Math 8 · 8.G.A.5 · G.SRT.A.3

AA similarity

Two triangles have two angles that match. Without measuring a single side, can you be sure they have the same shape?

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Geometric means in a right triangleGeometry Proofs about similar figuresGrades 9 to 11 Proportions in similar trianglesMath 8 Proving two triangles similarGeometry SSS and SAS similarityGeometry Triangle proportionality theoremGeometry +3 more

Where you are 10 in Similarity

Geometry11

Similarity10

Similar figures

Why this exists

To show two triangles are similar straight from the definition you need six facts: three pairs of congruent angles and three side ratios that agree. In most real problems some of those are out of reach. The height of a tree, the width of a river, a side hidden inside a diagram: these are the lengths you want to find, so they cannot be part of the evidence.

Angles are often much easier to come by. A corner two triangles share, two lines that cross, a line drawn parallel to another, the sun’s rays falling at the same slant on everything at once: each of these hands you a pair of congruent angles without a ruler. AA similarity says that two such pairs are enough. Once two angles match, the shape is settled, and every side ratio follows.

This one test is behind measuring heights with shadows, behind the triangle proportionality theorem, and behind the fact that the sine of an angle is a single number.

Definition

Two triangles are similar when one is a scaled copy of the other: matching angles congruent, and matching sides all in one ratio k, the scale factor. We write \triangle ABC \sim \triangle DEF, and the order of the letters is the matching: A with D, B with E, and C with F.

AA similarity (angle, angle). If two angles of one triangle are congruent to two angles of another triangle,

\angle A \cong \angle D \quad\text{and}\quad \angle B \cong \angle E

then the triangles are similar: \triangle ABC \sim \triangle DEF.

No side has to be measured. In the figure the two triangles share a 50^\circ angle and a 60^\circ angle, so they are similar, whatever their sizes.

Figure 1Two pairs of matching angles, and two sizes. The triangles are similar.

In the figures that follow, matching arcs mark congruent angles: one arc goes with one arc, two arcs with two arcs. The arcs, not the positions on the page, decide which vertex matches which.

Intuition

Take a segment AB and stand at each end. From A draw a ray at 50^\circ to the segment, and from B a ray at 60^\circ, both on the same side. The two rays lean toward each other, since 50^\circ + 60^\circ is less than 180^\circ, so they meet, and they meet at exactly one point C. Two angles and the side between them leave the triangle no freedom at all.

Now make the segment twice as long and draw the same two rays from its ends. Each new ray is parallel to the old one, and the rays meet twice as far away. The new triangle is the old one blown up by a factor of 2. Changing the side changed the size; the two angles alone decided the shape.

The third angle never needed checking. The angles of every triangle add to 180^\circ (Triangle angle sum), so once two angles are fixed the third has only one value left. Two matching angles really means three.

Results

AA similarity theorem. If two angles of one triangle are congruent to two angles of another triangle, the triangles are similar.

Once the test succeeds, everything similarity promises comes with it. The third angles are congruent, and the sides are in proportion, matched in the order of the similarity statement:

\frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA} = k

So one known pair of matching sides gives k, and k gives every other side.

AA asks only for angles, and these are the usual places a pair of congruent angles comes from:

Where the pair comes from	Why the two angles are congruent
a corner the two triangles share	it is one and the same angle
two lines crossing between the triangles	Vertical angles are congruent
a line parallel to a side	Corresponding angles or Alternate and same side angles
a right angle in each triangle	every right angle is 90^\circ
two angles of each triangle already known	the third is 180^\circ minus the other two

AA is one of three similarity tests. The other two work from sides, and are taught in SSS and SAS similarity:

Test	What it needs
AA	two pairs of congruent angles
SSS	three pairs of sides in one ratio
SAS	two pairs of sides in one ratio, and the angles between them congruent

Why it is true

The argument blows the first triangle up to the right size, then recognizes the result as a copy of the second.

Two figures are similar exactly when a similarity transformation, a dilation followed by rigid motions, carries one onto the other. So it is enough to find a dilation that turns \triangle ABC into a triangle congruent to \triangle DEF.

Suppose \angle A \cong \angle D and \angle B \cong \angle E. Let \displaystyle k = \frac{DE}{AB}, and dilate \triangle ABC from the point A with scale factor k. The point A stays where it is. B moves along ray AB to a point B' with AB' = k \cdot AB = DE, and C moves along ray AC to a point C'.

