Geometry · G.SRT.B.4

Triangle proportionality theorem

A line crosses a triangle parallel to one of its sides. How does it share out the other two sides?

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Triangle proportionality theorem

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Why this exists

Similar triangles compare a whole side with a whole side. Often the lengths you can actually get at are not whole sides. A fence runs across a triangular field parallel to one edge, and what you can pace out is the stretch from the corner to the fence and the stretch from the fence to the far edge. The lines of ruled paper cross a slanting pencil line, and what you see is the run of pieces between them.

This theorem is about the pieces. A line parallel to one side of a triangle cuts the other two sides in the same ratio, and the converse says that cutting them in the same ratio is exactly what makes a line parallel. Together they turn “parallel” into a statement about lengths, and a statement about lengths back into “parallel”. Take the triangle away and the same fact says that a family of parallel lines cuts every line across it in the same ratios, which is how a segment can be split into equal parts without measuring it.

Definition

Let \triangle ABC be a triangle, and let a line meet side \overline{AB} at D and side \overline{AC} at E, each strictly between the vertices. The line divides the two sides proportionally when the piece at A compares with the piece at the base in the same ratio on both sides:

\frac{AD}{DB} = \frac{AE}{EC}

Triangle proportionality theorem. If \overline{DE} \parallel \overline{BC}, then \overline{DE} divides \overline{AB} and \overline{AC} proportionally.

It is also called the side splitter theorem. In Figure 1 the arrowheads mark the two parallel segments, and the shaded \triangle ADE is the small triangle the parallel cuts off.

Figure 1\overline{DE} \parallel \overline{BC} cuts each of the other two sides into a piece at A and a piece at the base.

Each ratio compares two pieces of the same side: the piece touching A over the piece touching the base. The segment \overline{DE} is not a piece of either side, and it does not appear in this proportion.

Intuition

Shrink the triangle toward A. A dilation with center A and scale factor k, where 0 < k < 1, moves B to a point D on \overline{AB} with AD = k \cdot AB, and moves C to a point E on \overline{AC} with AE = k \cdot AC. A dilation sends a line that misses its center to a parallel line (what a dilation does to a line), so \overline{DE} \parallel \overline{BC}.

Now read off the pieces. On both sides the piece at A is the fraction k of the side, and the piece at the base is the rest, the fraction 1 - k:

\begin{aligned} \frac{AD}{DB} &= \frac{k \cdot AB}{(1 - k) \cdot AB} = \frac{k}{1 - k} \\ \frac{AE}{EC} &= \frac{k \cdot AC}{(1 - k) \cdot AC} = \frac{k}{1 - k} \end{aligned}

The lengths of the two sides cancel, which is why the sides need not be equal. Every parallel to \overline{BC} that crosses the triangle is one of these shrunken copies, one for each k, so every such parallel cuts both sides at the same fraction of the way down.

Ruled paper shows the same thing without a triangle. Its lines are parallel, and they cut a line drawn straight down the page into equal pieces. Lay a ruler across them at a slant and the pieces along the ruler come out longer, but still equal to each other. The slant changes the lengths; it cannot change their ratio.

Results

Triangle proportionality theorem. In \triangle ABC, if D lies on \overline{AB}, E lies on \overline{AC} and \overline{DE} \parallel \overline{BC}, then

\frac{AD}{DB} = \frac{AE}{EC}

Converse. If D lies strictly between A and B, E lies strictly between A and C, and \displaystyle \frac{AD}{DB} = \frac{AE}{EC}, then \overline{DE} \parallel \overline{BC}.

Equivalent forms. Each of these proportions implies the other two, so any of them can be used in the theorem or in its converse:

\frac{AD}{DB} = \frac{AE}{EC} \qquad \frac{AD}{AB} = \frac{AE}{AC} \qquad \frac{DB}{AB} = \frac{EC}{AC}

The second and third compare a piece with its whole side. The segment \overline{DE} enters only through the similar triangles: \triangle ADE \sim \triangle ABC, so

\frac{DE}{BC} = \frac{AD}{AB} = \frac{AE}{AC}

Parallel lines cut transversals proportionally. If three parallel lines cross a line s at P, Q, R and a line t at X, Y, Z, in that order, then

\frac{PQ}{QR} = \frac{XY}{YZ}

With more parallel lines, each pair of neighboring pieces on s is in the same ratio as the matching pair on t, so the pieces on t are the pieces on s all multiplied by one factor.

