Triangle Proportionality and Midsegments

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Learning goals

Distinguish part-to-part and part-to-whole ratios in a parallel triangle cut
Prove parallelism from proportional interior divisions using the side-splitter converse
Apply both conclusions of the midsegment theorem and match the central triangle correctly
Compare corresponding intervals where parallel lines cut two transversals
Construct equal parts and supplied ratios with compass and unmarked straightedge

A parallel cut gives two useful ratios

A parallel cut puts the side points at the same fraction of their whole sides. Suppose

\begin{aligned}AD&=8\\AB&=28\\AE&=6\\AC&=21.\end{aligned}

The whole-side fractions both equal $2/7$ . The leftover lengths are

\begin{aligned}DB&=20\\EC&=15.\end{aligned}

Therefore, the part-to-part ratios both equal $2/5$ . These are different ratios describing the same division.

One parallel cut gives equal whole-side fractions

Position

D lies between A and B. E lies between A and C. DE stays parallel to BC. AD / AB = 2/5. AE / AC = 2/5. AD / DB = 2/3. AE / EC = 2/3. AB is 12 units long. AC is 18 units long. BC is 21 units long. DE is 8.4 units long. The selected points are not midpoints.

D and E stay inside the two sides while DE remains parallel to BC. Compare whole-side fractions with ratios of the two parts.

Move the position slider, then select Midpoint. At $50\%$ , both side points are midpoints and the cut has half the base length. Without scripting, the figure shows one interior parallel cut with its ratios.

The side-splitter theorem, also called the triangle proportionality theorem, states the following. In a nondegenerate triangle $ABC$ , let $D$ lie between $A,B$ and $E$ between $A,C$ . If $DE\parallel BC$ , then

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

The corresponding part-to-whole equality is $AD/AB=AE/AC$ . Name the denominator before choosing which version to use.

Why a parallel cut divides the sides proportionally#

In the numerical example, the shared whole-side fraction $2/7$ leaves $5/7$ on each side. Dividing the used fraction by the leftover fraction gives $2/5$ , whether the whole side has length $28$ or $21$ .

In general, $DE\parallel BC$ gives $\angle ADE\cong\angle ABC$ by corresponding angles. The triangles also share the angle at $A$ , so AA similarity gives $\triangle ADE\sim\triangle ABC$ .

Let $k$ be their common small-to-whole fraction. Interior points ensure $0<k<1$ .

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

Dividing each first part by its leftover part gives the same ratio.

\begin{aligned}\frac{AD}{DB}&=\frac{k}{1-k},\\\frac{AE}{EC}&=\frac{k}{1-k}.\end{aligned}

Worked example 1 Keep the whole side out of a part-to-part ratio

Problem. In triangle $ABC$ , $D$ lies between $A,B$ , $E$ between $A,C$ , and $DE\parallel BC$ . The lengths are

\begin{aligned}AD&=4\text{ cm}\\DB&=7\text{ cm}\\AE&=6\text{ cm}.\end{aligned}

Find $EC$ and $AC$ .

The denominator DB names the leftover part, while AB names the whole left side.

Solution. Use first part over leftover part on each side.

\begin{aligned}\frac4{7}&=\frac6{EC},\\EC&=10.5\text{ cm},\\AC&=6+10.5,\\&=16.5\text{ cm}.\end{aligned}

As a check, the whole-side fractions are $4/11$ and $6/16.5$ , which are equal. The part-to-part ratio is $4/7$ , not $4/11$ .

Check your understanding

A parallel cut gives $AD:DB=3:8$ . Which is the fraction $AD/AB$ ?

3/8

8/11

3/11

11/3

Reverse the statement to prove parallelism

Equal interior division ratios can prove that a segment is parallel to the base. The side-splitter converse uses the same nondegenerate triangle and interior point conditions: if $AD/DB=AE/EC$ , then $DE\parallel BC$ .

