Geometry · G.SRT.B.4

Proving the triangle proportionality theorem

A line runs across a triangle parallel to one side. Why must it cut the other two sides in the same ratio, and why does a cut in the same ratio have to be parallel?

You need

Proving two triangles similarGeometry Triangle proportionality theoremGeometry

Where you are 4 in Proofs about similar figures

Geometry11

Geometric proof18

Proofs about similar figures4

Proving the Pythagorean theorem by rearrangement Proving two triangles similar Proving the Pythagorean theorem with similar triangles

Proving the triangle proportionality theorem

When you reach for it

Similar triangles give proportions between sides of triangles. Very often the lengths you want are not sides of any one triangle but pieces of sides: a triangle is cut by a line, and you know the pieces on one side and want the pieces on the other. The triangle proportionality theorem, also called the side splitter theorem, is the bridge. Reach for it when a figure has a line parallel to one side of a triangle and crossing the other two, and the question is about the pieces those two sides are cut into.

Reach for its converse in the other direction: when the Given talks about lengths, and the Prove says two lines are parallel.

Both are proved from similar triangles. The parallel line cuts off a small triangle that shares an angle with the whole triangle, and that similarity does all the work. What this lesson adds is the careful bookkeeping: which triangle is similar to which, under which correspondence, and the short piece of algebra that turns whole sides into pieces.

The procedure

The statement. Here is the theorem with every hypothesis written out.

Triangle proportionality theorem. In \triangle ABC, let D lie on side \overline{AB} and E on side \overline{AC}, each strictly between the two vertices. If \overline{DE} \parallel \overline{BC}, then

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

and also, for whole sides,

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

Figure 1\overline{DE} \parallel \overline{BC}, with D on side \overline{AB} and E on side \overline{AC}.

The first proportion is about the pieces the parallel line cuts, the second about the small triangle and the whole one. Both are proved below. “Strictly between” matters: it rules out D = B, where DB = 0 could not be divided by, and it fixes where the pieces sit.

A line through a point D strictly between A and B and parallel to \overline{BC} always does meet side \overline{AC} at a point strictly between A and C, so the theorem applies to every such line. It enters the triangle at D, cannot leave through \overline{BC}, which it never meets, and so leaves through \overline{AC}.

Name the similarity and its correspondence. The small triangle is \triangle ADE and the whole one is \triangle ABC. They share the vertex A, and D sits on the side toward B, E on the side toward C. So the claim is

\triangle ADE \sim \triangle ABC

with the correspondence A \leftrightarrow A, D \leftrightarrow B, E \leftrightarrow C. Two pairs of angles give it by AA similarity.

The shared angle. D lies on \overrightarrow{AB} and E on \overrightarrow{AC}, so \angle DAE and \angle BAC are the same angle, \angle A.
A pair of corresponding angles. Take \overleftrightarrow{AB} as a transversal of the parallel lines \overleftrightarrow{DE} and \overleftrightarrow{BC}. \angle ADE is formed by \overrightarrow{DA} and \overrightarrow{DE}; \angle ABC by \overrightarrow{BA} and \overrightarrow{BC}. The rays \overrightarrow{DA} and \overrightarrow{BA} point the same way along the transversal, and E and C lie on the same side of \overleftrightarrow{AB}, because E is between A and C. So the two angles sit in matching positions at the two crossings, and the Corresponding Angles Theorem makes them congruent.

Read the whole-side ratios off the similarity. Sides in positions 1 and 2, then 1 and 3, then 2 and 3:

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

This is the second half of the theorem, and it is the only one of the two that mentions \overline{DE}.

Turn whole sides into pieces. \overline{DB} and \overline{EC} are not sides of either triangle, so no similarity mentions them. The segment addition postulate brings them in: D is between A and B, so AB = AD + DB, and likewise AC = AE + EC. Clearing the fractions in \displaystyle \frac{AD}{AB} = \frac{AE}{AC} and substituting,

\begin{aligned} AD \cdot AC &= AE \cdot AB \\ AD\,(AE + EC) &= AE\,(AD + DB) \\ AD \cdot AE + AD \cdot EC &= AE \cdot AD + AE \cdot DB \\ AD \cdot EC &= AE \cdot DB \end{aligned}

The term AD \cdot AE is on both sides and cancels. Dividing by DB \cdot EC, which is not zero because D \ne B and E \ne C, gives \displaystyle \frac{AD}{DB} = \frac{AE}{EC}.

