Geometry · G.SRT.A.3 · G.SRT.B.5

SSS and SAS similarity

You can measure the three sides of two triangles but none of their angles. Is that enough to tell whether they have the same shape?

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AA similarityMath 8

Where you are 10 in Similarity

Geometry11

SSS and SAS similarity

Triangle proportionality theorem Triangle angle bisector theorem ▾ Show all 10 (4 more)

Why this exists

AA similarity settles the shape of a triangle from its angles: find two pairs of equal angles and the triangles are similar. But angles are often not what you have. A frame cut from timber, a field paced out on foot, a triangle drawn on a coordinate grid: each of these hands you lengths. To use AA you would first have to work the angles out, which is a long way round.

Lengths can do the job on their own, because a triangle is rigid. Pin three sticks together at their ends and the shape cannot flex, while four sticks joined in a loop can be pushed out of square. Three lengths fix a triangle completely, which is why SSS congruence works. This lesson is the version of that fact for shape rather than size: if three lengths fix a triangle, then three lengths in a fixed ratio fix its shape.

There is a second test with lengths in it. Two sticks hinged at a fixed angle also leave a triangle no freedom, since the third side only has to close the gap. That gives SAS similarity, the other half of this lesson.

Definition

Two triangles are similar when one is a scaled copy of the other: matching angles equal, and matching sides all in one ratio. That ratio is the scale factor k. This lesson gives two tests that prove triangles similar using their sides.

SSS similarity (side, side, side). If each side of one triangle is the same multiple k of the matching side of another,

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

then the triangles are similar: \triangle ABC \sim \triangle DEF. In Figure 1 each side of \triangle STU is \displaystyle \frac{3}{2} of the matching side of \triangle PQR, and Example 1 checks it.

Figure 1Each side of \triangle STU is \tfrac{3}{2} of the matching side of \triangle PQR.

SAS similarity (side, angle, side). If two sides of one triangle are the same multiple k of two sides of another, and the angles between those sides are congruent,

\frac{DE}{AB} = \frac{DF}{AC} = k \qquad\text{and}\qquad \angle A \cong \angle D

then \triangle ABC \sim \triangle DEF. The angle must be the one the two sides make with each other, the included angle. In Figure 2 it is the angle at A in one triangle and the angle at D in the other.

Figure 2Two pairs of sides in the ratio \tfrac{3}{2}, and the angles between them marked equal.

The order of the letters in \triangle ABC \sim \triangle DEF is the matching. A goes with D, B with E and C with F, so side AB goes with DE, BC with EF and CA with FD.

Intuition

Picture \triangle PQR in Figure 1 built from three rods of length 5, 8 and 10, pinned at the corners. Now cut three new rods, each one and a half times as long: 7.5, 12 and 15, and pin them together. The new triangle is bigger, but each rod is the same fraction of its neighbors as before, so the rods meet at the same angles. Multiplying every length by one number scales the picture, and a scaled picture keeps its shape.

Lengthen only two of the rods and the corners must change to close the triangle up. That is why all three ratios have to agree. One ratio out of line, however slightly, and the shape is different.

For SAS, think of two rods hinged at a corner and locked at a fixed angle. The gap between their far ends is then fixed too, so the third side has no choice. Lengthen both rods by the factor k and keep the hinge where it was, and the gap grows by the factor k as well. The whole triangle has been scaled.

Results

SSS similarity theorem. If the three sides of one triangle are proportional to the three sides of another, the triangles are similar.

SAS similarity theorem. If two sides of one triangle are proportional to two sides of another and the included angles are congruent, the triangles are similar.

Once either test succeeds, everything similarity promises comes with it: all three pairs of matching angles are congruent, and every pair of matching lengths is in the ratio k.

Two consequences get used all the time.

Matching by size. In similar triangles the shortest side of one matches the shortest side of the other, the middle with the middle and the longest with the longest, because multiplying by a positive k keeps lengths in the same order. So to test SSS from bare numbers, put each list in order and divide in order. If those three ratios disagree, no other matching can work.