Figure 2Dilate \triangle ABC from A until AB' = DE. The dashed image is congruent to \triangle DEF.

Now compare \triangle AB'C' with \triangle DEF, one part at a time.

The angle at A has not changed, because B' is still on ray AB and C' is still on ray AC. So the angle at A in \triangle AB'C' is congruent to \angle D.

A dilation sends a line that misses its center to a parallel line (what a dilation does to a line), so B'C' \parallel BC. Line AB crosses both, and \angle AB'C' and \angle ABC sit in matching positions at the two crossings, so they are congruent corresponding angles. That makes \angle AB'C' \cong \angle E.

The side between those two angles is AB', and AB' = DE.

Two angles and the side between them match, so \triangle AB'C' \cong \triangle DEF by ASA congruence. A sequence of rigid motions therefore carries \triangle AB'C' onto \triangle DEF; in the figure it includes a reflection, since \triangle DEF is drawn as a mirror image. The dilation followed by those rigid motions carries \triangle ABC onto \triangle DEF, so the triangles are similar. \blacksquare

Worked examples

The lesson is taught in two parts. Examples 1 to 5 belong to the first: using AA when the angles are given, and finding sides once two triangles are known to be similar. Examples 6 to 10 belong to the second: finding the two pairs of angles inside a figure, from a shared corner, parallel lines, right angles and the sun, and putting the result to work.

Example 1

Find the third angle, then match

In \triangle ABC, \angle A = 46^\circ and \angle B = 81^\circ. In \triangle PQR, \angle P = 53^\circ and \angle Q = 46^\circ. Are the triangles similar? If they are, write a similarity statement.

Solution

At first sight only one pair matches: \angle A and \angle Q are both 46^\circ. The angle sum gives the third angle of each triangle.

\angle C = 180^\circ - 46^\circ - 81^\circ = 53^\circ

\angle R = 180^\circ - 53^\circ - 46^\circ = 81^\circ

Now \angle A \cong \angle Q and \angle C \cong \angle P. That is two pairs of congruent angles, so the triangles are similar by AA. A similarity statement lists matching vertices in the same order. A goes with Q and C with P, so the remaining vertices B and R go together:

\triangle ABC \sim \triangle QRP

Check: \angle B = 81^\circ = \angle R, so the third pair agrees, as it must.

Example 2

When only one pair matches

In \triangle DEF, \angle D = 40^\circ and \angle E = 65^\circ. In \triangle XYZ, \angle X = 40^\circ and \angle Z = 70^\circ. Are the triangles similar?

Solution

Complete both lists of angles with the angle sum.

\angle F = 180^\circ - 40^\circ - 65^\circ = 75^\circ

\angle Y = 180^\circ - 40^\circ - 70^\circ = 70^\circ

The angles of \triangle DEF are 40^\circ, 65^\circ and 75^\circ; those of \triangle XYZ are 40^\circ, 70^\circ and 70^\circ. Similar triangles have all three pairs of angles congruent, and nothing in the second list can be paired with 65^\circ. So the triangles are not similar.

One pair of congruent angles, \angle D \cong \angle X, is not enough. With one angle fixed at 40^\circ, the remaining 140^\circ can be shared between the other two angles in many ways, and each way is a different shape.

Example 3

Missing sides from marked angles

In the figure, \angle A \cong \angle D and \angle B \cong \angle E, as the arcs show. AB = 4, BC = 6, CA = 5 and DE = 10. Find EF = x and FD = y.

Figure 3The arcs, not the positions on the page, say which vertex matches which.

Solution

There are two pairs of congruent angles, so \triangle ABC \sim \triangle DEF by AA, with A matching D, B matching E and C matching F. The second triangle is drawn flipped, so its position on the page is no guide; the arcs are.

The pair of matching sides with both lengths known is AB and DE, and it gives the scale factor:

k = \frac{DE}{AB} = \frac{10}{4} = \frac{5}{2}

EF matches BC and FD matches CA, so each is \displaystyle \frac{5}{2} of its partner:

EF = \tfrac{5}{2} \cdot 6 = 15, \qquad FD = \tfrac{5}{2} \cdot 5 = 12.5

So x = 15 and y = 12.5. Check: \displaystyle \frac{15}{6} and \displaystyle \frac{12.5}{5} are both 2.5, the same as \displaystyle \frac{10}{4}.