Here is the whole proof in two columns.

Given. In \triangle ABC, D is between A and B, E is between A and C, and \overline{DE} \parallel \overline{BC}.

Prove. \displaystyle \frac{AD}{DB} = \frac{AE}{EC} and \displaystyle \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}.

Statements	Reasons
1. \overline{DE} \parallel \overline{BC}; D between A, B; E between A, C	given
2. \angle ADE \cong \angle ABC	Corresponding Angles Theorem: line 1, transversal \overleftrightarrow{AB}
3. \angle DAE \cong \angle BAC	reflexive property of congruence: both are \angle A
4. \triangle ADE \sim \triangle ABC	AA similarity: lines 2 and 3
5. \displaystyle {\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}}	corresponding sides of similar triangles are proportional: line 4
6. {AB = AD + DB}, {AC = AE + EC}	segment addition postulate: line 1
7. {AD \cdot AC = AE \cdot AB}	multiplication property of equality: line 5
8. {AD \cdot EC = AE \cdot DB}	substitution of line 6 into line 7, distributive property, subtraction
9. \displaystyle {\frac{AD}{DB} = \frac{AE}{EC}}	division property of equality: line 8, with {DB \cdot EC \ne 0}

The same kind of algebra gives other true proportions, such as \displaystyle \frac{DB}{AB} = \frac{EC}{AC}. Every one of them compares a piece or the whole of \overline{AB} with the matching piece or the whole of \overline{AC}.

Running it backwards. Reversing the argument proves that a cut in the same ratio is a parallel cut, with the same hypotheses on where D and E lie.

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

Now no angle is known except \angle A, so the similarity has to come from SAS similarity, with \angle A as the included angle. That needs the two sides at A in \triangle ADE in proportion with the two sides at A in \triangle ABC, and the algebra above, reversed, supplies exactly that.

Statements	Reasons
1. D between A, B; E between A, C; \displaystyle {\frac{AD}{DB} = \frac{AE}{EC}}	given
2. {AD \cdot EC = AE \cdot DB}	multiplication property of equality: line 1
3. {AD \cdot AE + AD \cdot EC} {= AE \cdot AD + AE \cdot DB}	addition property of equality: line 2
4. {AD\,(AE + EC)} {= AE\,(AD + DB)}	distributive property: line 3
5. {AD \cdot AC = AE \cdot AB}	segment addition postulate: lines 1 and 4
6. \displaystyle {\frac{AD}{AB} = \frac{AE}{AC}}	division property of equality: line 5
7. \angle DAE \cong \angle BAC	reflexive property of congruence: both are \angle A
8. \triangle ADE \sim \triangle ABC	SAS similarity: lines 6 and 7
9. \angle ADE \cong \angle ABC	corresponding angles of similar triangles are congruent: line 8
10. \overline{DE} \parallel \overline{BC}	converse of the Corresponding Angles Theorem: line 9, transversal \overleftrightarrow{AB}

Line 6 pairs \overline{AD} with \overline{AB} and \overline{AE} with \overline{AC}, which is what makes the correspondence D \leftrightarrow B, E \leftrightarrow C in line 8. Line 5 is where “strictly between” does its work: AC = AE + EC holds only because E lies between A and C. Line 10 uses it once more: E and C are on the same side of \overleftrightarrow{AB}, so the angles of line 9 are in corresponding positions (proving lines parallel).

The theorem and its converse together say: for points strictly inside the two sides, \overline{DE} \parallel \overline{BC} if and only if \displaystyle \frac{AD}{DB} = \frac{AE}{EC} (a theorem and its converse).

Worked examples

Example 1

Pieces and a whole side

In \triangle PQR, S lies on \overline{PQ} and T on \overline{PR}, with \overline{ST} \parallel \overline{QR}. PS = 6, SQ = 4, PT = 9 and QR = 15. Find TR and ST.