The midpoint case. When D is the midpoint of \overline{AB} the common ratio is 1, so E is the midpoint of \overline{AC}, and DE = \tfrac12 BC. That special case is the midsegment theorem.

Why it is true

The full argument is its own lesson, proving the triangle proportionality theorem. Here is its outline.

The theorem. With \overline{AB} as a transversal of the parallels \overline{DE} and \overline{BC}, \angle ADE and \angle ABC are corresponding angles, so they are congruent (parallel lines and transversals). The triangles ADE and ABC also share \angle A. Two pairs of congruent angles make them similar by AA similarity, so their matching sides are in one ratio:

\frac{AB}{AD} = \frac{AC}{AE}

Subtract 1 from both sides, writing it as \displaystyle \frac{AD}{AD} on the left and \displaystyle \frac{AE}{AE} on the right:

\frac{AB - AD}{AD} = \frac{AC - AE}{AE}

By segment addition AB - AD = DB and AC - AE = EC, so \displaystyle \frac{DB}{AD} = \frac{EC}{AE}. Equal numbers have equal reciprocals, which gives \displaystyle \frac{AD}{DB} = \frac{AE}{EC}.

The converse. Run the algebra backwards. From \displaystyle \frac{AD}{DB} = \frac{AE}{EC}, take reciprocals and add 1 to both sides to reach \displaystyle \frac{AB}{AD} = \frac{AC}{AE}. Now the triangles ADE and ABC share the angle at A, and the sides around it are in one ratio, so they are similar by SAS similarity. Hence \angle ADE \cong \angle ABC, and congruent corresponding angles make \overline{DE} \parallel \overline{BC}.

Three parallel lines. In Figure 2, draw the diagonal \overline{PZ}; it crosses the middle line at G. In \triangle PRZ, the segment \overline{QG} lies on the middle line, so it is parallel to \overline{RZ}, and the theorem gives \displaystyle \frac{PQ}{QR} = \frac{PG}{GZ}. In \triangle ZXP, the segment \overline{YG} is parallel to \overline{XP}, and the theorem with Z as the shared vertex gives \displaystyle \frac{ZY}{YX} = \frac{ZG}{GP}, which turned over reads \displaystyle \frac{XY}{YZ} = \frac{PG}{GZ}. Both ratios equal \displaystyle \frac{PG}{GZ}, so they equal each other.

Figure 2The dashed diagonal splits the picture into two triangles, one for each transversal.

Worked examples

Example 1

Finding the fourth piece

In Figure 3, \overline{DE} \parallel \overline{BC}, with AD = 4, DB = 6 and AE = 5. Find x = EC.

Figure 3Three pieces are known. The parallel fixes the fourth.

Solution

The parallel divides the two sides proportionally, and the proportion pairs the piece at A over the piece at the base on each side:

\frac{AD}{DB} = \frac{AE}{EC} \qquad\text{so}\qquad \frac{4}{6} = \frac{5}{x}

Multiply both sides by 6x, which is allowed because x is a length and so is not zero:

\begin{aligned} 4x &= 30 \\ x &= 7.5 \end{aligned}

So EC = 7.5.

Check. \displaystyle \frac{5}{7.5} = \frac{2}{3} and \displaystyle \frac{4}{6} = \frac{2}{3}. The base piece is the longer one on both sides, as it should be when both ratios are the same number below 1.

Example 2

A whole side and a piece

In \triangle ABC, \overline{DE} \parallel \overline{BC}, with D on \overline{AB} and E on \overline{AC}. AD = 9, AB = 15, AC = 20 and BC = 30. Find AE, EC and DE.

Solution

A whole side is given rather than two pieces, so use the equivalent form that compares a piece with its whole side:

\frac{AD}{AB} = \frac{AE}{AC} \qquad\text{so}\qquad \frac{9}{15} = \frac{AE}{20}

Multiply by 20: \displaystyle AE = 20 \cdot \frac{9}{15} = 12. By segment addition, EC = 20 - 12 = 8.

\overline{DE} is not a piece of a side, so it comes from the similar triangles, \triangle ADE \sim \triangle ABC, whose matching sides are in the ratio \displaystyle \frac{AD}{AB}:

\frac{DE}{BC} = \frac{AD}{AB} \qquad\text{so}\qquad DE = 30 \cdot \frac{9}{15} = 18

Check. DB = 15 - 9 = 6, and \displaystyle \frac{9}{6} = \frac{12}{8} = \frac{3}{2}, so the pieces agree with the theorem. Using the piece ratio \displaystyle \frac{9}{6} for \overline{DE} would have given 45, longer than \overline{BC}, but a parallel nearer to A is a shrunken copy of \overline{BC} and must be shorter.

