Angle-Angle Similarity and the Triangle Angle Sum | Grade 8 Math | LevelKite

Two triangles are similar when one is a scaled copy of the other. Their corresponding (matching) angles are equal. Each side of the copy equals the matching side of the original times the same positive number, called the scale factor.

An angle is named by three points, with its vertex (corner point) in the middle. For example, $\angle BAC$ has its vertex at $A$ , and its sides run along $AB$ and $AC$ . You write $m\angle BAC$ for its measure in degrees.

The statement $\triangle ABC \sim \triangle DEF$ means triangle $ABC$ is similar to triangle $DEF$ . The order of the letters gives the matching: $A$ with $D$ , $B$ with $E$ , and $C$ with $F$ . Matching sides join matching vertices, so side $AB$ matches side $DE$ .

Checking all three angle pairs and all three side pairs takes a lot of work. Fortunately, triangles have a shortcut test, called a criterion, that uses only two angle pairs.

Key concept

The angle-angle (AA) criterion

If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. Each equal pair needs a stated reason. The measures can be given. The angles can be vertical angles, which are the opposite angles where two lines cross. Or a parallel-line angle fact can apply to lines marked parallel. How the angles look is never a reason, because the figures in this lesson are not drawn to scale.

The equal angles tell you which vertices match. Matching sides then have equal ratios, such as $\frac{DE}{AB} = \frac{EF}{BC}$ . Later in this lesson you will see why two angle pairs are enough.

m\angle BAC = m\angle EDF \text{ and } m\angle ABC = m\angle DEF \;\Rightarrow\; \triangle ABC \sim \triangle DEF

Not drawn to scale. The example below gives

m\angle BAC = m\angle EDF

and

m\angle ABC = m\angle DEF

. The picture is not the reason.

Worked example

In the figure above, you are given that $m\angle BAC = m\angle EDF$ and $m\angle ABC = m\angle DEF$ . All lengths are in centimeters: $AB = 6$ , $BC = 8$ and $DE = 9$ . Find $EF$ .

List the equal angle pairs and the reason for each. The problem gives both pairs, so you know they are equal without relying on the picture.
$m\angle BAC = m\angle EDF, \quad m\angle ABC = m\angle DEF$
Two angles of triangle ABC equal two angles of triangle DEF, so the triangles are similar by AA. Write the letters in matching order: A with D and B with E. That leaves C with F.
$\triangle ABC \sim \triangle DEF$
Side EF joins E and F, which match B and C, so EF matches BC. Side DE matches AB. In similar triangles, every matching pair gives the same ratio of new side over original side. That ratio is the scale factor, so set these two ratios equal.
$\frac{EF}{BC} = \frac{DE}{AB}$
Substitute the given lengths.
$\frac{EF}{8} = \frac{9}{6}$
Multiply both sides by 8. Doing the same thing to both sides keeps them equal. Because 8 is not zero, you can undo this step, so the solution stays the same.
$EF = 8 \cdot \frac{9}{6} = \frac{72}{6} = 12$

Answer

Side $EF$ is $12$ cm long. As a check, $\frac{12}{8} = 1.5$ and $\frac{9}{6} = 1.5$ .

The arrows mark side

AB

parallel to side

DE

. Not drawn to scale. The example below gives the lengths.

Worked example

In the figure above, segments $AE$ and $BD$ cross at $C$ . The arrows mark $AB \parallel DE$ , which means $AB$ is parallel to $DE$ . All lengths are in centimeters: $AB = 6$ , $AC = 4$ , $BC = 5$ and $DE = 9$ . Find $CE$ and $CD$ .

First pair: angles ACB and ECD are vertical angles, the opposite angles formed where lines AE and BD cross at C. Vertical angles are always equal.
$m\angle ACB = m\angle ECD$
Second pair: line AE crosses the parallel lines AB and DE. Angles BAC and DEC lie between the parallel lines, on opposite sides of AE. So they are alternate interior angles, which are equal because the lines are parallel.
$m\angle BAC = m\angle DEC$
Two pairs of equal angles make the triangles similar by AA. The pairs fix the matching: A goes with E (second pair) and C with C (first pair), so B goes with D.
$\triangle ABC \sim \triangle EDC$
CE joins C and E, which match C and A, so CE matches AC. ED matches AB. Set the two ratios equal, substitute, and multiply both sides by 4, which is not zero.
$\frac{CE}{AC} = \frac{ED}{AB} \;\Rightarrow\; \frac{CE}{4} = \frac{9}{6} \;\Rightarrow\; CE = 4 \cdot \frac{9}{6} = \frac{36}{6} = 6$
CD joins C and D, which match C and B, so CD matches BC. Use the same ratio and multiply both sides by 5.
$\frac{CD}{BC} = \frac{ED}{AB} \;\Rightarrow\; \frac{CD}{5} = \frac{9}{6} \;\Rightarrow\; CD = 5 \cdot \frac{9}{6} = \frac{45}{6} = 7.5$

Answer

Segment $CE$ is $6$ cm long and segment $CD$ is $7.5$ cm long.