Solution

The parallel cuts \overline{PQ} into \overline{PS} and \overline{SQ}, and \overline{PR} into \overline{PT} and \overline{TR}, with P the shared vertex. The segment proportion pairs the piece at P on one side with the piece at P on the other:

\begin{aligned} \frac{PS}{SQ} &= \frac{PT}{TR} \\ \frac{6}{4} &= \frac{9}{TR} \\ TR &= \frac{4 \cdot 9}{6} = 6 \end{aligned}

\overline{ST} is not a piece of either side, so it appears only in the whole-side ratios, from \triangle PST \sim \triangle PQR. By segment addition PQ = 6 + 4 = 10, so

\begin{aligned} \frac{ST}{QR} &= \frac{PS}{PQ} = \frac{6}{10} \\ ST &= \frac{6}{10} \cdot 15 = 9 \end{aligned}

Check. The third whole-side ratio agrees: PR = 9 + 6 = 15 and \displaystyle \frac{PT}{PR} = \frac{9}{15} = \frac{3}{5} = \frac{6}{10}. The segment ratio \displaystyle \frac{PS}{SQ} = \frac{3}{2} is a different number from \displaystyle \frac{ST}{QR} = \frac{3}{5}, which is why \overline{ST} has to be found from the whole sides.

Example 2

An unknown in both pieces

In \triangle JKL, M lies on \overline{JK} and N on \overline{JL}, with \overline{MN} \parallel \overline{KL}. JM = 8, MK = x - 1, JN = 10 and NL = x + 2. Find x and the two unknown pieces.

Solution

By the theorem, with J as the shared vertex,

\frac{JM}{MK} = \frac{JN}{NL} \qquad\text{so}\qquad \frac{8}{x - 1} = \frac{10}{x + 2}

Multiply both sides by (x - 1)(x + 2), which is allowed because both are lengths of pieces and so not zero:

\begin{aligned} 8(x + 2) &= 10(x - 1) \\ 8x + 16 &= 10x - 10 \\ 26 &= 2x \end{aligned}

So x = 13, MK = 12 and NL = 15. Check. \displaystyle \frac{8}{12} = \frac{2}{3} and \displaystyle \frac{10}{15} = \frac{2}{3}, and both pieces are positive, as lengths must be.

Example 3

Parallel or not

In \triangle XYZ, U lies on \overline{XY} and V on \overline{XZ}, each strictly between the vertices, with XU = 5 and UY = 7.5. Decide whether \overline{UV} \parallel \overline{YZ} when (a) XV = 4 and VZ = 6; (b) XV = 4 and VZ = 5.

Solution

The question is the converse’s, so compare the two ratios at X.

(a) \displaystyle \frac{XU}{UY} = \frac{5}{7.5} = \frac{2}{3} and \displaystyle \frac{XV}{VZ} = \frac{4}{6} = \frac{2}{3}. The ratios are equal and U and V are strictly inside the sides, so the converse gives \overline{UV} \parallel \overline{YZ}.

(b) \displaystyle \frac{XU}{UY} = \frac{2}{3} but \displaystyle \frac{XV}{VZ} = \frac{4}{5}. These differ, and that settles it through the theorem itself, not the converse: if \overline{UV} were parallel to \overline{YZ}, the theorem would force the two ratios to be equal. They are not, so \overline{UV} is not parallel to \overline{YZ}. This is the contrapositive of the theorem.

A “yes” needs the converse and a “no” needs the theorem. Each direction answers one of the two questions.

Example 4

Two parallels in one triangle

Given. In \triangle ABC, D lies on \overline{AB}, E on \overline{AC} and F on \overline{BC}, each strictly between the vertices; \overline{DE} \parallel \overline{BC} and \overline{EF} \parallel \overline{AB}.

Prove. \displaystyle \frac{AD}{DB} = \frac{BF}{FC}. Then, given AD = 3, DB = 5 and BC = 16, find BF, FC and DE.

Figure 2\overline{DE} \parallel \overline{BC} and \overline{EF} \parallel \overline{AB}.

Solution

Plan. Each parallel sits in the triangle in a different way, so the theorem is used twice, in two different triangles, and the vertex that plays the part of “A” changes.