Example 3

An unknown in one piece

In \triangle ABC, \overline{DE} \parallel \overline{BC}, with AD = x, DB = x + 4, AE = 6 and EC = 9. Find x and both pieces of \overline{AB}.

Solution

By the theorem,

\frac{x}{x + 4} = \frac{6}{9}

Multiply both sides by 9(x + 4); x + 4 is a length, so it is not zero:

\begin{aligned} 9x &= 6(x + 4) \\ 9x &= 6x + 24 \\ 3x &= 24 \\ x &= 8 \end{aligned}

So AD = 8 and DB = 12.

Check. \displaystyle \frac{8}{12} = \frac{2}{3} and \displaystyle \frac{6}{9} = \frac{2}{3}.

Example 4

Is the segment parallel?

In \triangle PQR, S lies on \overline{PQ} and T on \overline{PR}. PQ = 14, PS = 10, PR = 17.5 and PT = 12.5.

(a) Is \overline{ST} \parallel \overline{QR}?

(b) Keep S where it is, but suppose instead that PR = 21. Where on \overline{PR} must T be for \overline{ST} to be parallel to \overline{QR}?

Solution

(a) Proving two segments parallel is the converse’s job, and it needs the two ratios to match. The pieces come from segment addition: SQ = 14 - 10 = 4 and TR = 17.5 - 12.5 = 5. Then

\frac{PS}{SQ} = \frac{10}{4} = 2.5 \qquad\qquad \frac{PT}{TR} = \frac{12.5}{5} = 2.5

The ratios are equal and S and T lie strictly inside their sides, so by the converse \overline{ST} \parallel \overline{QR}.

(b) The wholes are given, so the form \displaystyle \frac{PS}{PQ} = \frac{PT}{PR} is quicker:

\frac{10}{14} = \frac{PT}{21} \qquad\text{so}\qquad PT = 21 \cdot \frac{10}{14} = 15

T must be 15 from P, which leaves TR = 6. At any other point of \overline{PR} the two ratios differ, and then the theorem itself rules out a parallel: if \overline{ST} were parallel to \overline{QR}, the ratios would have to be equal.

Check. \displaystyle \frac{15}{6} = 2.5 = \frac{10}{4}.

Example 5

Two roads across three streets

Elm, Oak and Pine Streets are straight and parallel. Two straight roads cross all three, as in Figure 4. Along the first road it is 240 m from Elm to Oak and 360 m from Oak to Pine. Along the second road it is 750 m from Elm to Pine. Find a and b, the two stretches of the second road.

Figure 4Two roads cross the same three parallel streets.

Solution

Three parallel lines cut the two roads proportionally, so the stretches on the second road are in the same ratio as those on the first:

\frac{a}{b} = \frac{240}{360} = \frac{2}{3}

So the 750 m splits in the ratio 2 : 3, that is, into five equal shares of 750 \div 5 = 150 m, two for a and three for b. So a = 300 m and b = 450 m.

Check. 300 + 450 = 750 and \displaystyle \frac{300}{450} = \frac{2}{3}. The second road is longer than the first, 750 m against 600 m, so the stretches themselves differ. Only their ratio is shared.

Example 6

A quadratic from the proportion

In \triangle ABC, \overline{DE} \parallel \overline{BC}, with AD = x, DB = 6, AE = 4 and EC = x - 5. Find x and all four pieces.

Solution

By the theorem,

\frac{x}{6} = \frac{4}{x - 5}

Multiply both sides by 6(x - 5), which is not zero because x - 5 is a length:

\begin{aligned} x(x - 5) &= 24 \\ x^2 - 5x - 24 &= 0 \\ (x - 8)(x + 3) &= 0 \end{aligned}

So x = 8 or x = -3. The equation does not know that x is a length, but the triangle does: x = -3 would make AD = -3, so it is rejected. With x = 8 the pieces are AD = 8, DB = 6, AE = 4 and EC = 3.

Check. \displaystyle \frac{8}{6} = \frac{4}{3}, and every piece is positive.