Why proportional interior divisions force a parallel#

First take

\begin{aligned}AD&=5\\DB&=8\\AC&=26.\end{aligned}

Dilate from $A$ by

\begin{aligned}k&=AD/AB,\\&=5/13.\end{aligned}

This sends $B$ to $D$ and $C$ to a point $E_0$ strictly inside $AC$ , because $0<k<1$ . The accepted line-image property sends $BC$ to the unique parallel through $D$ , so $DE_0\parallel BC$ . Now the direct theorem gives $AE_0:E_0C=5:8$ , so

\begin{aligned}AE_0&=10\\E_0C&=16.\end{aligned}

A given interior point $E$ dividing that same side into $10$ and $16$ must be this point $E_0$ .

For the general argument, dilate from $A$ by $k=AD/AB$ , which lies strictly between 0 and 1. The image of $B$ is $D$ , and the image $E_0$ of $C$ lies strictly inside segment $AC$ . Since $A$ is not on $BC$ , the accepted line-image property makes its image the unique parallel to $BC$ through $D$ . Thus $DE_0\parallel BC$ , and the direct theorem gives $E_0$ the same division ratio as the given $E$ . An interior division of one fixed segment with one given positive ratio is unique. To see this, let its whole length be $L>0$ and suppose first-part lengths $x,y$ lie strictly between $0$ and $L$ .

\begin{aligned}\frac{x}{L-x}&=\frac{y}{L-y},\\x(L-y)&=y(L-x),\\xL&=yL,\\x&=y.\end{aligned}

Equal distances from $A$ on the same ray identify the same point. Therefore $E_0=E$ , and the given segment $DE$ is the constructed parallel.

Worked example 2 Test the converse in two cases

Problem. In each case, $ABC$ is nondegenerate, $D$ lies between $A,B$ , and $E$ between $A,C$ . Decide whether $DE$ is parallel to $BC$ .

Case (a):

\begin{aligned}AD&=7\\DB&=9\\AE&=10.5\\EC&=13.5.\end{aligned}

Case (b):

\begin{aligned}AD&=7\\DB&=9\\AE&=10.5\\EC&=12.\end{aligned}

Solution. In case (a), the part-to-part ratios agree.

\frac{10.5}{13.5}=\frac79.

The converse proves parallelism.

In case (b), the right-side ratio is $7/8$ , while the left-side ratio is $7/9$ . They differ. The direct theorem says a parallel would force equal ratios, so $DE$ is not parallel to $BC$ .

Check your understanding

Why does the converse proof construct an actual parallel through D?

A Its intersection has a known ratio, so uniqueness identifies it with E B It assumes DE is parallel before proving it C A parallel must pass through every point of AC D It avoids needing interior point conditions

Midpoints produce a midsegment triangle

A midsegment joins the midpoints of two sides of a triangle. It has two conclusions: it is parallel to the third side and has half that side’s length.

Why a midsegment is parallel and half as long#

Take a triangle with sides

\begin{aligned}AB&=16\\AC&=22\\BC&=26.\end{aligned}

Its midpoint divisions give

\begin{aligned}AD&=DB,\\&=8\\AE&=EC,\\&=11.\end{aligned}

Both part-to-part ratios are $1$ , so the converse proves $DE\parallel BC$ . AA similarity then gives the half-size fraction, so $DE=13$ .

For any nondegenerate triangle, midpoints $D,E$ make both part-to-part ratios equal to $1$ . The converse makes $DE$ parallel to $BC$ . The resulting similarity has $AD/AB=1/2$ , so

DE=\frac12\,BC.

Also, dilation from $A$ by $1/2$ sends $B$ to its midpoint and $C$ to an interior point of $AC$ . Its line image is the parallel through that midpoint. The direct theorem then gives a $1:1$ division on $AC$ , so the intersection is its midpoint.

Worked example 3 Match the central triangle in the correct order

Problem. A triangle has

\begin{aligned}AB&=14\\AC&=18\\BC&=24.\end{aligned}

Let $D,E,F$ be the midpoints of $AB,AC,BC$ , respectively. Find the three sides of $DEF$ , write its correspondence with the outer triangle, and give the factor from the outer triangle to the central one.