\overline{DE} \parallel \overline{BC} cuts sides \overline{AB} and \overline{AC} of \triangle ABC, meeting at A. The pieces at A are \overline{AD} and \overline{AE}.
\overline{EF} \parallel \overline{AB} cuts sides \overline{CB} and \overline{CA} of the same triangle, meeting at C. Now the pieces at the shared vertex are \overline{CF} and \overline{CE}.

The ratio \displaystyle \frac{AE}{EC} appears in both, and links them.

Statements	Reasons
1. \overline{DE} \parallel \overline{BC}, \overline{EF} \parallel \overline{AB}; D, E, F strictly inside their sides	given
2. \displaystyle {\frac{AD}{DB} = \frac{AE}{EC}}	triangle proportionality theorem: \triangle ABC, line 1
3. \displaystyle {\frac{CF}{FB} = \frac{CE}{EA}}	triangle proportionality theorem: \triangle CBA, line 1
4. \displaystyle {\frac{BF}{FC} = \frac{AE}{EC}}	reciprocals of both sides of line 3
5. \displaystyle {\frac{AD}{DB} = \frac{BF}{FC}}	substitution: lines 2 and 4

The numbers. \displaystyle \frac{BF}{FC} = \frac{AD}{DB} = \frac{3}{5}, so BF = 3t and FC = 5t for some length t. F is between B and C, so 3t + 5t = 16, t = 2, and

BF = 6 \qquad FC = 10

\overline{DE} is not a piece of a side, so it comes from the whole-side ratio in \triangle ABC, with AB = 3 + 5 = 8:

\begin{aligned} \frac{DE}{BC} &= \frac{AD}{AB} = \frac{3}{8} \\ DE &= \frac{3}{8} \cdot 16 = 6 \end{aligned}

Check. DBFE has both pairs of opposite sides parallel, so it is a parallelogram and its opposite sides are congruent: DE should equal BF, and both are 6.

Example 5

Three parallel lines cut two transversals

Given. Parallel lines l, m and n. A transversal s meets them at A, B and C, with B between A and C; a second transversal t meets them at D, E and F. The six points are all different.

Prove. \displaystyle \frac{AB}{BC} = \frac{DE}{EF}. Then, given AB = 4, BC = 6 and DF = 15, find DE and EF.

Figure 3Three parallel lines and two transversals. The dashed \overline{AF} is drawn for the proof.

Solution

Plan. There is no triangle in the Given, so the proof has to make one. The segment \overline{AF} joins A on l to F on n, and it builds two triangles, one on each transversal, each with a parallel line cutting it.

First, where things lie. B is between A and C, so A and C are on opposite sides of m. The line l never meets m, so all of l, D included, is on A’s side; likewise all of n, F included, is on C’s side. So \overline{AF} crosses m at a point G strictly between A and F, and \overline{DF} crosses m at E, strictly between D and F.

In \triangle ACF, B is on side \overline{AC}, G on side \overline{AF}, and \overline{BG} lies on m, parallel to \overline{CF} on n. The theorem, with A as the shared vertex, gives \displaystyle \frac{AB}{BC} = \frac{AG}{GF}.
In \triangle FDA, E is on side \overline{FD}, G on side \overline{FA}, and \overline{EG} lies on m, parallel to \overline{DA} on l. The theorem, with F as the shared vertex, gives \displaystyle \frac{FE}{ED} = \frac{FG}{GA}, and taking reciprocals, \displaystyle \frac{DE}{EF} = \frac{AG}{GF}.

Both ratios equal \displaystyle \frac{AG}{GF}, so \displaystyle \frac{AB}{BC} = \frac{DE}{EF}. (Neither triangle is flat: F is not on s, since s meets n only at C, and D is not on \overleftrightarrow{AF}, since that line meets l only at A.)

The numbers. \displaystyle \frac{DE}{EF} = \frac{4}{6} = \frac{2}{3}, so DE = 2t and EF = 3t. E is between D and F, so 2t + 3t = 15, t = 3:

DE = 6 \qquad EF = 9

Check. \displaystyle \frac{6}{9} = \frac{2}{3} = \frac{4}{6}. This is the result parallel lines cut transversals proportionally. Notice that AC = 10 and DF = 15 differ: the pieces are in the same ratio, not the same length.