Example 7

Splitting a segment without measuring it

Segment \overline{AB} is 10 cm long. Show how to mark the point P on it with AP : PB = 2 : 3 without measuring \overline{AB}: only a straightedge, a compass to step off equal lengths, and a way to draw parallel lines. Then find AP.

Solution

Draw any ray from A that is not along \overline{AB}, and step off five equal lengths along it from A, of any size, ending at the marks M_1, M_2, M_3, M_4 and M_5. Join M_5 to B. Through M_2 draw the line parallel to \overline{M_5B}; it meets \overline{AB} at P, as in Figure 5.

Figure 5Five equal steps on any ray, one join, one parallel.

Why this works: in \triangle AM_5B, the line M_2P is parallel to the side \overline{M_5B} and crosses the other two sides, so the theorem applies with A as the shared vertex:

\frac{AP}{PB} = \frac{AM_2}{M_2M_5} = \frac{2}{3}

because \overline{AM_2} is two of the equal steps and \overline{M_2M_5} is three. The size of a step never enters, and neither does the slant of the ray.

So AP is two of five equal shares of 10 cm: AP = 4 cm and PB = 6 cm.

Drawing the parallel through every mark instead of one splits \overline{AB} into five equal parts, which is dividing a segment into equal parts, and partitioning a segment in a ratio does the same job with coordinates.

Boundaries

The line must be parallel. Any line that crosses two sides of a triangle cuts them into pieces, but the ratios match only when the line is parallel to the third side. The converse is what says “only when”.

The converse needs both points inside their sides. For a ratio other than 1, there are two points on the line AC that divide it in that ratio, one between A and C and one beyond C. Take AD = 6 and DB = 3, and AC = 6. The point E with AE = 4, EC = 2 gives the ratio 2, and so does the point E' beyond C with AE' = 12, E'C = 6. Only \overline{DE} is parallel to \overline{BC}: through D there is just one parallel to \overline{BC}, and it meets line AC at E.

\overline{DE} is not a piece. The pieces give \displaystyle \frac{AD}{DB}; the parallel segment gives \displaystyle \frac{DE}{BC} = \frac{AD}{AB}. These are different numbers whenever D is not at a vertex.

Outside the triangle the whole-side form survives. If the parallel meets the lines AB and AC beyond B and C, or on the far side of A, the triangles ADE and ABC are still similar, so \displaystyle \frac{AD}{AB} = \frac{AE}{AC} still holds with lengths measured from A. The pieces form needs rewriting, since D is no longer between A and B.

The transversals are free; the lines are not. For the three-lines result, the lines that are crossed must be parallel. The two transversals need not be parallel to each other, nor the same length. If they happen to be parallel, the pieces on them are opposite sides of parallelograms and so are equal in pairs, which is the ratio result with a factor of 1. A line parallel to the family crosses none of them and has no pieces.

Ratios, not differences. In Example 1 the pieces of \overline{AB} differ by 2 and the pieces of \overline{AC} by 2.5. The theorem fixes the ratio of the pieces and nothing else.

Where this goes next

The outline above is set out in full, as a two-column proof with every reason, in proving the triangle proportionality theorem. The pair of results is also a clean case of a theorem and its converse that are both true, each needing its own proof.

The construction in Example 7 is the start of geometric division: splitting a segment into any number of equal parts with a straightedge, a compass and parallels, and marking the point that divides a segment in a given ratio, which coordinates then do by arithmetic.

The theorem also unlocks a fact about angle bisectors. Draw a line through C parallel to the bisector of \angle A and the triangle proportionality theorem shows that the bisector cuts \overline{BC} in the ratio AB : AC. That is the triangle angle bisector theorem, the proportional companion of the fact that every point on an angle bisector is the same distance from both sides.

Exercises

Reading the proportion

In \triangle ABC, \overline{DE} \parallel \overline{BC}, with D on \overline{AB} and E on \overline{AC}. Which of these must be true? (a) \displaystyle \frac{AD}{DB} = \frac{AE}{EC} (b) \displaystyle \frac{AD}{AB} = \frac{DE}{BC} (c) \displaystyle \frac{AD}{DB} = \frac{DE}{BC} (d) \displaystyle \frac{DB}{AB} = \frac{EC}{AC}

Finding a piece

In Exercises 2 to 7, \overline{DE} \parallel \overline{BC} in \triangle ABC, with D on \overline{AB} and E on \overline{AC}.