Congruence is the case k = 1. With k = 1 the SSS and SAS similarity tests become the SSS and SAS congruence tests. Together with AA, the three similarity tests mirror the congruence tests:

Similarity test	What it needs	Matching congruence test
AA	two pairs of congruent angles	ASA and AAS
SSS	three pairs of sides in one ratio	SSS
SAS	two pairs of sides in one ratio, included angles congruent	SAS

Why it is true

Both proofs use one idea: blow the first triangle up to the right size, then recognize the result as a congruent copy of the second.

A dilation with scale factor k multiplies every length by k and keeps every angle (the reason is in what a dilation does to a line), and two figures are similar exactly when a dilation followed by rigid motions carries one onto the other (that is what a similarity transformation is). So it is enough to find a dilation that turns \triangle ABC into a triangle congruent to \triangle DEF.

Figure 3Dilate \triangle ABC from A until it is the size of \triangle DEF. The dashed image is congruent to \triangle DEF.

SSS. Suppose \displaystyle \frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA} = k. Dilate \triangle ABC from the point A with scale factor k. The point A stays where it is, B moves along ray AB to B', and C moves along ray AC to C'. Every length is multiplied by k:

\begin{aligned} AB' &= k \cdot AB = DE \\ B'C' &= k \cdot BC = EF \\ C'A &= k \cdot CA = FD \end{aligned}

So \triangle AB'C' and \triangle DEF have three pairs of equal sides, and by SSS congruence they are congruent: some sequence of rigid motions carries \triangle AB'C' onto \triangle DEF (in Figure 3, a half turn). The dilation followed by those rigid motions carries \triangle ABC onto \triangle DEF, so \triangle ABC \sim \triangle DEF.

SAS. Suppose \displaystyle \frac{DE}{AB} = \frac{DF}{AC} = k and \angle A \cong \angle D. Dilate from A by k as before, so that AB' = k \cdot AB = DE and AC' = k \cdot AC = DF. The points B' and C' lie on the same rays from A as B and C, so \angle B'AC' is the very same angle as \angle BAC, and that is congruent to \angle D. Two sides and the included angle match, so \triangle AB'C' \cong \triangle DEF by SAS congruence, and as before \triangle ABC \sim \triangle DEF.

Nothing here needs k to be more than 1: when k < 1 the dilation shrinks the triangle instead. Notice where each proof spends its hypothesis. SSS needs all three ratios because SSS congruence needs all three sides, and SAS needs the included angle because SAS congruence does. The similarity tests are the congruence tests, applied after a dilation.

Worked examples

Example 1

Checking three ratios

Are the triangles in Figure 1 similar? If they are, give the scale factor from \triangle PQR to \triangle STU and write the similarity statement with the vertices in matching order.

Solution

The triangles face different ways, so do not match sides by where they sit on the page. Match them by size, shortest with shortest:

	shortest	middle	longest
\triangle PQR	PQ = 5	QR = 8	RP = 10
\triangle STU	ST = 7.5	TU = 12	US = 15

Divide each side of \triangle STU by its partner:

\frac{7.5}{5} = \frac{3}{2}, \qquad \frac{12}{8} = \frac{3}{2}, \qquad \frac{15}{10} = \frac{3}{2}

All three ratios are \displaystyle \frac{3}{2}, so by SSS similarity the triangles are similar, with scale factor \displaystyle \frac{3}{2}.

Now read off the vertex matching. A vertex is where two sides meet. Q is where PQ and QR meet; their partners ST and TU meet at T, so Q goes with T. P is where PQ and RP meet; their partners ST and US meet at S, so P goes with S. That leaves R with U:

\triangle PQR \sim \triangle STU

So \angle P \cong \angle S, \angle Q \cong \angle T and \angle R \cong \angle U, and not one angle was measured.

Example 2

When one ratio is out of line

One triangle has sides 6, 9 and 12. Another has sides 8, 12 and 15. Are they similar?

Solution

Both lists are already in order, so divide in order:

\frac{8}{6} = \frac{4}{3}, \qquad \frac{12}{9} = \frac{4}{3}, \qquad \frac{15}{12} = \frac{5}{4}

The first two ratios agree and the third does not, so the triangles are not similar. The matching in order is the only one that could have worked (see Results), so there is nothing else to try. For a scale factor of \displaystyle \frac{4}{3} the longest side would have to be \displaystyle \frac{4}{3} \cdot 12 = 16, not 15. Two ratios out of three are not enough: the shape depends on all three sides.