Example 4

Angles written in terms of x

In \triangle ABC, \angle A = 52^\circ and \angle C = 61^\circ. In \triangle DEF, \angle D = (x + 12)^\circ, \angle E = (2x - 13)^\circ and \angle F = (x + 21)^\circ. Find x, and decide whether the triangles are similar.

Solution

The three angles of \triangle DEF add to 180^\circ:

(x + 12) + (2x - 13) + (x + 21) = 180

Collecting like terms gives 4x + 20 = 180, so 4x = 160 and x = 40. Substituting,

\angle D = 52^\circ, \quad \angle E = 67^\circ, \quad \angle F = 61^\circ

Since \angle D \cong \angle A and \angle F \cong \angle C, AA gives \triangle ABC \sim \triangle DEF. Check with the third pair: \angle B = 180^\circ - 52^\circ - 61^\circ = 67^\circ, which is \angle E.

Example 5

Sides written in terms of x

In \triangle PQR and \triangle STU, \angle P \cong \angle S and \angle Q \cong \angle T. PQ = 6, QR = x + 2, ST = 9 and TU = 2x - 1. Find x, QR and TU.

Solution

By AA, \triangle PQR \sim \triangle STU, so PQ matches ST and QR matches TU. Every pair of matching sides is in the ratio

k = \frac{ST}{PQ} = \frac{9}{6} = \frac{3}{2}

so \displaystyle TU = \frac{3}{2} \cdot QR:

2x - 1 = \tfrac{3}{2}(x + 2)

Multiplying both sides by 2 clears the fraction: 4x - 2 = 3x + 6, so x = 8. Then QR = 10 and TU = 15. Check: \displaystyle \frac{15}{10} = \frac{3}{2}.

Example 6

A segment parallel to one side

In \triangle ABC, D lies on AB and E on AC, with DE \parallel BC. AD = 6, DB = 10 and DE = 9. Find BC.

Figure 4A segment parallel to one side cuts off a small triangle similar to the whole.

Solution

The small \triangle ADE and the whole \triangle ABC share the corner A. Look for two pairs of congruent angles.

The angle at A belongs to both triangles, so it is one pair. Line AB crosses the parallel lines DE and BC, and \angle ADE and \angle ABC sit in matching positions at the two crossings, so they are congruent Corresponding angles. That is the second pair, and by AA

\triangle ADE \sim \triangle ABC

with D matching B and E matching C. The side of the large triangle that matches AD is the whole of AB, and by segment addition AB = AD + DB = 6 + 10 = 16. So

k = \frac{AB}{AD} = \frac{16}{6} = \frac{8}{3}

and BC, which matches DE, is \displaystyle \frac{8}{3} \cdot 9. So BC = 24.

The same parallel lines also make \angle AED \cong \angle ACB, the double arcs in the figure, but AA needed only two pairs.

Example 7

Two triangles meeting at a point

Segments AD and BC cross at E, and AB \parallel CD. AB = 8, CD = 12, AE = 6 and BE = 5. Find ED and CE.

Figure 5Parallel sides make alternate interior angles in the two triangles.

Solution

Line AD crosses the parallel lines AB and CD, so \angle BAE and \angle CDE are Alternate and same side angles and are congruent (one arc each in the figure). Line BC crosses the same parallels, so \angle ABE \cong \angle DCE as well (two arcs). By AA

\triangle ABE \sim \triangle DCE

with A matching D, B matching C and E matching E. The third pair, the two angles at E, are Vertical angles, which agrees.

The known matching pair is AB and DC:

k = \frac{DC}{AB} = \frac{12}{8} = \frac{3}{2}

ED matches AE, and CE matches BE. So \displaystyle ED = \frac{3}{2} \cdot 6 and \displaystyle CE = \frac{3}{2} \cdot 5, which gives ED = 9 and CE = 7.5.

The similarity statement, not the picture, says which sides go together: ED is the partner of AE because E matches E and D matches A.

Example 8

Heights from shadows

A stick 1.4 m tall stands upright on level ground and casts a shadow 2 m long. At the same moment a tree nearby casts a shadow 9 m long. How tall is the tree?

Figure 6Parallel sun rays meet the ground at the same angle.

Solution

Each object, its shadow and the sun’s ray past its top make a right triangle. Two assumptions turn the scene into geometry: the stick and the tree both stand vertically on level ground, and the sun is so far away that its rays arrive parallel to each other.