AD = 3, DB = 9 and AE = 4. Find EC.
AD = 10, DB = 4 and EC = 6. Find AE.
AD = 7, AB = 21 and AC = 27. Find AE and EC.
AE = 4.5, EC = 3 and AB = 20. Find AD and DB.
DB = 5, AB = 12 and AE = 14. Find EC.
AD = 6, DB = 9 and BC = 20. Find DE.

Using algebra

The same arrangement. Find x and the pieces that depend on it.

AD = x, DB = 10, AE = 6 and EC = 15.
AD = 2x + 1, DB = 8, AE = 15 and EC = 12.
AD = x + 3, DB = x, AE = 12 and EC = 8.
AD = x, DB = x + 5, AE = 4 and EC = 6.
AD = x, DB = 8, AE = 3 and EC = x - 2.

Parallel or not

D lies strictly between A and B, and E strictly between A and C. Decide whether \overline{DE} \parallel \overline{BC}, and give the reason.

AD = 6, DB = 4, AE = 9 and EC = 6.
AD = 5, DB = 3, AE = 8 and EC = 5.
AB = 18, AD = 12, AC = 24 and AE = 16.
AD = 4, DB = x, AE = 6 and EC = 9. Find the value of x that makes \overline{DE} \parallel \overline{BC}.

Parallel lines and transversals

Three parallel lines cut one transversal into pieces 6 and 10, and a second transversal into pieces 9 and y, in matching order. Find y.
Three parallel lines cut one transversal into pieces 8 and 12. On a second transversal the distance from the first line to the third is 25. Find the two pieces on the second transversal.
Four parallel lines cut one transversal into pieces 3, 4 and 5, in order. The first piece on a second transversal is 4.5. Find the other two.
Three parallel lines cut one transversal into pieces x + 1 and 2x, and a second into pieces 9 and 15, in matching order. Find x and both pieces on the first transversal.

Reasoning

Show that if \displaystyle \frac{AD}{DB} = \frac{AE}{EC}, then \displaystyle \frac{AD}{AB} = \frac{AE}{AC}.
In \triangle ABC, \overline{DE} \parallel \overline{BC} with AD = 4, DB = 6 and BC = 15. Mira writes \displaystyle \frac{DE}{BC} = \frac{AD}{DB} and gets DE = 10. Explain what is wrong, and find DE.
Explain how the converse shows that the segment joining the midpoints of two sides of a triangle is parallel to the third side.
In \triangle ABC, a line through a point D of \overline{AB} is parallel to \overline{AC} and meets \overline{BC} at F. Which proportion does the theorem give, and why?
D lies between A and B with AD = 8 and DB = 4, and AC = 9. Find both points E on line AC with \displaystyle \frac{AE}{EC} = 2, and explain why only one of them makes \overline{DE} parallel to \overline{BC}.
Parallel lines cut one transversal into equal pieces. Explain why they cut every other transversal into equal pieces too.

