line joining their ends. Each such triangle is right-angled. Along one nonvertical straight line, these right triangles have the same acute angles, so they are similar. Their corresponding rises and runs scale equally, leaving the ratio unchanged.
A line passes through \((1,2)\) and \((7,5)\). Draw (or imagine) two slope triangles, one with run 2 and one with run 6. What are their signed rises and slopes?
Step 1: From the two points, the total rise is \(5-2=3\) and the run is \(7-1=6\), so \(m=3/6=1/2\).
Step 2: A run of 2 is one-third of 6, so the rise must be 1. Its slope is \(1/2\).
Step 3: With run 6 and rise 3, the slope is again \(3/6=1/2\). Both triangles trace the same straight line.
On a line sloping downward, a triangle with a horizontal run of 4 has a signed rise of −3. A larger triangle has a horizontal run of 12. Find its signed rise and both slopes.
Step 1: The small slope is \(m=-3/4\).
Step 2: The larger run is \(12/4=3\) times the small run. Similarity multiplies the rise by 3 as well: \(-3\cdot3=-9\).
Step 3: The large slope is \(-9/12=-3/4\), just like the small slope.
Common error: writing \(3/4\) drops the minus sign. A downward segment has a negative signed rise when you move right.
A line passes through \((-3,7)\) and \((6,1)\). Find its slope, then describe a second, smaller slope triangle with integer side changes.
\[m=\frac{1-7}{6-(-3)}=\frac{-6}{9}=-\frac23.\]Reduce both signed changes by a factor of 3: a run of 3 and a fall of 2 forms a second slope triangle. Its slope is \(-2/3\).
Check your work — Go 15–18 answer key
- Using equal horizontal and vertical grid units, slope \(\frac12\); rise/run examples \(\frac12=\frac24=\frac36\). Plot/construction not shown.
- Using equal grid units, slope \(3\); examples \(\frac31=\frac62=\frac93\). Plot/construction not shown.
- Using equal grid units, slope \(-\frac14\); examples \(\frac{-1}{4}=\frac{-2}{8}=\frac{-3}{12}\). Plot/construction not shown.
- Using equal grid units, slope \(-\frac32\); examples \(\frac{-3}{2}=\frac{-6}{4}=\frac{-9}{6}\). Plot/construction not shown.
Additional Learning Resources
| Resource and direct link | What it covers / why use it | Scope / What to skip |
|---|---|---|
| Khan Academy — Proof: Parallel lines divide triangle sides proportionally | Worked proof closely matching the difference-of-segments argument in the main task. | Close match. The full proof is suitable for Main 4–6. |
| Khan Academy — Applying the triangle proportionality theorem | A practical example of multiple parallel lines crossing roads, much like Set 7–10. | Close match. Watch the example involving intersecting parallel roads; no trigonometry is needed. |
| Paul’s Online Math Notes — Algebra, 3.2: Lines | Written reference with worked examples for slope, signed rise/run, and the two-point slope formula (Go 15–18). | Some extra material. Focus on the opening slope definition and initial worked slope examples; skip later point-slope and perpendicular-line equations for today. |
| Paul’s Online Math Notes — Lines practice problems | Extra practice for computing the slope from pairs of points, with worked solution links. | Some extra material. Try problems 1–2 on slope; later equation-writing and line-classification problems are outside today’s focus. |
| Khan Academy — Proving slope is constant using similarity | A fuller geometric explanation of why every slope triangle on a line has the same rise/run. | Close match. Optional deeper proof for Go 15–18. |
The resource links require an internet connection. The lesson text and diagrams work offline; mathematical typesetting requires an internet connection. No closely matched Professor Leonard lecture on this specific theorem was confirmed, so no broader lecture is assigned.
Teacher review — Substantive packet issues
1. Parallelism is essential but inconsistently specified in the Set figures (7, 9, 11–14; pp. 23–24). Some intended parallel lines are not explicitly identified by arrow marks or statements. Without parallelism, the conclusions of similar triangles and proportional segments do not follow. Suggested correction: state the relevant pairs of parallel lines as given, especially for the real-world photograph in question 12.
No further serious mathematical errors or ambiguities were identified. The main task uses approximate measurement as an intermediate check and then explicitly requests an algebraic proof; that sequence is not itself a mathematical error.
Section 6.4 packet-to-lesson coverage
| Packet assignment | Expected skill | Where it is taught |
|---|---|---|
| Main 1–3 · p. 20 | AA similarity of triangles made by parallel lines; correct proportionality statements with three nested triangles | In class |
| Main 4–6 and proof · p. 21 | Proportional segments lying between parallels; numerical check and general algebraic proof | In class |
| Ready 1–4 · p. 22 | Pythagorean theorem, unknown leg or hypotenuse, radicals | Section 2 |
| Ready 5–6 · p. 22 | Set up proportions from similar right triangles; match sides and solve | Section 2 |
| Set 7–8 · p. 23 | Find matching segments and write three equal ratios | Section 3 |
| Set 9–10 · p. 23 | Solve for unknown portions cut off by parallel lines | Section 3 |
| Set 11–12 · p. 23; 13–14 · p. 24 | Find parallel pairs; write similarly ordered triangle names | Section 4 |
| Go 15–18 · p. 24 | Different-sized right slope triangles, signed rise/run, same slope | Section 5 |
All illustrations and worked numerical examples are newly created; they are not measurements of the MVP packet’s printed drawings. This lesson stops with proportions from parallel lines and slope triangles; the applications and more involved angle problems of Section 6.5 are not introduced.
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