Both triangles have a right angle where the object meets the ground. The ground line crosses the two parallel rays, so the angles the rays make with the ground are corresponding angles, and congruent. That is two pairs, so by AA the triangles are similar, with the stick matching the tree and the stick’s shadow matching the tree’s shadow.

k = \frac{9}{2} = 4.5

The tree’s height h is 4.5 times the stick’s, so h = 4.5 \cdot 1.4 and the tree is 6.3 m tall.

This way of measuring has its own lesson, measuring what you cannot reach.

Example 9

The altitude of a right triangle

In \triangle ABC, \angle ACB = 90^\circ, AC = 12, BC = 16 and AB = 20. D is the point on AB with CD \perp AB. Find AD and CD.

Figure 7The altitude from the right angle splits the triangle into two smaller triangles.

Solution

CD cuts the big triangle into two smaller ones. Compare \triangle ADC with \triangle ACB. The angle at A is in both. \angle ADC = 90^\circ because CD \perp AB, and \angle ACB = 90^\circ is given. Two pairs, so by AA

\triangle ADC \sim \triangle ACB

with A matching A, D matching C and C matching B. The right angle of the small triangle, at D, goes with the right angle of the big one, at C.

The sides follow from the order of the letters. AD matches AC, DC matches CB, and AC matches AB, so

\frac{AD}{AC} = \frac{DC}{CB} = \frac{AC}{AB} = \frac{12}{20} = \frac{3}{5}

Therefore \displaystyle AD = \frac{3}{5} \cdot 12 and \displaystyle CD = \frac{3}{5} \cdot 16, which gives AD = 7.2 and CD = 9.6.

Check with Pythagoras in \triangle ADC: 7.2^2 + 9.6^2 = 51.84 + 92.16 = 144 = 12^2.

The same argument with the angle at B shows \triangle CDB \sim \triangle ACB as well, so all three triangles in the figure are similar. That fact is the start of geometric means in a right triangle and of a proof of the Pythagorean theorem.

Example 10

Every right triangle with a 30^\circ angle

A right triangle has a 30^\circ angle and a hypotenuse of 13. How long is the side opposite the 30^\circ angle?

Solution

Start with one triangle where the answer can be seen. Take an equilateral triangle with sides 2; each of its angles is 60^\circ. Drop the segment from its top corner perpendicular to the base. The two halves are right triangles with the same hypotenuse 2 and a shared leg, so they are congruent by HL. That means the segment meets the base at its midpoint and splits the top angle into two 30^\circ angles. Each half is a right triangle with angles 30^\circ, 60^\circ and 90^\circ, hypotenuse 2, and side opposite the 30^\circ angle equal to half the base, which is 1.

The given triangle also has a right angle and a 30^\circ angle. By AA it is similar to the half triangle, with hypotenuse matching hypotenuse:

k = \frac{13}{2}

The side opposite 30^\circ matches the side of length 1, so it is \displaystyle \frac{13}{2} \cdot 1, which is 6.5.

Nothing here depended on 13. In every right triangle with a 30^\circ angle, the side opposite that angle is half the hypotenuse. AA makes the ratio of two sides of a right triangle depend only on its angle, never on its size, which is exactly what lets the sine of an angle be a single number.

Boundaries

Triangles only. AA works because two angles of a triangle and the side between them fix the whole triangle. Polygons with more sides are not so stiff. A square and a long rectangle both have four right angles, so every angle matches, yet they are not the same shape: for a quadrilateral the side ratios have to be checked as well.

Figure 8Every angle matches, and the shapes still differ.

Shape, never size. AA says nothing about how big the triangles are, so on its own it can never show two triangles congruent: a small and a large equilateral triangle have all three angles congruent. Congruence needs one pair of matching sides as well, which is ASA or AAS (why AAA and SSA fail).

The flat plane. The proof uses two facts of flat geometry: the angles of a triangle add to 180^\circ, and a triangle can be blown up by a dilation without changing its angles. Both rest on the parallel postulate. On the surface of a sphere neither holds. There a triangle’s angles add to more than 180^\circ, and two triangles with the same three angles are already the same size, so a larger copy with the same angles does not exist.

Measuring problems. When AA is applied to a real scene, the conclusion is only as good as the assumptions that produced the angle pairs: an upright post, level ground, parallel rays. A post that leans makes a different triangle.