Solution. Apply the midsegment theorem to each pair of midpoints.

\begin{aligned}DE&=\frac12 BC,\\&=12,\\EF&=\frac12 AB,\\&=7,\\FD&=\frac12 AC,\\&=9.\end{aligned}

Thus $DE$ matches $CB$ , $EF$ matches $BA$ , and $FD$ matches $AC$ . Their endpoints give $D$ to $C$ , $E$ to $B$ , and $F$ to $A$ . The ordered statement is

\triangle DEF\sim\triangle CBA.

The factor from the outer triangle $CBA$ to $DEF$ is $1/2$ ; the reverse factor is $2$ . Writing $DEF$ to $ABC$ would mismatch these unequal sides.

The central triangle is turned relative to the outer one. Its corresponding order is CBA, not ABC.

Check your understanding

D is the midpoint of AB, E is the midpoint of AC, and DE is 17 units long. What is BC?

8.5

17

19

34

Parallel cross-lines and two transversals

Three distinct parallel lines meet two distinct transversals at $A,B,C$ and $P,Q,R$ , in the same cross-line order. The corresponding intercepted intervals satisfy

\frac{AB}{BC}=\frac{PQ}{QR}.

The ratios compare intervals between the same pairs of parallel lines, in the same order.

Why parallel cross-lines give proportional intervals#

For intersecting transversals, first imagine their intersection $O$ outside the three cross-lines. Distances $OA,OB,OC$ of $3, 11, 17$ and matching distances $OP,OQ,OR$ of $4.5, 16.5, 25.5$ all have factor $1.5$ . Subtracting gives intervals $8, 6$ on the first transversal and $12, 9$ on the second, still with that factor.

In general, extend the transversals to their intersection $O$ . Cross-section triangles have matching angles from the parallel lines and a shared or vertical angle at $O$ . AA gives one common factor for the distances from $O$ on the two transversals. Subtract endpoint distances for intervals on one side of $O$ ; add them when an interval crosses $O$ . Both operations preserve the common factor, so corresponding intercepted intervals are proportional. A cross-line through $O$ has zero endpoint distances and need not be used to form a triangle.

When O lies inside an intercepted interval, add the two vertex distances. A common factor survives addition as well as subtraction.

For parallel transversals, suppose the first strip intercepts $6$ units and the next $9$ on the left. A diagonal in each strip creates two triangles with two equal angle pairs from the parallels and a shared included side. ASA congruence forces right-hand intercepts of $6$ and $9$ as well.

If the transversals are parallel, there is no intersection $O$ . Consider one strip between two cross-lines, labeled $A,P$ on the first and $B,Q$ on the second. Draw $AQ$ . The parallels are

\begin{aligned}AB\parallel PQ\\AP\parallel BQ.\end{aligned}

Therefore, alternate interior angles give

\begin{aligned}\angle BAQ&\cong\angle PQA,\\\angle BQA&\cong\angle PAQ.\end{aligned}

The included side $AQ$ is shared. Thus $\triangle ABQ\cong\triangle QPA$ by ASA, giving $AB=PQ$ .

Shared diagonal AQ and two parallel-line angle pairs give ASA. No quadrilateral property is needed.

Apply the same argument in the next strip to obtain $BC=QR$ . Both corresponding interval factors are $1$ , so the ratio equality holds for parallel transversals too.

Worked example 4 Keep the cross-line order fixed

Problem. Three parallel lines meet transversals at $A,B,C$ and $P,Q,R$ in corresponding order. Suppose

\begin{aligned}AB&=5.5\text{ mm}\\BC&=8.8\text{ mm}\\PQ&=7.5\text{ mm}.\end{aligned}

Find $QR$ .

Solution. The first interval on one transversal corresponds to the first interval on the other, and the second to the second.

\begin{aligned}\frac{QR}{8.8}&=\frac{7.5}{5.5},\\QR&=12\text{ mm}.\end{aligned}

This compares intervals, not distances from an unstated vertex. The transversal lines need not themselves be parallel.

