Similarity and AA

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Learning goals

Identify corresponding vertices, angles, and sides in an ordered similarity statement
Explain AA similarity using a dilation followed by an earlier congruence criterion
Calculate missing lengths using consistently ordered ratios of corresponding sides
Justify angle correspondences in nested triangles and triangles across intersecting lines

Matching shape through transformations

A triangle can change size and position while keeping its shape. Start with a triangle whose base is $4$ units. Dilate it by $1.5$ , then turn and slide the copy. Its base becomes $6$ units, while the angles keep their measures. A reflection may also be needed to match a copy facing the other way.

Two figures are similar if a positive dilation, together with a sequence of rigid motions, maps one figure exactly onto the other. A rigid motion is a translation, rotation, or reflection. The dilation can have factor $1$ , so congruent figures are included among similar figures.

The dilation properties from the previous lesson tell us that lengths scale by one positive factor and angle measures stay unchanged. Rigid motions preserve both lengths and angle measures.

Keep two angles fixed while size and orientation change

Scale factor Turn Slide right Reflection

The matching is A to P, B to Q, and C to R. The scale factor from ABC to PQR is 1.5. AB is 4 units long. PQ is 6 units long. Every image side is 1.5 times its matching original side. Angles A and P each measure about 71.6°. Angles B and Q each measure 45°. The copy turns 90° counterclockwise. The copy slides right by 3 steps. The copy retains its starting side of the base before it turns. The corresponding lengths differ, so the triangles are not congruent. The matching vertices are not all aligned.

A matches P, B matches Q, and C matches R throughout. A shared-location label combines names without changing that matching. Each slide step moves the copy two units right.

Change the scale factor while keeping the other controls still. The displayed angles stay fixed and every corresponding side uses the same factor.

To align the triangles, set the scale to $1$ , the turn to $0^\circ$ , the slide to $0$ , and the reflection to “retained.” Then vary just the turn or slide. The triangles remain congruent even while their positions differ. Reset restores the initial rotated enlargement. Without scripting, the diagram and readout display that initial state.

Put the correspondence in the names

A correspondence specifies which point of the first figure matches each point of the second. It follows the transformation, not the page position. If $A$ matches $P$ , $B$ matches $Q$ , and $C$ matches $R$ , write

\triangle ABC\sim\triangle PQR.

The symbol $\sim$ means “is similar to.” Read the names in order. For these triangles, the angle measures and length ratios satisfy

\begin{aligned}m\angle A&=m\angle P,\\m\angle B&=m\angle Q,\\m\angle C&=m\angle R.\end{aligned}

\begin{aligned}\frac{PQ}{AB}&=k,\\\frac{QR}{BC}&=k,\\\frac{PR}{AC}&=k.\end{aligned}

Here $k$ is the scale factor from $ABC$ to $PQR$ : image length divided by original length. These side lengths are proportional, meaning that their matching ratios are equal. Reversing the mapping replaces $k$ by $\frac 1k$ . At $k=1$ , corresponding lengths are equal, so the triangles are congruent.

Worked example 1 Follow the order instead of the picture

Problem. It is given that $\triangle ABC\sim\triangle QRP$ . The lengths $AB$ , $QR$ , and $BC$ are $6$ cm, $9$ cm, and $8$ cm. Identify the side matching $BC$ , find its length, and give the scale factor in each direction.

Solution. The statement pairs $A$ with $Q$ , $B$ with $R$ , and $C$ with $P$ . Thus $BC$ matches $RP$ .

Use the direction from $ABC$ to $QRP$ throughout the calculation.

\begin{aligned}k&=\frac{QR}{AB},\\k&=\frac96,\\&=1.5,\\RP&=1.5\cdot8,\\&=12\text{ cm}.\end{aligned}

The reverse factor, from $QRP$ to $ABC$ , is $\frac 23$ . A larger second triangle must give a factor above $1$ in the forward direction and below $1$ in reverse.

Check your understanding

Given $\triangle JKL\sim\triangle VUW$ , which side corresponds to $JL$ ?

VU

VW

UW

UL

Why two angles are enough for triangles

Matching two triangle angles guarantees similarity, even when no side lengths have been given. This is the angle-angle criterion, or AA similarity: if two angles of one nondegenerate triangle equal two corresponding angles of another, the triangles are similar.

The triangle-angle sum makes the third angles equal too. That observation alone does not explain why side lengths must be proportional. The missing step is a dilation followed by congruence.

Why AA establishes similarity#

First take triangles with corresponding angles $36^\circ$ and $79^\circ$ . Let the side between those angles be $5$ units in $ABC$ and $8$ units in $PQR$ .

Dilate $ABC$ by $\frac 85$ . The side between the two selected angles becomes $8$ units long, and the two angles stay $36^\circ$ and $79^\circ$ . The dilated triangle and $PQR$ now have two equal angles and the equal included side. By the earlier ASA congruence criterion, rigid motions can make them coincide. The resizing and rigid motions together give similarity.

Scale the included side to match, preserve the two angles, then use ASA congruence. The right panel shows the resulting alignment.

Nothing depended on the particular numbers. In general, suppose

\begin{aligned}\angle A\cong\angle P\\\angle B\cong\angle Q.\end{aligned}

All triangle sides are positive. Choose

k=\frac{PQ}{AB}.

Dilate $ABC$ by this factor about $A$ , calling the image $A'B'C'$ . Length scaling gives

\begin{aligned}A'B'&=\frac{PQ}{AB}\cdot AB,\\&=PQ.\end{aligned}

Dilation preserves the two angles, so

\begin{aligned}\angle A'\cong\angle P\\\angle B'\cong\angle Q.\end{aligned}

The matching side $A'B'$ lies between these angles. Thus $\triangle A'B'C'\cong\triangle PQR$ by ASA. There is a sequence of rigid motions mapping the dilated triangle onto $PQR$ , which proves similarity by its definition.