Where this goes next

AA is the test reached for most often, because angles are the easiest evidence to find. Spotting the two pairs in a crowded figure is the subject of proving two triangles similar, and writing the argument out in full is proofs about similar figures. Once two triangles are known to be similar, proportions in similar triangles sets up and solves the equations between their sides, as Examples 3, 5 and 6 did.

Example 6 is the first step of the triangle proportionality theorem, which says a line parallel to one side splits the other two in the same ratio. Example 9 continues into geometric means in a right triangle. Example 10 is where trigonometry begins: because every right triangle with a given acute angle has the same shape, the ratio of its sides belongs to the angle alone, and that ratio is the sine. When the evidence is lengths instead of angles, the tests are SSS and SAS similarity, and the scale factor AA produces tells how perimeters and areas compare in scale factor and perimeter, area and volume.

Exercises

Exercises 1 to 9 go with the first part of the lesson (Examples 1 to 5), Exercises 10 to 20 with the second (Examples 6 to 10), and the reasoning group draws on both.

Part 1: similar or not

For each pair, find the missing angles and decide whether the triangles are similar. If they are, write a similarity statement with the vertices in matching order.

\triangle ABC with \angle A = 35^\circ and \angle B = 80^\circ; \triangle DEF with \angle D = 80^\circ and \angle E = 65^\circ.
\triangle GHI with \angle G = 42^\circ and \angle H = 73^\circ; \triangle JKL with \angle J = 42^\circ and \angle K = 75^\circ.
\triangle PQR with \angle Q = 90^\circ and \angle P = 28^\circ; \triangle STU with \angle T = 90^\circ and \angle U = 62^\circ.
\triangle ABC with AB = AC and \angle A = 50^\circ; \triangle DEF with DE = DF and \angle E = 65^\circ.
In \triangle ABC, \angle A = (x + 10)^\circ, \angle B = (2x)^\circ and \angle C = (3x - 10)^\circ. In \triangle DEF, \angle D = 80^\circ and \angle E = 40^\circ. Find x, then decide.

Part 1: finding sides

In \triangle ABC and \triangle DEF, \angle A \cong \angle D and \angle B \cong \angle E. AB = 6, BC = 9, CA = 12 and DE = 8. Find EF and FD.
In \triangle PQR and \triangle XYZ, \angle P \cong \angle X and \angle Q \cong \angle Y. PQ = 15, QR = 12 and XY = 10. Find YZ.
In \triangle JKL and \triangle MNO, \angle J \cong \angle M and \angle K \cong \angle N. JK = 4, KL = x + 3, MN = 8 and NO = 3x - 1. Find x, KL and NO.
In \triangle ABC and \triangle DEF, \angle A \cong \angle D and \angle C \cong \angle F. AB = 5 and DE = 7.5, and the perimeter of \triangle ABC is 18. Find the perimeter of \triangle DEF.

Part 2: parallel lines and a shared corner

In Exercises 10 to 13, D lies on side AB of \triangle ABC and E lies on side AC.

DE \parallel BC, AD = 3, AB = 12 and DE = 5. Find BC.
DE \parallel BC, AE = 4, EC = 6 and BC = 15. Find DE.
DE \parallel BC, AD = x, DB = 5, DE = 6 and BC = 9. Find x.
\angle AED \cong \angle ABC, AB = 12, AC = 9 and AD = 6. Find AE.

In Exercises 14 and 15, AB \parallel CD and segments AD and BC cross at E, as in Example 7.

AE = 5, ED = 8, AB = 10 and BE = 6. Find CD and CE.
AE = x, ED = x + 6, BE = 4 and CE = 7. Find x and ED.

Part 2: right triangles, shadows and mirrors

At the same moment, a 2.5 m post casts a 4 m shadow and a building casts a 36 m shadow on the same level ground. How tall is the building?
Kofi is 1.5 m tall and his shadow is 1.2 m long. At the same moment the shadow of a flagpole near him is 8 m long. How tall is the flagpole?
Arjun lays a small mirror flat on level ground 20 m from the foot of a tree. He walks backwards from the mirror, in line with the tree, until he sees the top of the tree in it. He is then 2.5 m from the mirror, and his eyes are 1.7 m above the ground. Light leaves a flat mirror at the same angle to it as it arrives. How tall is the tree?
In \triangle ABC, \angle ACB = 90^\circ, AC = 9 and AB = 15. D is the point on AB with CD \perp AB. Find BC, AD and CD.
A right triangle has a 60^\circ angle and a hypotenuse of 14. Find both legs.