Example 6

The angle bisector cuts the opposite side in the ratio of the other two sides

Given. In \triangle ABC, \overrightarrow{AD} bisects \angle BAC, with D on \overline{BC}.

Prove. \displaystyle \frac{BD}{DC} = \frac{AB}{AC}. Then, given AB = 9, AC = 6 and BC = 10, find BD and DC.

Figure 4\overrightarrow{AD} bisects \angle BAC. The dashed lines are drawn for the proof.

Solution

Plan. The Prove wants a ratio of the pieces of \overline{BC}, and the theorem gives such a ratio only when a parallel line does the cutting. There is no parallel in the Given, so build one: through C, the line parallel to \overline{AD}. Then \overline{AD} becomes a side splitter of a larger triangle, and the bisector’s equal angles make one side of that triangle equal to AC.

Building E. By the parallel postulate there is exactly one line through C parallel to \overleftrightarrow{AD}. It meets \overleftrightarrow{BA}: if it did not, \overleftrightarrow{AD} and \overleftrightarrow{AB} would be two different lines through A both parallel to it, which the postulate forbids. Call the meeting point E. Since D is between B and C, the points B and C lie on opposite sides of \overleftrightarrow{AD}, and the new line never crosses \overleftrightarrow{AD}, so E is on C’s side, opposite B. So A, where \overleftrightarrow{BE} crosses \overleftrightarrow{AD}, lies between B and E.

Two angle pairs.

\angle BAD \cong \angle AEC. With \overleftrightarrow{BE} as the transversal of \overleftrightarrow{AD} \parallel \overleftrightarrow{EC}, both angles have a ray pointing along it toward B, and D and C lie on the same side of it. They are corresponding angles.
\angle DAC \cong \angle ACE. With \overleftrightarrow{AC} as the transversal, D lies on B’s side of it and E on the other side, since A is between B and E. They are alternate interior angles.

The bisector makes \angle BAD \cong \angle DAC, so \angle AEC \cong \angle ACE, and in \triangle ACE the sides opposite those angles are equal by the converse of the base angles theorem: AE = AC.

The proportion. In \triangle BCE, A is strictly between B and E, D strictly between B and C, and \overline{AD} \parallel \overline{EC}. The theorem, with B as the shared vertex, gives

\frac{BD}{DC} = \frac{BA}{AE} = \frac{AB}{AC}

the last step because AE = AC.

The numbers. \displaystyle \frac{BD}{DC} = \frac{9}{6} = \frac{3}{2}, so BD = 3t, DC = 2t and 3t + 2t = 10. So t = 2, BD = 6 and DC = 4. The longer side \overline{AB} gets the longer piece. This is the triangle angle bisector theorem.

Boundaries

The line must be parallel to the third side. The theorem is about a line crossing the two sides that meet at a vertex and parallel to the side opposite that vertex. A line crossing two sides at some other slant cuts off a triangle that may or may not be similar to the whole, and the pieces need not be in proportion.

The converse needs D and E inside the sides. In Figure 5, D is on side \overline{AB} with \displaystyle \frac{AD}{DB} = \frac{4}{2} = 2, and E is on \overleftrightarrow{AC} past C, with AE = 6 and \displaystyle \frac{AE}{EC} = \frac{6}{3} = 2. The ratios match and \overline{DE} is not parallel to \overline{BC}. The proof of the converse breaks at line 5, since AC = AE + EC is false when E is past C. The one parallel to \overline{BC} through D meets \overline{AC} at G, where \displaystyle \frac{AG}{GC} = 2, so AG = 2. It meets \overleftrightarrow{AC} nowhere else, and E \ne G, so \overline{DE} is not that parallel.

Figure 5AC = CE = 3, so \displaystyle \frac{AD}{DB} = \frac{AE}{EC} = 2, yet \overline{DE} is not parallel to \overline{BC}. The parallel through D is the dashed \overline{DG}.

Past the vertices. A line parallel to \overline{BC} can also cross \overleftrightarrow{AB} and \overleftrightarrow{AC} beyond B and C, or beyond A on the far side. Measured as lengths, with no signs, the same proportions still hold, and the proof is the same similarity with the pieces added or subtracted differently (Exercise 21 does the first case). For the converse the positions must then be given as well: D and E both beyond B and C, or both on the far side of A. Handling all positions in one statement needs signed ratios, which this lesson does not use.

