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Triangle Proportionality and Midsegments: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two denominators, two different comparisons

In nondegenerate triangle $ABC$ , $D$ lies between $A,B$ , $E$ between $A,C$ , and $DE\parallel BC$ . The lengths are
$\begin{aligned}AD&=9\\DB&=16\\AC&=40.\end{aligned}$
(a) Find $AE$ and $EC$ .

(b) Give $AD/AB$ and $AD/DB$ , and explain why they are different.

Label the whole side before setting up the proportion.

Text description of this figure

The parallel segment DE cuts interior points on AB and AC. Given lengths AD9, DB16 and AC40 are labeled; the drawing is schematic.
1. Hint 1
  Name the whole side before choosing a proportion.
2. Hint 2
  The whole-side fraction transfers to the other whole side. Subtract only after finding its first part.
Answer

(a)
$\begin{aligned}AE&=14.4\\EC&=25.6.\end{aligned}$
(b)
$\begin{aligned}AD/AB&=9/25\\AD/DB&=9/16.\end{aligned}$

Full solution

The whole side is
$\begin{aligned}AB&=9+16,\\&=25.\end{aligned}$
AA gives the same part-to-whole fraction on both sides.
$\begin{aligned}\frac{AE}{40}&=\frac 9{25},\\AE&=14.4.\end{aligned}$

The remaining part is
$\begin{aligned}EC&=40-14.4,\\&=25.6.\end{aligned}$

The two comparisons have different denominators.
$\begin{aligned}\frac{AD}{AB}&=\frac 9{25},\\\frac{AD}{DB}&=\frac 9{16}.\end{aligned}$

The first compares the first part with the whole; the second compares it with the leftover part.

Answer

(a)
$\begin{aligned}AE&=14.4\\EC&=25.6.\end{aligned}$
(b)
$\begin{aligned}AD/AB&=9/25\\AD/DB&=9/16.\end{aligned}$

Key idea

A ratio is determined by both its numerator and its denominator.
Problem 2 Choose the value that makes a parallel

In nondegenerate triangle $ABC$ , $D$ lies between $A,B$ and $E$ between $A,C$ . The lengths are
$\begin{aligned}AD&=2x+1\\DB&=9\\AE&=10\\EC&=15.\end{aligned}$
Find the positive value of $x$ that makes $DE\parallel BC$ . Name the result that justifies the parallel.
1. Hint 1
  The converse compares corresponding first-part to leftover-part ratios.
2. Hint 2
  Set the ratio containing the expression equal to the ratio on AC, then check that all lengths are positive.
Answer

$x=2.5$ ; the side-splitter converse gives $DE\parallel BC$ .

Full solution

For the parallel, the interior division ratios must agree.
$\begin{aligned}\frac{2x+1}{9}&=\frac{10}{15},\\2x+1&=6,\\x&=2.5.\end{aligned}$

At this value, $AD=6>0$ , and both part-to-part ratios are $2/3$ .
With the stated nondegenerate triangle and interior points, the side-splitter converse proves $DE\parallel BC$ .

Answer

$x=2.5$ ; the side-splitter converse gives $DE\parallel BC$ .

Key idea

Equal interior division ratios can prove parallelism rather than merely use it.
Problem 3 A parallel through one midpoint

In nondegenerate triangle $ABC$ , $D$ is the midpoint of $AB$ and $AC=30$ . A parallel to $BC$ through $D$ meets $AC$ at an interior point $E$ .

(a) Find $AE$ and $EC$ and identify the position of $E$ on $AC$ .

(b) If
$\begin{aligned}BC&=5u+4\\DE&=3u-2\end{aligned}$
, find $u$ , $BC$ , and $DE$ .
1. Hint 1
  Use the direct side-splitter theorem with the known equal parts of AB.
2. Hint 2
  Once both endpoints are midpoints, use both conclusions of the midsegment theorem.
Answer

(a)
$\begin{aligned}AE&=EC,\\&=15\end{aligned}$
, so $E$ is the midpoint of $AC$ . (b)
$\begin{aligned}u&=8\\BC&=44\\DE&=22.\end{aligned}$

Full solution

Since $D$ is a midpoint, $AD/DB=1$ .
Side-splitter gives $AE/EC=1$ , so the two parts of the side of length $30$ are equal.
Hence
$\begin{aligned}AE&=EC,\\&=15\end{aligned}$
, and the interior point $E$ is its midpoint.