Reasoning

Explain why any two equilateral triangles are similar.
Explain why two right triangles are similar as soon as an acute angle of one is congruent to an acute angle of the other.
A square with sides 3 and a rectangle 5 long and 2 wide each have four right angles. Show that they are not similar, and explain why the reasoning behind AA does not carry over to quadrilaterals.
In trapezoid ABCD, AB \parallel CD, and the diagonals AC and BD cross at E. Prove that \dfrac{AE}{CE} = \dfrac{AB}{CD}.
Two triangles have two pairs of congruent angles, and one side of the first is congruent to the matching side of the second. Use AA to explain why the triangles are congruent.

Answers

Similar: \triangle ABC \sim \triangle FDE. The third angles are \angle C = 180^\circ - 35^\circ - 80^\circ = 65^\circ and \angle F = 180^\circ - 80^\circ - 65^\circ = 35^\circ. So \angle A \cong \angle F, \angle B \cong \angle D and \angle C \cong \angle E.
Not similar. \angle I = 180^\circ - 42^\circ - 73^\circ = 65^\circ and \angle L = 180^\circ - 42^\circ - 75^\circ = 63^\circ. The angles are 42^\circ, 73^\circ, 65^\circ against 42^\circ, 75^\circ, 63^\circ: only one pair matches.
Similar: \triangle PQR \sim \triangle STU. \angle R = 180^\circ - 90^\circ - 28^\circ = 62^\circ and \angle S = 180^\circ - 90^\circ - 62^\circ = 28^\circ, so \angle P \cong \angle S, \angle Q \cong \angle T and \angle R \cong \angle U.
Similar: \triangle ABC \sim \triangle DEF. In an isosceles triangle the angles opposite the equal sides are equal. So in \triangle ABC, \displaystyle \angle B = \angle C = \frac{180^\circ - 50^\circ}{2} = 65^\circ, and in \triangle DEF, \angle F = \angle E = 65^\circ and \angle D = 180^\circ - 130^\circ = 50^\circ.
x = 30, and \triangle ABC \sim \triangle EFD. The angles add to 6x = 180, so x = 30, giving \angle A = 40^\circ, \angle B = 60^\circ, \angle C = 80^\circ. In \triangle DEF, \angle F = 180^\circ - 80^\circ - 40^\circ = 60^\circ. So A matches E, B matches F and C matches D.
EF = 12 and FD = 16. By AA the triangles are similar with \displaystyle k = \frac{DE}{AB} = \frac{8}{6} = \frac{4}{3}, so \displaystyle EF = \frac{4}{3} \cdot 9 and \displaystyle FD = \frac{4}{3} \cdot 12.
YZ = 8. By AA, \triangle PQR \sim \triangle XYZ with \displaystyle k = \frac{XY}{PQ} = \frac{10}{15} = \frac{2}{3}, and YZ matches QR: \displaystyle \frac{2}{3} \cdot 12 = 8.
x = 7, so KL = 10 and NO = 20. By AA, \displaystyle k = \frac{MN}{JK} = \frac{8}{4} = 2, so 3x - 1 = 2(x + 3), which gives 3x - 1 = 2x + 6 and x = 7.
27. By AA, \triangle ABC \sim \triangle DEF with \displaystyle k = \frac{7.5}{5} = 1.5. Every side of \triangle DEF is 1.5 times its partner, so the perimeter is 1.5 \cdot 18 = 27.
BC = 20. The shared angle at A and the corresponding angles \angle ADE \cong \angle ABC give \triangle ADE \sim \triangle ABC by AA, with \displaystyle k = \frac{AB}{AD} = \frac{12}{3} = 4. So BC = 4 \cdot 5 = 20.
DE = 6. As in Exercise 10, \triangle ADE \sim \triangle ABC. Here AC = 4 + 6 = 10, so \displaystyle \frac{DE}{BC} = \frac{AE}{AC} = \frac{4}{10} and DE = 0.4 \cdot 15 = 6.
x = 10. \triangle ADE \sim \triangle ABC with AB = x + 5, so \displaystyle \frac{x}{x + 5} = \frac{6}{9} = \frac{2}{3}. Cross-multiplying, 3x = 2x + 10, so x = 10.
AE = 8. The angle at A is shared and \angle AED \cong \angle ABC, so by AA \triangle AED \sim \triangle ABC, with E matching B and D matching C. Then \displaystyle \frac{AE}{AB} = \frac{AD}{AC}, so \displaystyle AE = \frac{12 \cdot 6}{9} = 8.
CD = 16 and CE = 9.6. Alternate interior angles give \triangle ABE \sim \triangle DCE by AA, with \displaystyle k = \frac{ED}{AE} = \frac{8}{5}. So \displaystyle CD = \frac{8}{5} \cdot 10 = 16 and \displaystyle CE = \frac{8}{5} \cdot 6 = 9.6.