Check your understanding

Three parallel cross-lines cut two transversals at A,B,C and P,Q,R in order. The first and second strip lengths are

\begin{aligned}AB&=4.5\\BC&=6.0\end{aligned}

; $PQ=7.5.$ What is the second-strip length QR?

4.5

6.0

10.0

12.0

Divide a segment with compass and straightedge

Equal compass steps on an auxiliary ray can transfer equal fractions to any given segment. The compass fixes the step length; it does not measure a quotient with a ruler. Use the earlier segment and angle-copy constructions and the parallel-line construction to draw the required parallels.

For a $2:3$ division, mark five equal compass steps on an auxiliary ray, join the last point to $B$ , and construct the parallel through the second step. The first two of the five auxiliary steps become $2/5$ of $AB$ , leaving $3/5$ on the other side.

Equal compass steps become exact fractions on AB

Stage

Stage 5 of 5. Name its intersection with AB as X. The five auxiliary step lengths are equal. Equal-radius circles and a copied chord retain the angle-copy evidence. The line through P2 is parallel to P5B. AX / AB = 2/5. AX / XB = 2/3.

This construction divides AB in the ratio 2 to 3 using an auxiliary ray, five equal compass steps, and an angle-copied parallel.

The figure opens at its completed stage so the construction remains meaningful on paper. Step back to see the auxiliary ray, equal compass openings, last-point join, angle copy, and intersection added in order. To obtain a positive integer ratio $p:q$ , use $p+q$ equal steps and draw the parallel through step $p$ .

To divide $AB$ into $n$ equal parts, draw a ray from $A$ not on line $AB$ . Mark $n$ consecutive equal compass steps $P_1,\ldots,P_n$ , join $P_nB$ , and construct parallels through the earlier step points. In triangle $AP_nB$ , the point at step $i$ is $i/n$ of the auxiliary whole, so its intersection with $AB$ is $i/n$ of that whole too. Consecutive intersections therefore give equal parts.

Worked example 5 Construct four equal parts without a marked ruler

Problem. Using compass and unmarked straightedge, divide a given segment $AB$ into four equal parts. Describe the construction and explain the ratio made by the third division point.

Solution. Draw an auxiliary ray from $A$ away from line $AB$ . Keep one compass opening to mark four equal steps ending at $P_4$ . Join $P_4$ to $B$ . Copy the appropriate angle to construct parallels through $P_1,P_2,P_3$ , meeting $AB$ at $X_1,X_2,X_3$ .

Retain the equal compass steps and the three constructed parallels.

The side-splitter theorem in triangle $AP_4B$ gives the whole-side fractions $1/4$ , $2/4$ , and $3/4$ . Their differences are all $1/4$ , so the four consecutive parts of $AB$ are equal. The third point has $AX_3:X_3B=3:1$ , not $3:4$ .

A supplied ratio can also come from two given positive segments, rather than an integer step count. Copy the first given length $a$ as $AU$ and the second $b$ as $UV$ , consecutively on an auxiliary ray. Join $VB$ and construct the parallel through $U$ , meeting $AB$ at $X$ .

Copy the supplied lengths themselves. A single and a double tick distinguish the two lengths.

The side-splitter theorem gives

\begin{aligned}AX/XB&=AU/UV,\\&=a/b.\end{aligned}

This constructs the ratio of the supplied segments; it does not assume that every numerical real length can be constructed.

Check your understanding

An auxiliary ray has seven equal steps, its last point is joined to B, and a parallel through step 3 meets AB at X. What is AX:XB?

3:7

3:4

4:3

7:3

Triangle Proportionality and Midsegments

Learning goals

A parallel cut gives two useful ratios

Reverse the statement to prove parallelism

Midpoints produce a midsegment triangle

Parallel cross-lines and two transversals

Divide a segment with compass and straightedge

Common mistakes

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Multiple Choice Questions (MCQ)

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