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Similarity and AA: 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 Read the matching in order

It is given that $\triangle ABC\sim\triangle FDE$ .

(a) Name the vertex corresponding to $B$ and the side corresponding to $AC$ .

(b) Complete the ratio statement using the matching side of $BC$ :
$\frac{FD}{AB}=\frac{\square}{BC}.$
1. Hint 1
  Read matching positions in the two triangle names.
2. Hint 2
  A side matches the side joining the images of its two endpoints.
Answer

Part (a). $B$ matches $D$ ; $AC$ matches $FE$ . Part (b).
$\frac{FD}{AB}=\frac{DE}{BC}.$

Full solution

(a) The matching is $A$ to $F$ , $B$ to $D$ , and $C$ to $E$ .
Therefore the matching side of $AC$ joins $F$ and $E$ .

(b) The endpoints of $BC$ match $D$ and $E$ .
The numerator must be $DE$ , keeping second-triangle lengths over first-triangle lengths.

Answer

Part (a). $B$ matches $D$ ; $AC$ matches $FE$ . Part (b).
$\frac{FD}{AB}=\frac{DE}{BC}.$

Key idea

The vertex order determines every side ratio, regardless of page position.
Problem 2 Complete two angle sets

In triangle $ABC$ , angles $A$ and $B$ measure $47^\circ$ and $56^\circ$ . In triangle $DEF$ , angles $D$ and $E$ measure $56^\circ$ and $77^\circ$ .

(a) Find angles $C$ and $F$ .

(b) Prove the triangles similar and write a valid ordered similarity statement.
1. Hint 1
  Use the angle sum separately in each triangle.
2. Hint 2
  Pair equal angle measures before ordering the triangle names.
Answer

Part (a).
$\begin{gathered}m\angle C=77^\circ\\m\angle F=47^\circ\end{gathered}$
Part (b).
$\triangle ABC\sim\triangle FDE$
by AA.

Full solution

(a) Subtract the known angles from 180 degrees.
$\begin{aligned}m\angle C&=180^\circ-47^\circ-56^\circ,\\&=77^\circ,\\m\angle F&=180^\circ-56^\circ-77^\circ,\\&=47^\circ.\end{aligned}$

(b) Match $A$ with $F$ , $B$ with $D$ , and $C$ with $E$ .
Any two of these angle pairs establish AA similarity.
Reading the first triangle as $ABC$ requires the second order $FDE$ .

The completed angles determine the order FDE.

Answer

Part (a).
$\begin{gathered}m\angle C=77^\circ\\m\angle F=47^\circ\end{gathered}$
Part (b).
$\triangle ABC\sim\triangle FDE$
by AA.

Key idea

Complete missing angles before deciding the correspondence.
Problem 3 Use one direction for every side

It is given that $\triangle ABC\sim\triangle QRP$ . The lengths $AB$ , $BC$ , and $CA$ are $8$ cm, $12$ cm, and $7$ cm, and $QR=10$ cm.

(a) Find the scale factor from $ABC$ to $QRP$ .

(b) Find $RP$ and $PQ$ .
1. Hint 1
  The first two letters in each triangle name give one matching side pair.
2. Hint 2
  Use that factor for the other two side pairs without reversing either ratio.
Answer

Part (a).
$k=\frac 54.$
Part (b).
$\begin{gathered}RP=15\text{ cm}\\PQ=8.75\text{ cm}\end{gathered}$

Full solution

(a) Side $AB$ matches $QR$ .
$\begin{aligned}k&=\frac{10}{8},\\&=\frac 54.\end{aligned}$

(b) Side $BC$ matches $RP$ , and $CA$ matches $PQ$ .
$\begin{aligned}RP&=\tfrac 54\cdot 12,\\&=15\text{ cm},\\PQ&=\tfrac 54\cdot 7,\\&=8.75\text{ cm}.\end{aligned}$

Both image sides are larger than their original sides, consistent with the factor above one.