Everything here rests on the parallel postulate. The first step is the Corresponding Angles Theorem, which is where the postulate enters (proofs about parallel lines).

Where this goes next

With the ratio equal to 1, the converse says that the segment joining the midpoints of two sides is parallel to the third side, and the whole-side ratio makes it half as long: the midsegment of a triangle (Exercise 19). The converse is the standard way to prove lines parallel from lengths alone.

Example 5 is the theorem spread across a band of parallel lines, and it is what makes dividing a segment into equal parts with a straightedge and compass work: equal steps along one ray, joined back by parallels, give equal steps along the segment. Example 6, the angle bisector theorem, is the model for proofs that build a parallel line the Given did not supply. The same move of reading proportions off a similarity, with a different pair of triangles, proves the Pythagorean theorem and the power of a point.

Exercises

Reading the proportions

In Exercises 1 to 4, D lies strictly between A and B, E strictly between A and C, and \overline{DE} \parallel \overline{BC}, as in Figure 1.

AD = 4, DB = 6 and AE = 5. Find EC.
AD = 3, AB = 12 and BC = 20. Find DE.
AE = 8, EC = 12 and AB = 25. Find AD and DB.
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}; (e) \displaystyle \frac{AB}{AD} = \frac{AC}{AE}.

Unknowns in the pieces

With the same figure and hypotheses, find x and every unknown piece.

AD = x, DB = 6, AE = 4, EC = 8.
AD = x + 1, DB = 2x, AE = 6, EC = 10.
AD = x, DB = x + 5, AE = 6, EC = 9.
AD = x, DB = 4, AE = 9, EC = x.

Parallel or not

In \triangle XYZ, U lies on \overleftrightarrow{XY} and V on \overleftrightarrow{XZ}. Decide whether \overline{UV} \parallel \overline{YZ}, and give the reason.

U and V are strictly inside the sides; XU = 6, UY = 4, XV = 9, VZ = 6.
U and V are strictly inside the sides; XU = 6, UY = 4, XV = 8, VZ = 6.
U and V are strictly inside the sides; XU = 5, XY = 15, XV = 4, XZ = 12.
U is strictly inside \overline{XY} with XU = 2 and UY = 4. V is on the far side of X from Z, with X between V and Z, XV = 3 and XZ = 3.

Transversals and bisectors

Three parallel lines cut a transversal into pieces of 5 and 7, in that order. They cut a second transversal into pieces of which the first is 8. Find the second.
Three parallel lines cut one transversal into pieces x and 12, and another into pieces 6 and 9, in matching order. Find x.
In \triangle ABC, AB = 12, AC = 8 and BC = 15. The bisector of \angle A meets \overline{BC} at D. Find BD and DC.
In \triangle PQR, the bisector of \angle P meets \overline{QR} at S. PQ = 10, PR = 14 and QS = 5. Find SR and QR.

Writing the proof

Given. In \triangle JKL, M is strictly between J and K, N strictly between J and L, and \overline{MN} \parallel \overline{KL}. Prove. JM \cdot NL = JN \cdot MK.
Given. ABCD is a trapezoid with \overline{AB} \parallel \overline{DC}; E is strictly between A and D, F strictly between B and C, and \overline{EF} \parallel \overline{AB}. Prove. \displaystyle \frac{AE}{ED} = \frac{BF}{FC}.
Given. In \triangle ABC, M is the midpoint of \overline{AB} and N the midpoint of \overline{AC}. Prove. \overline{MN} \parallel \overline{BC} and \displaystyle MN = \frac{1}{2}BC.

Reasoning

In Figure 1, suppose AD = DB = 2 and BC = 10. Find \displaystyle \frac{AD}{DB} and \displaystyle \frac{DE}{BC}, and explain why \overline{DE} never appears in the segment proportion.
A line parallel to \overline{BC} meets \overleftrightarrow{AB} at D and \overleftrightarrow{AC} at E, with B between A and D and C between A and E. Prove that \displaystyle \frac{AB}{BD} = \frac{AC}{CE}.
Prove the converse a second way, indirectly. Suppose D and E are strictly inside the sides with \displaystyle \frac{AD}{DB} = \frac{AE}{EC}, and let the parallel to \overline{BC} through D meet \overline{AC} at G. Show that G = E.