Example 3

Two sides and the angle between them

In Figure 2, \angle A \cong \angle D. Show that \triangle ABC \sim \triangle DEF, and find x = EF.

Solution

The marked angle at A sits between sides AB and AC, and the marked angle at D sits between DE and DF, so these are included angles and SAS is the test to try. Compare the sides that form them:

\frac{DE}{AB} = \frac{6}{4} = \frac{3}{2}, \qquad \frac{DF}{AC} = \frac{9}{6} = \frac{3}{2}

The ratios are equal and the included angles are congruent, so by SAS similarity \triangle ABC \sim \triangle DEF with scale factor \displaystyle \frac{3}{2}.

EF is the side opposite the marked angle in \triangle DEF, so its partner is BC, the side opposite the marked angle in \triangle ABC:

x = EF = \tfrac{3}{2} \cdot BC = \tfrac{3}{2} \cdot 5 = 7.5

So EF = 7.5. The test itself never used BC; it was needed only to find EF afterwards.

Example 4

Solving for a variable

\triangle ABC has AB = 5, BC = 7 and CA = 9. \triangle DEF has DE = 10, EF = 2x and FD = x + 11. Find x so that \triangle ABC \sim \triangle DEF.

Solution

The statement \triangle ABC \sim \triangle DEF fixes the matching: AB with DE, BC with EF, CA with FD. The one pair with no unknown gives the scale factor:

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

Each other side of \triangle DEF must be twice its partner. From EF = 2 \cdot BC:

2x = 14 \qquad\Longrightarrow\qquad x = 7

From FD = 2 \cdot CA:

x + 11 = 18 \qquad\Longrightarrow\qquad x = 7

Both conditions give x = 7, so it works: the sides of \triangle DEF are 10, 14 and 18, exactly twice 5, 7 and 9. The second equation is a real check, not a formality. Two equations in one unknown need not agree, and if they had given different values, no x would make the triangles similar.

Example 5

A triangle inside a triangle

In Figure 4, D lies on \overline{AB} with AD = 5 and DB = 7, and E lies on \overline{AC} with AE = 6 and EC = 4. Also BC = 11. Find DE.

Figure 4\triangle ADE sits inside \triangle ABC and shares the angle at A.

Solution

\triangle ADE and \triangle ABC share the angle at A, and in each triangle it is the included angle: between AD and AE in one, between AB and AC in the other. So SAS is the test to try. The whole sides are AB = 5 + 7 = 12 and AC = 6 + 4 = 10.

The first matching to try puts AD with AB and AE with AC:

\frac{AD}{AB} = \frac{5}{12}, \qquad \frac{AE}{AC} = \frac{6}{10} = \frac{3}{5}

These are not equal, so that matching fails. The angle at A is shared either way, so swap the partners, AD with AC and AE with AB:

\frac{AD}{AC} = \frac{5}{10} = \frac{1}{2}, \qquad \frac{AE}{AB} = \frac{6}{12} = \frac{1}{2}

Equal, with the included angle shared, so by SAS similarity \triangle ADE \sim \triangle ACB with scale factor \displaystyle \frac{1}{2}. In that statement D sits under C and E under B, so DE matches CB:

DE = \tfrac{1}{2} \cdot CB = \tfrac{1}{2} \cdot 11 = 5.5

The matching also says \angle ADE \cong \angle ACB: the angle at D equals the angle at C, not the one at B. The small triangle is a flipped copy of the large one. Had the first matching worked instead, \overline{DE} would have been parallel to \overline{BC}, which is the setting of the triangle proportionality theorem.

Example 6

Two triangles meeting at a point

In Figure 5, \overline{AC} and \overline{BD} cross at E, with AE = 4, EC = 10, BE = 6, ED = 15 and AB = 5. Show that \triangle AEB \sim \triangle CED, find CD, and show that \overline{AB} \parallel \overline{CD}.