Now $DE$ is a midsegment, so its length is half of $BC$ .
$\begin{aligned}2(3u-2)&=5u+4,\\6u-4&=5u+4,\\u&=8.\end{aligned}$

Substitute to find the positive lengths.
$\begin{aligned}BC&=5(8)+4,\\&=44,\\DE&=3(8)-2,\\&=22.\end{aligned}$

Answer

(a)
$\begin{aligned}AE&=EC,\\&=15\end{aligned}$
, so $E$ is the midpoint of $AC$ . (b)
$\begin{aligned}u&=8\\BC&=44\\DE&=22.\end{aligned}$

Key idea

A midpoint plus a parallel supplies the second midpoint.
Problem 4 Match the central triangle by its sides

A triangle has
$\begin{aligned}AB&=22\\AC&=28\\BC&=36.\end{aligned}$
The points $D,E,F$ are the midpoints of $AB,AC,BC$ respectively.

(a) Find $DE$ , $EF$ , and $FD$ .

(b) Write the natural ordered similarity from $DEF$ to the outer triangle and state its factor in that direction.
1. Hint 1
  For each central side, name the outer side to which it is parallel.
2. Hint 2
  Find the vertex corresponding to D from the two outer sides matching DE and DF.
Answer

(a)
$\begin{aligned}DE&=18\\EF&=11\\FD&=14.\end{aligned}$
(b) $\triangle DEF\sim\triangle CBA$ , with central-to-outer factor $2$ .

Full solution

Apply the midsegment theorem to each pair of midpoints.
$\begin{aligned}DE&=\tfrac 12 BC,\\&=18,\\EF&=\tfrac 12 AB,\\&=11,\\FD&=\tfrac 12 AC,\\&=14.\end{aligned}$

The side pairs are $DE$ with $CB$ , $EF$ with $BA$ , and $FD$ with $AC$ .
Their shared endpoints give $D$ to $C$ , $E$ to $B$ , and $F$ to $A$ .
Thus $\triangle DEF\sim\triangle CBA$ by SSS similarity.

The requested factor goes from the central triangle to the outer one:
$\begin{aligned}CB/DE&=36/18,\\&=2.\end{aligned}$
The reverse direction has factor $1/2$ .

Use the side opposite each midpoint pair to establish the matching.

Answer

(a)
$\begin{aligned}DE&=18\\EF&=11\\FD&=14.\end{aligned}$
(b) $\triangle DEF\sim\triangle CBA$ , with central-to-outer factor $2$ .

Key idea

The central triangle reverses the natural outer vertex order.
Problem 5 Transversals with no common vertex

Three parallel cross-lines meet one transversal at $A,B,C$ and another at $P,Q,R$ , in corresponding order. The two transversals are themselves parallel. Given
$\begin{aligned}AB&=9\\BC&=14\end{aligned}$
, find $PQ$ and $QR$ .

Justify equality of the corresponding intervals using earlier ASA congruence. Do not assume a property of quadrilaterals.
1. Hint 1
  Draw AQ across the first strip. The opposite interval endpoints determine two triangles.
2. Hint 2
  Use the two parallel families to match the angles around AQ, which is shared.
Answer

$\begin{aligned}PQ&=9\\QR&=14.\end{aligned}$

Full solution

In the first strip,
$\begin{aligned}AB\parallel PQ\\AP\parallel BQ.\end{aligned}$
The diagonal $AQ$ gives equal angles
$\begin{aligned}\angle BAQ&=\angle PQA\\\angle BQA&=\angle PAQ.\end{aligned}$

Side $AQ$ is common and lies between those angles.
Hence $\triangle ABQ\cong\triangle QPA$ by ASA, so
$\begin{aligned}AB&=QP,\\&=9.\end{aligned}$

Apply the same argument in the next strip, using diagonal $BR$ , to obtain
$\begin{aligned}BC&=QR,\\&=14.\end{aligned}$
No intersection of the transversals or quadrilateral theorem is needed.

Two parallel families and a shared included side give ASA.

Answer

$\begin{aligned}PQ&=9\\QR&=14.\end{aligned}$

Key idea

When parallel transversals never meet, a strip diagonal provides an ASA argument.
Problem 6 An interval that crosses the vertex

Two transversals meet at $O$ . Their points occur in order $A,O,B,C$ and $P,O,Q,R$ , with $AP\parallel BQ\parallel CR$ . The lengths are
$\begin{aligned}OA&=2.8\\OB&=4.2\\OP&=4\\OQ&=6\\BC&=7.\end{aligned}$
(a) Find $AB$ and $PQ$ .