Answers

EC = 7.5. \displaystyle \frac{AD}{DB} = \frac{AE}{EC} gives \displaystyle \frac{4}{6} = \frac{5}{EC}, so \displaystyle EC = \frac{6 \cdot 5}{4}.
DE = 5. \displaystyle \frac{DE}{BC} = \frac{AD}{AB} = \frac{3}{12} = \frac{1}{4}, so \displaystyle DE = \frac{20}{4}.
AD = 10 and DB = 15. \displaystyle \frac{AD}{DB} = \frac{8}{12} = \frac{2}{3}, so AD = 2t, DB = 3t and 5t = 25, t = 5.
(a), (b), (d) and (e). (a) is the segment proportion and (b) the whole-side ratios. (e) is the reciprocal of \displaystyle \frac{AD}{AB} = \frac{AE}{AC}. (d) is \displaystyle 1 - \frac{AD}{AB} = 1 - \frac{AE}{AC}, since DB = AB - AD and EC = AC - AE. (c) is false in general: \overline{DB} is not a side of either similar triangle (Exercise 20 has numbers).
x = 3. \displaystyle \frac{x}{6} = \frac{4}{8}, so \displaystyle x = 6 \cdot \frac{1}{2}.
x = 5, AD = 6, DB = 10. \displaystyle \frac{x + 1}{2x} = \frac{6}{10} gives 10x + 10 = 12x. Check. \displaystyle \frac{6}{10} = \frac{6}{10}.
x = 10, AD = 10, DB = 15. \displaystyle \frac{x}{x + 5} = \frac{6}{9} gives 9x = 6x + 30. Check. \displaystyle \frac{10}{15} = \frac{2}{3} = \frac{6}{9}.
x = 6, so AD = 6 and EC = 6. \displaystyle \frac{x}{4} = \frac{9}{x} gives x^2 = 36, and x = -6 is rejected because a length is positive. Check. \displaystyle \frac{6}{4} = \frac{9}{6} = \frac{3}{2}.
Parallel. \displaystyle \frac{XU}{UY} = \frac{6}{4} = \frac{3}{2} and \displaystyle \frac{XV}{VZ} = \frac{9}{6} = \frac{3}{2}, with both points inside the sides, so the converse applies.
Not parallel. \displaystyle \frac{6}{4} = \frac{3}{2} but \displaystyle \frac{8}{6} = \frac{4}{3}. If the segments were parallel the theorem would make the ratios equal, so by the contrapositive they are not.
Parallel. UY = 15 - 5 = 10 and VZ = 12 - 4 = 8, so \displaystyle \frac{XU}{UY} = \frac{5}{10} and \displaystyle \frac{XV}{VZ} = \frac{4}{8}, both \displaystyle \frac{1}{2}, and the converse applies. (Equally, \displaystyle \frac{XU}{XY} = \frac{XV}{XZ} = \frac{1}{3} gives SAS similarity directly.)
Not parallel, although \displaystyle \frac{XU}{UY} = \frac{2}{4} = \frac{1}{2} and \displaystyle \frac{XV}{VZ} = \frac{3}{6} = \frac{1}{2}, with VZ = 3 + 3. The converse does not apply, since V is not between X and Z. The parallel to \overline{YZ} through U meets \overline{XZ} at a point G with \displaystyle \frac{XG}{GZ} = \frac{1}{2}, so XG = 1, on the other side of X from V. One line through U meets \overleftrightarrow{XZ} only once, so \overline{UV} is not that parallel.
\displaystyle \frac{56}{5} = 11.2. \displaystyle \frac{5}{7} = \frac{8}{y}, so \displaystyle y = \frac{7 \cdot 8}{5}.
x = 8. \displaystyle \frac{x}{12} = \frac{6}{9}, so \displaystyle x = 12 \cdot \frac{2}{3}.
BD = 9 and DC = 6. \displaystyle \frac{BD}{DC} = \frac{AB}{AC} = \frac{12}{8} = \frac{3}{2}, so BD = 3t, DC = 2t and 5t = 15, t = 3.
SR = 7 and QR = 12. \displaystyle \frac{QS}{SR} = \frac{PQ}{PR} = \frac{10}{14}, so \displaystyle SR = \frac{14 \cdot 5}{10} = 7 and QR = 5 + 7.