Figure 5Two triangles meeting at a point, with a pair of vertical angles between them.

Solution

\angle AEB and \angle CED are vertical angles, so they are congruent. Each is the included angle between the two sides that meet at E: AE and BE in one triangle, CE and DE in the other. Match A with C (they lie on one line through E) and B with D:

\frac{CE}{AE} = \frac{10}{4} = \frac{5}{2}, \qquad \frac{DE}{BE} = \frac{15}{6} = \frac{5}{2}

The ratios are equal, so by SAS similarity \triangle AEB \sim \triangle CED with scale factor \displaystyle \frac{5}{2}. CD matches AB:

CD = \tfrac{5}{2} \cdot AB = \tfrac{5}{2} \cdot 5 = 12.5

Similarity also gives matching angles: \angle EAB \cong \angle ECD. These two angles sit at A and C, on opposite sides of the line AC and between the lines AB and CD, so they are alternate interior angles. Equal alternate interior angles force the lines to be parallel, so \overline{AB} \parallel \overline{CD}.

Matching the other way, A with D, would compare \displaystyle \frac{DE}{AE} = \frac{15}{4} with \displaystyle \frac{CE}{BE} = \frac{10}{6}, which are not equal. The vertical angle pair decides which sides are partners.

Example 7

Similar triangles on a grid

\triangle ABC has vertices A(0, 0), B(3, 0) and C(1, 2), and \triangle PQR has vertices P(2, 1), Q(2, 7) and R(-2, 3). Show that the triangles are similar, find the scale factor, and name the angle of \triangle PQR that is congruent to \angle ACB.

Solution

Find every side with the distance formula:

\begin{aligned} AB &= 3 \\ BC &= \sqrt{(1-3)^2 + (2-0)^2} = \sqrt{8} = 2\sqrt{2} \\ CA &= \sqrt{1^2 + 2^2} = \sqrt{5} \end{aligned}

and

\begin{aligned} PQ &= 6 \\ QR &= \sqrt{(-2-2)^2 + (3-7)^2} = \sqrt{32} = 4\sqrt{2} \\ RP &= \sqrt{(2+2)^2 + (1-3)^2} = \sqrt{20} = 2\sqrt{5} \end{aligned}

In order of size, \displaystyle \sqrt{5} < 2\sqrt{2} < 3 (about 2.24, 2.83 and 3) and \displaystyle 2\sqrt{5} < 4\sqrt{2} < 6. Divide in order:

\frac{2\sqrt{5}}{\sqrt{5}} = 2, \qquad \frac{4\sqrt{2}}{2\sqrt{2}} = 2, \qquad \frac{6}{3} = 2

By SSS similarity the triangles are similar with scale factor 2. The partners are CA with RP, BC with QR and AB with PQ. C is where BC and CA meet, and their partners QR and RP meet at R, so C goes with R; in the same way A goes with P and B with Q. So \triangle ABC \sim \triangle PQR, and \angle ACB \cong \angle PRQ.

\triangle PQR is not just a slid copy: \overline{AB} is horizontal and its partner \overline{PQ} is vertical, so the copy has been turned as well as enlarged. SSS does not care.

Boundaries

The angle in SAS must be the included one. In Figure 6, both triangles ABC_1 and ABC_2 have AB = 10, BC = 6 and \angle A = 30^\circ, yet they are plainly different shapes. Two sides and an angle that is not between them (SSA) do not fix a triangle, so there is no SSA test for similarity, just as there is none for congruence. Exercise 24 proves these two are not similar.

Figure 6Two different triangles with AB = 10, BC = 6 and \angle A = 30^\circ.

All three ratios, not two. Two pairs of sides in proportion, with nothing said about the angle between them, decide nothing (Example 2).

A ratio, not a difference. The triangles with sides 3, 4, 5 and 5, 6, 7 differ by 2 on every side, but \displaystyle \frac{5}{3} \ne \frac{6}{4}, so they are not similar. Adding the same amount stretches a short side by a bigger fraction than a long one.