(b) Find $QR$ and explain why subtraction alone would not give the first pair of intervals.
1. Hint 1
  Check whether O lies inside each intercepted interval.
2. Hint 2
  Find the second-to-first vertex-distance factor, then apply it to a corresponding interval.
Answer

(a)
$\begin{aligned}AB&=7\\PQ&=10.\end{aligned}$
(b) $QR=10$ ; intervals crossing $O$ require addition.

Full solution

The point $O$ lies inside both first intervals.
Add their two endpoint distances.
$\begin{aligned}AB&=2.8+4.2,\\&=7,\\PQ&=4+6,\\&=10.\end{aligned}$

AA between the triangles from $O$ gives one common second-to-first factor.
$\begin{aligned}\frac{OP}{OA}&=\frac 4{2.8},\\&=\frac{10}{7},\\\frac{OQ}{OB}&=\frac 6{4.2},\\&=\frac{10}{7}.\end{aligned}$

That factor also applies to corresponding intervals between the next cross-lines, so
$\begin{aligned}QR&=(10/7)\cdot 7,\\&=10.\end{aligned}$

A difference gives the gap only for two endpoints on the same ray from $O$ .
Here each first interval crosses $O$ , so its length is a sum; multiplying both summands by one factor preserves that factor.

The location of O decides whether a gap is a sum or a difference.

Answer

(a)
$\begin{aligned}AB&=7\\PQ&=10.\end{aligned}$
(b) $QR=10$ ; intervals crossing $O$ require addition.

Key idea

Use addition across the vertex and subtraction along one ray.
Problem 7 Seven equal parts without ruler division

On paper, draw a segment $AB$ and actually divide it into seven equal parts using a compass and an unmarked straightedge. Carry out the auxiliary-ray and parallel construction, retaining its compass marks, then explain why the seven parts are equal.

Call the successive division points $X_1,\ldots,X_6$ , starting from $A$ . What is $AX_5:X_5B$ ?
1. Hint 1
  Put seven equal compass steps on an auxiliary ray rather than trying to measure AB divided by seven.
2. Hint 2
  Join the last step to B, then transfer the earlier steps with constructed parallels.
Answer

Use seven equal auxiliary steps and parallels through the first six; $AX_5:X_5B=5:2$ .

Full solution

Draw an auxiliary ray from $A$ that does not lie on $AB$ .
With one fixed positive compass opening, mark consecutive points $P_1,\ldots,P_7$ .
Join $P_7$ to $B$ .

Use the earlier angle-copy construction to draw through each $P_i$ , for $i=1,\ldots,6$ , a parallel to $P_7B$ .
Let it meet $AB$ at $X_i$ .

Similarity in the auxiliary triangle gives
$AX_i/AB=i/7.$
Consecutive whole-side fractions differ by $1/7$ , including the first and last pieces, so all seven parts are equal.

Five parts lie before $X_5$ and two after it.
Thus
$AX_5:X_5B=5:2.$
The construction never requires reading a quotient from a ruler.

Each consecutive whole-side fraction increases by one seventh.

Answer

Use seven equal auxiliary steps and parallels through the first six; $AX_5:X_5B=5:2$ .

Key idea

Equal compass steps transfer to equal fractions through parallels.
Problem 8 Reverse a supplied ratio

You are given actual segments of lengths
$\begin{aligned}a&=7\text{ mm}\\b&=10\text{ mm}\end{aligned}$
, together with segment $AB$ . A proposed construction copies $AU=a$ and then $UV=b$ on an auxiliary ray, joins $VB$ , and constructs $UX\parallel VB$ .

The target is $AX:XB=10:7$ . Explain the error, then give a corrected compass-and-unmarked-straightedge construction without measuring the quotient.
1. Hint 1
  Test the proposed order with simple unequal lengths before using letters.
2. Hint 2
  The first auxiliary part becomes the first part of the requested ratio.
Answer

The proposal gives $7:10$ . Copy the supplied $10$ mm segment first and the $7$ mm segment second, then use the same parallel construction to obtain $10:7$ .

Full solution

The proposal transfers the consecutive auxiliary parts in their original order, so it gives
$AX:XB=7:10.$
That differs from the target $10:7$ .

For the correction, copy the given length $b=10$ mm as $AU$ , then copy the given length $a=7$ mm as $UV$ on the same auxiliary ray, with $A,U,V$ in order.
Join $V$ to $B$ and construct the parallel through $U$ , meeting $AB$ at $X$ .