Line (1): M between J and K, N between J and L, \overline{MN} \parallel \overline{KL} (given). (2): \displaystyle \frac{JM}{MK} = \frac{JN}{NL} (triangle proportionality theorem: line 1, shared vertex J). (3): JM \cdot NL = JN \cdot MK (multiplication property of equality: multiply both sides of line 2 by MK \cdot NL).
Line (1): \overline{AB} \parallel \overline{DC}, \overline{EF} \parallel \overline{AB}, with E and F strictly inside \overline{AD} and \overline{BC} (given). (2): \overline{EF} \parallel \overline{DC} (two lines parallel to the same line are parallel to each other: line 1). (3): \overleftrightarrow{AB}, \overleftrightarrow{EF} and \overleftrightarrow{DC} are three parallel lines cut by the transversals \overleftrightarrow{AD} and \overleftrightarrow{BC}, with E between A and D (lines 1 and 2). (4): \displaystyle \frac{AE}{ED} = \frac{BF}{FC} (parallel lines cut transversals proportionally, Example 5: line 3). The six points A, E, D, B, F, C are all different, as Example 5 requires, because E and F are strictly inside the legs.
Line (1): M and N are the midpoints (given). (2): AM = MB and AN = NC, with M between A and B and N between A and C (definition of midpoint: line 1). (3): \displaystyle \frac{AM}{MB} = 1 = \frac{AN}{NC} (division: line 2). (4): \overline{MN} \parallel \overline{BC} (converse of the triangle proportionality theorem: lines 2 and 3). (5): \displaystyle \frac{MN}{BC} = \frac{AM}{AB} = \frac{1}{2} (whole-side ratios of the theorem: line 4, with AB = 2 \cdot AM by segment addition). (6): \displaystyle MN = \frac{1}{2}BC (multiplication: line 5).
\displaystyle \frac{AD}{DB} = 1 and \displaystyle \frac{DE}{BC} = \frac{1}{2}, since \displaystyle \frac{DE}{BC} = \frac{AD}{AB} = \frac{2}{4}, so DE = 5. The two ratios differ. The similarity compares \overline{DE} only with sides of \triangle ADE and \triangle ABC, and \overline{DB} is a side of neither: it enters only through AB = AD + DB, which turns \displaystyle \frac{AD}{AB} = \frac{AE}{AC} into \displaystyle \frac{AD}{DB} = \frac{AE}{EC}. No such step exists for \overline{DE}, which is not a piece of \overline{BC}.
\angle ABC \cong \angle ADE by the Corresponding Angles Theorem, with transversal \overleftrightarrow{AD}: \overrightarrow{BA} and \overrightarrow{DA} point the same way, and C and E lie on the same side of \overleftrightarrow{AB}. The angle at A is shared, so \triangle ABC \sim \triangle ADE by AA, and \displaystyle \frac{AB}{AD} = \frac{AC}{AE}. Here B is between A and D, so AD = AB + BD, and likewise AE = AC + CE. Clearing fractions: AB\,(AC + CE) = AC\,(AB + BD), so AB \cdot CE = AC \cdot BD, and dividing by BD \cdot CE gives \displaystyle \frac{AB}{BD} = \frac{AC}{CE}.
D is strictly between A and B, so the parallel through D meets \overline{AC} at a point G strictly between A and C (The procedure), and the theorem gives \displaystyle \frac{AG}{GC} = \frac{AD}{DB} = \frac{AE}{EC}. Call this ratio r and write b = AC. G is between A and C, so GC = b - AG and AG = r\,(b - AG), giving \displaystyle AG = \frac{rb}{1 + r}. E is also between A and C, so EC = b - AE and the same algebra gives \displaystyle AE = \frac{rb}{1 + r}. So AG = AE, and G and E lie on \overrightarrow{AC} at the same distance from A: G = E. So \overline{DE} is the parallel \overline{DG}.