Triangles only. A square with side 1 and a rhombus with side 2 and a 60^\circ angle have all four sides in the ratio 2, but they are not similar. A quadrilateral is not rigid, so its sides do not fix its angles, and no SSSS test exists.

The lengths must make a triangle. The numbers 2, 3 and 6 are not the sides of any triangle, because 2 + 3 < 6 breaks the triangle inequality. The tests compare triangles that exist.

Equal ratios need the right partners. A correct ratio with the wrong partners proves nothing. Examples 5 and 6 each had one matching that worked and one that did not, and only the vertex matching from the working one may be used afterwards.

Where this goes next

The tests in this lesson end with a similarity statement, and the statement is where the work usually starts. Once two triangles are known to be similar, any missing length follows from one ratio, which is the subject of proportions in similar triangles. Choosing between AA, SSS and SAS from what a figure gives you, and writing the argument out, is proving two triangles similar and, more generally, proofs about similar figures.

SAS similarity is also the engine behind several named results. A line across two sides of a triangle cuts off a small triangle matching the whole the straight way, with its two sides along the big ones in the same ratio, exactly when the line is parallel to the third side (Example 5 shows a crossed matching that is similar without being parallel), which is the triangle proportionality theorem and its converse, and the segment joining two midpoints is the special case \displaystyle k = \frac{1}{2}, the triangle midsegment theorem (Exercise 23).

For right triangles, the right angle is always included between the two legs, so two right triangles with legs in the same ratio are similar. The ratio of the legs alone fixes the shape, and naming such ratios is where right triangle trigonometry begins. And since every length scales by k, perimeters scale by k too, while areas scale by k^2, which is scale factor and area.

Exercises

Three pairs of sides

The side lengths of two triangles are given. Decide whether the triangles are similar, and if they are, give the scale factor from the first to the second.

3, 5, 6 and 9, 15, 18
4, 6, 9 and 6, 9, 13
8, 10, 14 and 12, 15, 21
2.5, 4, 5 and 10, 16, 20
\triangle ABC has AB = 6, BC = 10 and CA = 12. \triangle XYZ has XY = 15, YZ = 9 and ZX = 18. Are they similar? If so, write the similarity statement with the vertices in matching order.

Two sides and the angle between them

\triangle PQR has PQ = 8, PR = 10 and m\angle P = 40^\circ. \triangle STU has ST = 12, SU = 15 and m\angle S = 40^\circ. Are the triangles similar? Give the scale factor and the similarity statement.
Exercise 6 again, but with m\angle S = 50^\circ. Can you conclude by SAS that the triangles are similar? Explain.
One right triangle has legs 6 and 8, and another has legs 9 and 12. Explain why they are similar, and use the scale factor to find the hypotenuse of the second from the hypotenuse of the first.
In \triangle ABC and \triangle DEF, \angle A \cong \angle D, AB = 5, AC = 7, BC = 6, DE = 7 and DF = 9.8. Show the triangles are similar and find EF.

Finding missing lengths

\triangle ABC \sim \triangle DEF with AB = 4, BC = 6, CA = 7 and DE = 10. Find EF and FD.
One triangle has sides 6, 8 and 12, and another has sides 9, x + 5 and 18, matched in that order. Find x so that the triangles are similar.
\triangle ABC \sim \triangle DEF with AB = 6, BC = x + 1, CA = 9, DE = 4, EF = 6 and FD = y. Find x and y.
\triangle ABC \sim \triangle DEF with AB = x, BC = 9, DE = 4 and EF = x. Find x.

Triangles inside and across triangles

D lies on \overline{AB} and E on \overline{AC}, with AD = 3, DB = 6, AE = 4, EC = 8 and DE = 6. Show that \triangle ADE \sim \triangle ABC and find BC.
In \triangle ABC, AB = 10, AC = 8 and BC = 12. D lies on \overline{AB} with AD = 4, and E lies on \overline{AC} with AE = 5. Show that \triangle ADE \sim \triangle ACB and find DE.
\overline{AC} and \overline{BD} cross at E, with AE = 6, EC = 9, BE = 8, ED = 12 and AB = 7. Find CD, and say what the similarity tells you about \overline{AB} and \overline{CD}.
\overline{AC} and \overline{BD} cross at E, with AE = x, EC = 12, BE = 4 and ED = 6. Find the value of x that makes \triangle AEB \sim \triangle CED by SAS similarity.