Answers

(a), (b) and (d). (a) is the theorem. (b) holds because \triangle ADE \sim \triangle ABC, so matching sides are in one ratio. (d) follows from (b), since \displaystyle \frac{DB}{AB} = 1 - \frac{AD}{AB} = 1 - \frac{AE}{AC} = \frac{EC}{AC}. (c) is false: \displaystyle \frac{DE}{BC} equals \displaystyle \frac{AD}{AB}, a different number; with AD = 4 and DB = 6, \displaystyle \frac{AD}{DB} = \frac{2}{3} but \displaystyle \frac{DE}{BC} = \frac{4}{10}.
EC = 12. \displaystyle \frac{3}{9} = \frac{4}{EC}, so 3 \cdot EC = 36.
AE = 15. \displaystyle \frac{10}{4} = \frac{AE}{6}, so \displaystyle AE = \frac{60}{4}.
AE = 9 and EC = 18. \displaystyle \frac{AD}{AB} = \frac{7}{21} = \frac{AE}{27}, so AE = 9, and EC = 27 - 9.
AD = 12 and DB = 8. \displaystyle \frac{AD}{DB} = \frac{4.5}{3} = \frac{3}{2}, so AB = 20 splits into five shares of 4: AD = 3 \cdot 4 and DB = 2 \cdot 4.
EC = 10. AD = 12 - 5 = 7, so \displaystyle \frac{7}{5} = \frac{14}{EC} and 7 \cdot EC = 70.
DE = 8. AB = 6 + 9 = 15, and \displaystyle \frac{DE}{BC} = \frac{AD}{AB} = \frac{6}{15}, so \displaystyle DE = 20 \cdot \frac{6}{15}.
x = 4. \displaystyle \frac{x}{10} = \frac{6}{15}, so 15x = 60.
x = 4.5, so AD = 10. \displaystyle \frac{2x + 1}{8} = \frac{15}{12} = \frac{5}{4}, so 2x + 1 = 10.
x = 6, so AD = 9 and DB = 6. \displaystyle \frac{x + 3}{x} = \frac{12}{8}, so 8(x + 3) = 12x and 24 = 4x.
x = 10, so AD = 10 and DB = 15. \displaystyle \frac{x}{x + 5} = \frac{4}{6}, so 6x = 4x + 20.
x = 6, so AD = 6 and EC = 4. \displaystyle \frac{x}{8} = \frac{3}{x - 2}, so x(x - 2) = 24, which is (x - 6)(x + 4) = 0. The root x = -4 would make AD negative, so it is rejected.
Parallel. \displaystyle \frac{6}{4} = \frac{3}{2} and \displaystyle \frac{9}{6} = \frac{3}{2}; the ratios are equal and the points are inside the sides, so the converse applies.
Not parallel. \displaystyle \frac{5}{3} and \displaystyle \frac{8}{5} differ, since 5 \cdot 5 = 25 but 3 \cdot 8 = 24. If \overline{DE} were parallel, the theorem would force them to be equal.
Parallel. \displaystyle \frac{AD}{AB} = \frac{12}{18} = \frac{2}{3} and \displaystyle \frac{AE}{AC} = \frac{16}{24} = \frac{2}{3}. Equivalently DB = 6, EC = 8 and \displaystyle \frac{12}{6} = \frac{16}{8} = 2.
x = 6. \displaystyle \frac{4}{x} = \frac{6}{9}, so 6x = 36.
y = 15. \displaystyle \frac{6}{10} = \frac{9}{y}, so 6y = 90.
10 and 15. The pieces are in the ratio 8 : 12 = 2 : 3, so 25 splits into five shares of 5.
6 and 7.5. Every piece on the second transversal is \displaystyle \frac{4.5}{3} = 1.5 times its partner, so the others are 4 \cdot 1.5 and 5 \cdot 1.5.
x = 5, and the pieces are 6 and 10. \displaystyle \frac{x + 1}{2x} = \frac{9}{15} = \frac{3}{5}, so 5x + 5 = 6x.
Take reciprocals: \displaystyle \frac{DB}{AD} = \frac{EC}{AE}. Add 1 to both sides: \displaystyle \frac{DB + AD}{AD} = \frac{EC + AE}{AE}, and by segment addition this is \displaystyle \frac{AB}{AD} = \frac{AC}{AE}. Take reciprocals again: \displaystyle \frac{AD}{AB} = \frac{AE}{AC}.
DE = 6. \overline{DE} is not a piece of a side; it matches \overline{BC} in the similar triangles ADE and ABC, where the matching ratio is \displaystyle \frac{AD}{AB}, not \displaystyle \frac{AD}{DB}. AB = 10, so \displaystyle DE = 15 \cdot \frac{4}{10} = 6.
If D and E are the midpoints of \overline{AB} and \overline{AC}, then AD = DB and AE = EC, so \displaystyle \frac{AD}{DB} = 1 = \frac{AE}{EC}. Both points are strictly inside their sides, so the converse gives \overline{DE} \parallel \overline{BC}. This is the parallel half of the midsegment theorem.
\displaystyle \frac{BD}{DA} = \frac{BF}{FC}. The line DF is parallel to the side \overline{AC} and crosses the other two sides, \overline{BA} and \overline{BC}, which meet at B. That is the theorem with B as the shared vertex: piece at B over piece at the far side, on both sides.
AE = 6 with EC = 3, and AE = 18 with EC = 9. With AE = e: between A and C, \displaystyle \frac{e}{9 - e} = 2 gives e = 6; beyond C, \displaystyle \frac{e}{e - 9} = 2 gives e = 18. Only the first gives a parallel. By the theorem, the one parallel to \overline{BC} through D meets \overline{AC} where \displaystyle \frac{AE}{AC} = \frac{AD}{AB} = \frac{8}{12}, that is at AE = 6, so the line to the point 18 from A is a different line and cannot be parallel.
On the first transversal every pair of neighboring pieces has ratio 1. Parallel lines cut transversals proportionally, so the matching pair on any other transversal also has ratio 1, which means those pieces are equal too.