Watch out: matching sides by position. In the example above, $CB$ sits on the right of the top triangle and $CE$ sits on the right of the bottom one. Pairing them looks natural: $\frac{CE}{5} = \frac{9}{6}$ gives $CE = 7.5$ cm. But that pairing claims $E$ matches $B$ , and no angle fact says $m\angle DEC = m\angle ABC$ . The fact you do have is $m\angle DEC = m\angle BAC$ . So $E$ matches $A$ , and $CE$ pairs with $AC$ . With $CE = 7.5$ , that pair's ratio would be $\frac{7.5}{4} = 1.875$ , not the scale factor $\frac{9}{6} = 1.5$ . The bottom triangle is a turned-around, larger copy of the top one, so position on the page is not a reason. Let the equal angles set the matching.

Why are two angle pairs enough for similarity? The answer rests on a fact about every triangle: its three angles add to $180^\circ$ . Here is an argument that shows why this is true for any triangle $ABC$ .

The argument adds one line. In the figure below, line $DE$ passes through corner $C$ and is marked parallel to the opposite side $AB$ . Point $D$ is on $A$ 's side of $C$ , and $E$ is on $B$ 's side. The argument then uses two facts. First, when a line crosses two parallel lines, the alternate interior angles it makes are equal. Second, a straight angle measures $180^\circ$ . A straight angle is an angle whose two sides point in opposite directions along one line.

Line

DE

passes through

C

, and the arrows mark it parallel to

AB

. Angles

DCA

,

ACB

and

BCE

sit side by side at

C

. Not drawn to scale.

Worked example

In the figure above, $m\angle CAB = 52^\circ$ and $m\angle CBA = 61^\circ$ , and line $DE$ passes through $C$ parallel to $AB$ . Find $m\angle DCA$ , $m\angle BCE$ and $m\angle ACB$ . Then add the three angles of triangle $ABC$ .

Line AC crosses the parallel lines DE and AB. Angles DCA and CAB lie between those lines, on opposite sides of AC. So they are alternate interior angles, which are equal.
$m\angle DCA = m\angle CAB = 52^\circ$
Line BC crosses the same parallel lines, so angles BCE and CBA are alternate interior angles and are equal.
$m\angle BCE = m\angle CBA = 61^\circ$
D, C and E lie on one line, so angles DCA, ACB and BCE together fill the straight angle DCE, which measures 180 degrees. Substitute the two measures you just found.
$52^\circ + m\angle ACB + 61^\circ = 180^\circ$
Add 52 and 61 to get 113. Then subtract 113 degrees from both sides. Subtracting the same amount from both sides keeps them equal.
$113^\circ + m\angle ACB = 180^\circ \;\Rightarrow\; m\angle ACB = 180^\circ - 113^\circ = 67^\circ$

Answer

Angle $DCA$ measures $52^\circ$ , angle $BCE$ measures $61^\circ$ and angle $ACB$ measures $67^\circ$ . So the angles of triangle $ABC$ add to $52^\circ + 67^\circ + 61^\circ = 180^\circ$ .

Nothing in those steps depended on $52^\circ$ or $61^\circ$ . The same steps work for any triangle $ABC$ with line $DE$ through $C$ parallel to $AB$ :

$m\angle DCA = m\angle CAB$ , because line $AC$ makes these alternate interior angles with the parallel lines.
$m\angle BCE = m\angle CBA$ , because line $BC$ makes these alternate interior angles with the parallel lines.
$m\angle DCA + m\angle ACB + m\angle BCE = 180^\circ$ , because the three angles fill straight angle $DCE$ .
Replacing the first and last terms using steps 1 and 2 gives $m\angle CAB + m\angle ACB + m\angle CBA = 180^\circ$ .

Every triangle has such a line through a corner, so the angles of every triangle add to $180^\circ$ . That is why AA needs only two pairs. Suppose two angles of one triangle equal two angles of another. Then each third angle is $180^\circ$ minus the same total, so the third angles are equal too.

Takeaway: Two triangles are similar when two angles of one equal two angles of the other. Each pair needs a stated reason: given, vertical angles, or lines marked parallel. Let the equal angles, not positions on the page, decide which vertices match. Then set the ratios of matching sides equal. A line through one corner, parallel to the opposite side, places copies of the other two angles beside the third angle so that together they fill a straight angle. That is why a triangle's three angles always add to $180^\circ$ .