Triangles on a grid

A(0, 0), B(4, 0), C(0, 3) and P(1, 1), Q(9, 1), R(1, 7). Show that \triangle ABC \sim \triangle PQR and give the scale factor.
A(0, 0), B(2, 1), C(1, 3) and P(0, 0), Q(-2, 4), R(-6, 2). Show that \triangle ABC \sim \triangle PQR and give the scale factor.

Reasoning

Explain why, when testing SSS from two lists of lengths, matching shortest with shortest, middle with middle and longest with longest is the only matching that can work.
Show that any two equilateral triangles are similar.
Show that two isosceles triangles with equal angles between their equal sides are similar.
D and E are the midpoints of sides \overline{AB} and \overline{AC} of \triangle ABC. Show that \triangle ADE \sim \triangle ABC, that \displaystyle DE = \frac{1}{2} BC, and that \overline{DE} \parallel \overline{BC}.
In Figure 6, \triangle ABC_1 and \triangle ABC_2 have AB = 10, BC = 6 and \angle A = 30^\circ in common. Find AC_1 and AC_2, show that the two triangles are not similar, and say what this means for a “side, side, angle” test.
Give an example of two quadrilaterals whose four sides are all in the same ratio but which are not similar, and explain why the SSS idea does not carry over.

Answers

Similar, scale factor 3. \displaystyle \frac{9}{3} = \frac{15}{5} = \frac{18}{6} = 3.
Not similar. \displaystyle \frac{6}{4} = \frac{3}{2} and \displaystyle \frac{9}{6} = \frac{3}{2}, but \displaystyle \frac{13}{9} \ne \frac{3}{2}. For a scale factor of \displaystyle \frac{3}{2} the longest side would have to be 13.5.
Similar, scale factor \displaystyle \frac{3}{2}. \displaystyle \frac{12}{8} = \frac{15}{10} = \frac{21}{14} = \frac{3}{2}.
Similar, scale factor 4. \displaystyle \frac{10}{2.5} = \frac{16}{4} = \frac{20}{5} = 4.
Similar, scale factor \displaystyle \frac{3}{2}, and \triangle ABC \sim \triangle ZYX. In order of size the sides are AB = 6, BC = 10, CA = 12 and YZ = 9, XY = 15, ZX = 18, each ratio \displaystyle \frac{3}{2}. B is where AB and BC meet, and their partners YZ and XY meet at Y, so B goes with Y; likewise A goes with Z and C with X.
Similar, scale factor \displaystyle \frac{3}{2}, \triangle PQR \sim \triangle STU. \displaystyle \frac{ST}{PQ} = \frac{12}{8} = \frac{3}{2} and \displaystyle \frac{SU}{PR} = \frac{15}{10} = \frac{3}{2}, and the included angles \angle P and \angle S are both 40^\circ, so SAS similarity applies.
No. SAS similarity needs the angles between the proportional sides to be congruent, and here they are 40^\circ and 50^\circ. The two equal ratios alone do not fix the shape: opening the hinge to a different angle changes the third side and the other two angles.
The hypotenuse of the second is 15. In each triangle the right angle is the included angle between the legs, and \displaystyle \frac{9}{6} = \frac{12}{8} = \frac{3}{2}, so the triangles are similar by SAS with scale factor \displaystyle \frac{3}{2}. The first hypotenuse is \displaystyle \sqrt{6^2 + 8^2} = 10, so the second is \displaystyle \frac{3}{2} \cdot 10 = 15. Check: \displaystyle \sqrt{9^2 + 12^2} = 15.
EF = 8.4. \displaystyle \frac{DE}{AB} = \frac{7}{5} and \displaystyle \frac{DF}{AC} = \frac{9.8}{7} = \frac{7}{5}, and the congruent angles at A and D are included, so the triangles are similar by SAS with scale factor \displaystyle \frac{7}{5}. EF matches BC, so \displaystyle EF = \frac{7}{5} \cdot 6 = 8.4.