Answer

Part (a).
$k=\frac 54.$
Part (b).
$\begin{gathered}RP=15\text{ cm}\\PQ=8.75\text{ cm}\end{gathered}$

Key idea

The scale direction stays fixed even when side names wrap around the triangle.
Problem 4 Finish AA with a nonincluded side

Nondegenerate triangles $ABC$ and $PQR$ satisfy
$\begin{aligned}\angle A\cong\angle P\\\angle B\cong\angle Q.\end{aligned}$
The angle measures at $A$ and $B$ are $31^\circ$ and $72^\circ$ . Also,
$\begin{aligned}AC&=9\\PR&=15.\end{aligned}$
Describe a dilation of $ABC$ and name an earlier congruence criterion that justify $\triangle ABC\sim\triangle PQR$ . Explain why the chosen side works even though it is not between the two given angle vertices.
1. Hint 1
  Make the matching sides AC and PR equal after the dilation.
2. Hint 2
  The two given angle pairs survive the dilation. Recall the congruence criterion with a nonincluded side.
Answer

Dilate $ABC$ by $\frac 53$ ; the dilated triangle is congruent to $PQR$ by AAS, so $\triangle ABC\sim\triangle PQR$ .

Full solution

Dilate $ABC$ by image length over original length for the chosen side.
$\begin{aligned}k&=\frac{15}{9},\\&=\frac 53.\end{aligned}$

Call the image $A'B'C'$ .
The side $A'C'$ has length $15$ , matching $PR$ .
The angle pairs $A'$ to $P$ and $B'$ to $Q$ remain equal because dilation preserves angles.

The side $A'C'$ is not between the selected vertices $A'$ and $B'$ .
Thus the two angle pairs and this nonincluded side give AAS congruence.
A rigid-motion match after the dilation establishes similarity.

Alternatively, deriving the equal third angles first allows ASA with the chosen side $A'C'$ ; that argument is also valid when its extra angle-sum step is supplied.

Scale the chosen side, then use AAS congruence.

Answer

Dilate $ABC$ by $\frac 53$ ; the dilated triangle is congruent to $PQR$ by AAS, so $\triangle ABC\sim\triangle PQR$ .

Key idea

An AA justification can use AAS when the chosen matching side is nonincluded.
Problem 5 Find the leftover segment

In triangle $ABC$ , $D$ lies between $A$ and $B$ , $E$ lies between $A$ and $C$ , and $DE\parallel BC$ . The lengths are
$\begin{aligned}AD&=7\\DB&=5\\AE&=8.4.\end{aligned}$
Prove a similarity statement, then find $EC$ .

Use the given point orders and parallelism; do not measure the drawing.

Text description of this figure

D lies between A and B, E lies between A and C, and DE is parallel to BC. AD is 7, DB is 5, and AE is 8.4; EC is the requested unknown. The drawing is schematic.
1. Hint 1
  Identify an angle shared by the whole triangle and the small triangle.
2. Hint 2
  Use a corresponding-angle pair from the parallel segments for AA.
3. Hint 3
  Find the whole side AC before subtracting AE.
Answer

$\triangle ADE\sim\triangle ABC$ by AA; $EC=6$ .

Full solution

The small and large triangles share angle $A$ .
Since $DE\parallel BC$ , corresponding angles give $\angle ADE\cong\angle ABC$ .
Therefore $\triangle ADE\sim\triangle ABC$ by AA.

First add the two parts of the left side.
$\begin{aligned}AB&=7+5,\\&=12.\end{aligned}$

Use small-side lengths over whole-triangle lengths.
$\begin{aligned}\frac{8.4}{AC}&=\frac 7{12},\\AC&=14.4.\end{aligned}$

The requested segment is the leftover part of $AC$ .
$\begin{aligned}EC&=14.4-8.4,\\&=6.\end{aligned}$

Answer

$\triangle ADE\sim\triangle ABC$ by AA; $EC=6$ .

Key idea

A ratio of whole matching sides does not directly give a leftover segment.
Problem 6 Add across the intersection

Points $U,O,W$ are collinear in that order, and $V,O,X$ are collinear in that order. It is given that
$\begin{aligned}UV\parallel WX\\UO&=8\\WO&=20\\VO&=6.\end{aligned}$
(a) Prove $\triangle UOV\sim\triangle WOX$ .

(b) Find the full length $VX$ .

The requested whole length crosses O.