x = 8, so ED = 14. \triangle ABE \sim \triangle DCE, so \displaystyle \frac{AE}{DE} = \frac{BE}{CE}: \displaystyle \frac{x}{x + 6} = \frac{4}{7}. Cross-multiplying, 7x = 4x + 24, so x = 8.
The building is 22.5 m tall. Both stand upright on level ground (a pair of right angles) and the sun’s rays are parallel (a pair of corresponding angles), so the triangles are similar by AA with \displaystyle k = \frac{36}{4} = 9. The height is 9 \cdot 2.5 = 22.5.
The flagpole is 10 m tall. By AA as in Exercise 16, \displaystyle k = \frac{8}{1.2} = \frac{20}{3}, and \displaystyle \frac{20}{3} \cdot 1.5 = 10.
The tree is 13.6 m tall. Arjun and the tree both stand upright on level ground, so each makes a right angle with it. At the mirror, the ray from the treetop and the ray to his eyes make equal angles with the ground. By AA his triangle (height 1.7, distance 2.5) is similar to the tree’s (height h, distance 20), with \displaystyle k = \frac{20}{2.5} = 8, so h = 8 \cdot 1.7 = 13.6.
BC = 12, AD = 5.4 and CD = 7.2. By Pythagoras, \displaystyle BC = \sqrt{15^2 - 9^2} = 12. As in Example 9, \triangle ADC \sim \triangle ACB, so \displaystyle AD = \frac{AC}{AB} \cdot AC = \frac{9}{15} \cdot 9 = 5.4 and \displaystyle CD = \frac{AC}{AB} \cdot BC = \frac{9}{15} \cdot 12 = 7.2.
The legs are 7 and \displaystyle 7\sqrt{3}. The triangle’s angles are 90^\circ, 60^\circ and 30^\circ, so by AA it is similar to half of an equilateral triangle with sides 2 (Example 10), whose legs are 1 and \displaystyle \sqrt{2^2 - 1^2} = \sqrt{3}. The scale factor is \displaystyle \frac{14}{2} = 7, so the leg opposite 30^\circ is 7 and the leg opposite 60^\circ is \displaystyle 7\sqrt{3}.
The three sides of an equilateral triangle are equal, so its three angles are equal (the angles opposite equal sides of a triangle are equal). Three equal angles adding to 180^\circ are 60^\circ each. So any two equilateral triangles have two pairs of congruent angles, both 60^\circ, and AA makes them similar.
Each triangle has a right angle, and all right angles are congruent: that is one pair. The congruent acute angles are a second pair. Two pairs of congruent angles, so the triangles are similar by AA. (The remaining acute angles agree as well, each being 90^\circ minus the given one.)
If they were similar, every side of the rectangle would be the same multiple k of a side of the square, which is 3. Then all four sides of the rectangle would equal 3k, but they are 2 and 5; equivalently the ratios \displaystyle \frac{2}{3} and \displaystyle \frac{5}{3} disagree. AA works for triangles because two angles and the side between them fix the whole triangle, so matching angles leave only the size free. A quadrilateral’s angles do not fix it: a rectangle can be stretched in one direction alone, keeping all four right angles while its shape changes.
Diagonal AC crosses the parallel sides AB and CD, so \angle BAE \cong \angle DCE (alternate interior angles). Diagonal BD crosses them too, so \angle ABE \cong \angle CDE. By AA, \triangle ABE \sim \triangle CDE, with A matching C, B matching D and E matching E. Matching sides are in one ratio, so \displaystyle \frac{AE}{CE} = \frac{AB}{CD}.
By AA the triangles are similar, so every pair of matching sides is in the same ratio k. One pair of matching sides is congruent, so that ratio is 1, and every side of one triangle equals its partner in the other. All three pairs of sides are congruent, so the triangles are congruent by SSS. This is the ASA and AAS congruence tests seen through similarity.