The side-splitter theorem in triangle $AVB$ gives
$\begin{aligned}AX/XB&=AU/UV,\\&=10/7.\end{aligned}$
Copying the two supplied segments is enough; no measured decimal quotient is needed.

Correct the order on the auxiliary ray before drawing the parallel.

Answer

The proposal gives $7:10$ . Copy the supplied $10$ mm segment first and the $7$ mm segment second, then use the same parallel construction to obtain $10:7$ .

Key idea

The order of the copied lengths determines the order of the final ratio.
Problem 9 Find the first invalid numbered line

In nondegenerate triangle $ABC$ , $D,E$ are interior side points and $DE\parallel BC$ . Given
$\begin{aligned}AD&=3.5\\DB&=10.5\\AE&=6\end{aligned}$
, consider this attempt.

Line 1: $AB=14$ .

Line 2: $AD/AB=AE/EC$ .

Line 3: $3.5/14=6/EC$ .

Line 4: $EC=24$ .

Identify the first invalid line, explain the mismatched comparison, and find the correct $EC$ .

Compare the first and leftover parts on both sides.

Text description of this figure

The schematic parallel cut labels AD3.5, DB10.5 and AE6; EC is unknown. No computed side length is drawn.
1. Hint 1
  Classify each ratio in Line 2 as part-to-whole or part-to-part.
2. Hint 2
  Use matching denominators, then decide whether a subtraction is still needed.
Answer

Line 2 is the first invalid line. The correct length is $EC=18$ .

Full solution

Line 1 correctly adds the two parts.
Line 2 is the first error: $AD/AB$ is part-to-whole, while $AE/EC$ is part-to-part.

Use part-to-part on both sides instead.
$\begin{aligned}\frac{3.5}{10.5}&=\frac 6{EC},\\\frac 13&=\frac 6{EC},\\EC&=18.\end{aligned}$

Alternatively, whole-side fractions give
$\begin{aligned}AE/AC&=3.5/14,\\&=1/4\end{aligned}$
, so
$\begin{aligned}AC&=24\\EC&=24-6,\\&=18.\end{aligned}$
The value in Line 4 is the whole side $AC$ , not its leftover part.

Answer

Line 2 is the first invalid line. The correct length is $EC=18$ .

Key idea

Matching fraction shapes is essential before cross-multiplying.
Problem 10 Does one midpoint force every joining segment?

Refute the claim: a segment from the midpoint of one triangle side to any interior point of another side must be a midsegment.

Give a concrete nondegenerate triangle and interior points that satisfy the stated midpoint condition but fail the conclusion. Explain the failure using this lesson's ratios.
1. Hint 1
  Keep the first endpoint at a midpoint but divide the second side into unequal positive parts.
2. Hint 2
  Choose outer side lengths satisfying the triangle inequality, then compare the two part-to-part ratios.
Answer

One counterexample has
$\begin{aligned}AB&=14\\AC&=20\\BC&=24\end{aligned}$
, with
$\begin{aligned}AD&=DB,\\&=7\\AE&=8\\EC&=12.\end{aligned}$
Then $DE$ is not a midsegment and is not parallel to $BC$ .

Full solution

Take a triangle with
$\begin{aligned}AB&=14\\AC&=20\\BC&=24.\end{aligned}$
All lengths are positive and satisfy the triangle inequality, so this is a possible nondegenerate triangle.
Place $D$ halfway along $AB$ and $E$ eight units from $A$ along $AC$ .

Then
$\begin{aligned}AD&=DB,\\&=7\\AE&=8\\EC&=12.\end{aligned}$
The first endpoint is a midpoint; the second is interior but is not a midpoint.

The part-to-part ratios differ.
$\begin{aligned}\frac{AD}{DB}&=1,\\\frac{AE}{EC}&=\frac 23.\end{aligned}$

If $DE$ were parallel to $BC$ , side-splitter would make the ratios equal, a contradiction.
Thus this segment fails both the second-midpoint requirement and the claimed parallelism.

One actual counterexample defeats the universal claim.

Answer

One counterexample has
$\begin{aligned}AB&=14\\AC&=20\\BC&=24\end{aligned}$
, with
$\begin{aligned}AD&=DB,\\&=7\\AE&=8\\EC&=12.\end{aligned}$
Then $DE$ is not a midsegment and is not parallel to $BC$ .

Key idea

A midsegment requires two midpoints; one alone is insufficient.

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