Text description of this figure

U, O, W are collinear in order, V, O, X are collinear in order, and UV is parallel to WX. UO is 8, WO is 20, and VO is 6. The unknown OX is one part of the requested whole VX. The drawing is schematic.
1. Hint 1
  The angles at O are vertical angles.
2. Hint 2
  The parallel outer segments give another angle pair; keep the middle vertex O in its matching position.
3. Hint 3
  First find OX, then use the point order V, O, X.
Answer

Part (a). AA gives $\triangle UOV\sim\triangle WOX$ . Part (b).
$VX=21.$

Full solution

(a) Vertical angles give $\angle UOV\cong\angle WOX$ .
With $UW$ as a transversal of $UV\parallel WX$ , alternate interior angles give $\angle VUO\cong\angle XWO$ .
Thus the stated correspondence follows by AA.

(b) The matched sides are $VO$ and $XO$ , in the same direction as $UO$ and $WO$ .
$\begin{aligned}\frac{XO}{6}&=\frac{20}{8},\\XO&=15.\end{aligned}$

Because $O$ lies between $V$ and $X$ , add the two parts.
$\begin{aligned}VX&=6+15,\\&=21.\end{aligned}$

Answer

Part (a). AA gives $\triangle UOV\sim\triangle WOX$ . Part (b).
$VX=21.$

Key idea

The correspondence determines the ratio; the point order determines whether to add.
Problem 7 The second acute angle has a different name

Triangles $ABC$ and $DEF$ are right triangles with right angles at $B$ and $E$ . The angle at $A$ is $28^\circ$ and the angle at $D$ is $62^\circ$ . Also,
$\begin{aligned}AB&=13\text{ cm}\\EF&=19.5\text{ cm}\\BC&=s\text{ cm}.\end{aligned}$
Here, $s>0$ .

(a) Justify the correspondence and write a correct similarity statement.

(b) Find $ED$ .
1. Hint 1
  Find each triangle’s remaining acute angle.
2. Hint 2
  Match equal angle measures, including the right angles, before using the side lengths.
Answer

Part (a).
$\triangle ABC\sim\triangle FED$
by AA. Part (b).
$ED=1.5s\text{ cm}.$

Full solution

(a) The first triangle has $C=62^\circ$ , and the second has
$F=28^\circ.$
Thus $A$ matches $F$ , $B$ matches $E$ , and $C$ matches $D$ .
The right angles and either acute-angle pair establish AA.

(b) Side $AB$ matches $FE$ , so the factor from $ABC$ to $FED$ is $1.5$ .
Side $BC$ matches $ED$ .
$ED=1.5s\text{ cm}.$

Find the other acute angles before matching the sides.

Answer

Part (a).
$\triangle ABC\sim\triangle FED$
by AA. Part (b).
$ED=1.5s\text{ cm}.$

Key idea

Equal angle measures control the correspondence even when different acute angles are initially given.
Problem 8 Reverse the direction after solving

It is given that $\triangle ABC\sim\triangle DEF$ . The lengths are
$\begin{aligned}AB&=2x+1\\DE&=3x-1\\BC&=9\\EF&=12\end{aligned}$
, and all side lengths are positive.

(a) Find $x$ .

(b) State the scale factor from $DEF$ to $ABC$ , and use it to check the lengths of $DE$ and $AB$ .
1. Hint 1
  The numerical corresponding pair BC and EF fixes one scale ratio.
2. Hint 2
  Apply that same ratio to AB and DE before reversing the direction in part (b).
Answer

Part (a).
$x=7.$
Part (b). The factor from $DEF$ to $ABC$ is $\frac 34$ ;
$\begin{aligned}DE&=20\\AB&=15.\end{aligned}$

Full solution

(a) The factor from $ABC$ to $DEF$ is $12/9$ , or $4/3$ .
$\begin{aligned}\frac{3x-1}{2x+1}&=\frac 43,\\3(3x-1)&=4(2x+1),\\9x-3&=8x+4,\\x&=7.\end{aligned}$

(b) Substitute to get
$\begin{aligned}DE&=20\\AB&=15.\end{aligned}$
Reversing the direction gives factor $3/4$ .
$\frac 34\cdot 20=15.$

The check agrees with the original matching and with the positive-length condition.