Those rigid motions preserve the scaled side lengths. Therefore every side of $PQR$ is $k$ times its matching side of $ABC$ , which also justifies using proportional sides after AA has established the similarity.

Worked example 2 Find the angle matching before the length ratio

Problem. Triangle $ABC$ has angles

\begin{aligned}A&=38^\circ\\B&=64^\circ.\end{aligned}

Triangle $DEF$ has angles

\begin{aligned}D&=78^\circ\\E&=38^\circ.\end{aligned}

The side lengths $AB$ and $EF$ are $7$ cm and $10.5$ cm. Prove the triangles similar, write a correct ordered statement, and find the factor from the first triangle to the second.

Solution. Use the angle sum to find the remaining angles.

\begin{aligned}m\angle C&=180^\circ-38^\circ-64^\circ,\\&=78^\circ,\\m\angle F&=180^\circ-78^\circ-38^\circ,\\&=64^\circ.\end{aligned}

The matching is $A$ to $E$ , $B$ to $F$ , and $C$ to $D$ . Two of these angle pairs establish AA, so

\triangle ABC\sim\triangle EFD.

Now the ratio uses the side $EF$ corresponding to $AB$ .

\begin{aligned}k&=\frac{10.5}{7},\\&=1.5.\end{aligned}

The angle matches determine the name order, even though the second triangle begins with a different letter.

Check your understanding

Two nondegenerate triangles have two pairs of equal corresponding angles. After dilating one triangle to match the side between those angles, which earlier result finishes the similarity justification?

A The triangle-angle sum alone B The definition of a midpoint C ASA congruence D Equality of every original side length

A small triangle inside a larger one

Overlapping triangles can still have a fixed correspondence. Name the complete small triangle and complete large triangle before forming a ratio.

Worked example 3 Use parallels to establish the two angle pairs

Problem. In $\triangle ABC$ , $D$ lies between $A$ and $B$ , and $E$ lies between $A$ and $C$ . It is given that

\begin{aligned}DE\parallel BC\\AD&=9\text{ cm}\\AB&=15\text{ cm}\\AE&=7.2\text{ cm}.\end{aligned}

Prove $\triangle ADE\sim\triangle ABC$ and find $AC$ .

The point orders and parallel segments are given. Drawn lengths are schematic.

Solution. Rays $AD$ and $AB$ are the same ray, as are $AE$ and $AC$ . Thus the small and large triangles share the angle at $A$ .

With $AB$ as a transversal of $DE$ and $BC$ , corresponding angles give $\angle ADE\cong\angle ABC$ . These two angle matches prove $\triangle ADE\sim\triangle ABC$ by AA, with $A$ to $A$ , $D$ to $B$ , and $E$ to $C$ .

Keep small-triangle lengths over large-triangle lengths.

\begin{aligned}\frac{AE}{AC}&=\frac{AD}{AB},\\\frac{7.2}{AC}&=\frac9{15},\\AC&=12\text{ cm}.\end{aligned}

The known $7.2$ cm is the small side $AE$ , not the leftover segment $EC$ .

Check your understanding

In the nested configuration above, which ratio has the same small-to-large direction as $AD/AB$ ?

AC/AE

AE/EC

DB/AB

DE/BC

Triangles across crossing lines

Two triangles can face opposite ways and still have matching angles. At an intersection, vertical angles give one pair; another pair must come from a stated fact.

Worked example 4 Keep the shared intersection in the middle

Problem. Points $A,O,C$ are collinear in that order, and $B,O,D$ are collinear in that order. The segments $AB$ and $CD$ are parallel. The lengths $AO$ , $CO$ , and $BO$ are $9$ cm, $15$ cm, and $6.6$ cm. Prove $\triangle AOB\sim\triangle COD$ and find $DO$ .

The crossing lines give vertical angles at O, and the marked parallel segments supply a second pair.

Solution. Angles $AOB$ and $COD$ are vertical angles, so they are equal. Since $AB\parallel CD$ and $AC$ is a transversal, $\angle BAO\cong\angle DCO$ by alternate interior angles.

AA now gives the stated order: $A$ matches $C$ , $O$ matches $O$ , and $B$ matches $D$ . Use second-triangle lengths over first-triangle lengths.

\begin{aligned}\frac{DO}{BO}&=\frac{CO}{AO},\\\frac{DO}{6.6}&=\frac{15}{9},\\DO&=11\text{ cm}.\end{aligned}

The left-hand position of $B$ does not make it correspond to whichever right-hand point looks closest. The angle matches determine the correspondence.

Worked example 5 Recognize when the evidence stops short

Problem. Two triangles each have a $55^\circ$ angle. Is that fact enough to prove them similar? Justify your answer.

Solution. No. One triangle can have angles $55^\circ$ , $70^\circ$ , and $55^\circ$ . Another can have angles $55^\circ$ , $60^\circ$ , and $65^\circ$ . Both sets are positive and sum to $180^\circ$ , so they describe possible triangles, but the full angle sets do not match.

One common angle can coexist with different remaining angles.

AA needs equal angles at two distinct vertices in each triangle. A single angle, even one marked in both pictures, supplies only one pair.

Check your understanding

In a crossing-lines configuration, only the vertical angles at the intersection are known to be equal. What can be concluded about the two triangles?

A More information is needed for AA B They must be similar by AA C They must be congruent D They cannot be similar

Similarity and AA

Learning goals

Matching shape through transformations

Put the correspondence in the names

Why two angles are enough for triangles

A small triangle inside a larger one

Triangles across crossing lines

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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