Answer

Part (a).
$x=7.$
Part (b). The factor from $DEF$ to $ABC$ is $\frac 34$ ;
$\begin{aligned}DE&=20\\AB&=15.\end{aligned}$

Key idea

Reverse every corresponding ratio together when reversing a similarity mapping.
Problem 9 Test a claim about isosceles triangles

Consider the claim “Any two nondegenerate isosceles triangles are similar.” Decide whether the claim is true. If it is false, give two possible angle sets that refute it and explain why they work.
1. Hint 1
  An isosceles triangle has two equal base angles, but its vertex angle can vary.
2. Hint 2
  Choose different vertex angles and use the triangle-angle sum to determine the base angles.
Answer

The claim is false. One counterexample uses angle sets 52°, 64°, 64° and 86°, 47°, 47°; each is isosceles, but their angle sets do not match.

Full solution

Take one triangle with vertex angle $52^\circ$ .
Its two equal base angles each measure $64^\circ$ .
$\frac{180^\circ-52^\circ}{2}=64^\circ.$

Take another with vertex angle $86^\circ$ .
Its two base angles each measure $47^\circ$ .
$\frac{180^\circ-86^\circ}{2}=47^\circ.$

Each angle set is positive, sums to $180^\circ$ , and has equal base angles, so each describes a nondegenerate isosceles triangle.
No complete angle matching is possible between the sets.
Similarity would preserve all angle measures, so these triangles refute the claim.

Each triangle is isosceles, while their complete angle sets differ.

Answer

The claim is false. One counterexample uses angle sets 52°, 64°, 64° and 86°, 47°, 47°; each is isosceles, but their angle sets do not match.

Key idea

A shared triangle type does not guarantee matching angle measures.
Problem 10 Find the missing condition at a crossing

Points $A,O,C$ are collinear in that order, and $B,O,D$ are collinear in that order. No parallelism of $AB$ and $CD$ is given. The angles at $A$ and $D$ measure $43^\circ$ and $58^\circ$ .

Find the measure of $\angle AOB$ that makes the ordered similarity $\triangle AOB\sim\triangle COD$ valid. Prove that this measure supplies the required angle pairs, and explain why vertical angles alone did not determine the correspondence.

The drawing supplies point order; the statement supplies the angle measures.

Text description of this figure

A, O, C and B, O, D are collinear in the specified orders. The angle at A is 43 degrees and the angle at D is 58 degrees. No parallel arrows are present because no parallelism is given. Drawn angles are schematic.
1. Hint 1
  Read the requested order: A must match C, O must match O, and B must match D.
2. Hint 2
  For that order, the angle at B must have the given measure of the angle at D.
3. Hint 3
  Use the angle sum in each triangle and the equality of the vertical angles at O.
Answer

$m\angle AOB=79^\circ$ . Then the angles at $A,C$ are $43^\circ$ , at $B,D$ are $58^\circ$ , and at $O$ are $79^\circ$ , establishing the ordered similarity by AA.

Full solution

For the requested order, $B$ must correspond to $D$ .
Thus the angle at $B$ must be $58^\circ$ .
The first triangle then requires
$\begin{aligned}m\angle AOB&=180^\circ-43^\circ-58^\circ,\\&=79^\circ.\end{aligned}$

The angle $COD$ is vertical to $AOB$ , so it also measures $79^\circ$ .
The angle at $C$ in the second triangle is then
$\begin{aligned}m\angle C&=180^\circ-79^\circ-58^\circ,\\&=43^\circ.\end{aligned}$

Angles at $A$ and $C$ now match, and the angles at $O$ match.
AA gives $\triangle AOB\sim\triangle COD$ .
The value is necessary for the stated order because its $B$ to $D$ match forced the first angle-sum calculation.

Vertical angles supplied only the $O$ to $O$ pair.
They supplied no equality between the other angle pairs until the required intersection measure was found.

Answer

$m\angle AOB=79^\circ$ . Then the angles at $A,C$ are $43^\circ$ , at $B,D$ are $58^\circ$ , and at $O$ are $79^\circ$ , establishing the ordered similarity by AA.

Key idea

A requested correspondence can impose an angle condition that the picture alone does not supply.

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