EF = 15 and FD = 17.5. The scale factor is \displaystyle \frac{DE}{AB} = \frac{10}{4} = \frac{5}{2}, so \displaystyle EF = \frac{5}{2} \cdot 6 and \displaystyle FD = \frac{5}{2} \cdot 7.
x = 7. The known pairs give \displaystyle \frac{9}{6} = \frac{18}{12} = \frac{3}{2}, so \displaystyle x + 5 = \frac{3}{2} \cdot 8 = 12.
x = 8 and y = 6. The scale factor from \triangle DEF to \triangle ABC is \displaystyle \frac{AB}{DE} = \frac{6}{4} = \frac{3}{2}. So \displaystyle x + 1 = \frac{3}{2} \cdot 6 = 9, and \displaystyle 9 = \frac{3}{2} y.
x = 6. Matching ratios are equal, so \displaystyle \frac{AB}{DE} = \frac{BC}{EF}, that is \displaystyle \frac{x}{4} = \frac{9}{x}. Then x^2 = 36, so x = 6 or x = -6, and a length cannot be negative. Check: \displaystyle \frac{6}{4} = \frac{9}{6} = \frac{3}{2}.
BC = 18. AB = 9 and AC = 12, so \displaystyle \frac{AD}{AB} = \frac{3}{9} = \frac{1}{3} and \displaystyle \frac{AE}{AC} = \frac{4}{12} = \frac{1}{3}. The angle at A is shared and included in both, so \triangle ADE \sim \triangle ABC by SAS with scale factor \displaystyle \frac{1}{3}, and BC = 3 \cdot DE = 18.
DE = 6. \displaystyle \frac{AD}{AC} = \frac{4}{8} = \frac{1}{2} and \displaystyle \frac{AE}{AB} = \frac{5}{10} = \frac{1}{2}, with the angle at A shared and included, so \triangle ADE \sim \triangle ACB by SAS. DE matches CB, so \displaystyle DE = \frac{1}{2} \cdot 12 = 6. (Matching AD with AB fails: \displaystyle \frac{4}{10} \ne \frac{5}{8}.)
CD = 10.5, and \overline{AB} \parallel \overline{CD}. \displaystyle \frac{CE}{AE} = \frac{9}{6} = \frac{3}{2} and \displaystyle \frac{DE}{BE} = \frac{12}{8} = \frac{3}{2}, and \angle AEB \cong \angle CED as vertical angles, so \triangle AEB \sim \triangle CED by SAS with scale factor \displaystyle \frac{3}{2}, giving \displaystyle CD = \frac{3}{2} \cdot 7. Then \angle EAB \cong \angle ECD, a pair of alternate interior angles, so the lines are parallel.
x = 8. With A matched to C and B to D, SAS needs \displaystyle \frac{AE}{CE} = \frac{BE}{DE}, that is \displaystyle \frac{x}{12} = \frac{4}{6}, so x = 8. The vertical angles at E supply the included angles.
Similar, scale factor 2. AB = 4, AC = 3, \displaystyle BC = \sqrt{4^2 + 3^2} = 5 and PQ = 8, PR = 6, \displaystyle QR = \sqrt{8^2 + 6^2} = 10. Each side of \triangle PQR is twice its partner, with A, B, C matching P, Q, R.
Similar, scale factor 2. \displaystyle AB = \sqrt{5}, \displaystyle BC = \sqrt{5}, \displaystyle CA = \sqrt{10} and \displaystyle PQ = \sqrt{20} = 2\sqrt{5}, \displaystyle QR = \sqrt{20} = 2\sqrt{5}, \displaystyle RP = \sqrt{40} = 2\sqrt{10}. Each side of \triangle PQR is twice its partner, so \triangle ABC \sim \triangle PQR by SSS.
If the triangles are similar with scale factor k > 0, each side of the second is k times its partner. When a < b, also ka < kb, so the partner of a shorter side is shorter. The order of the sides is kept, so the shortest must go with the shortest and so on. Any other matching would pair sides out of order and could not give one common ratio.
Call the sides of one a, a, a and of the other b, b, b. All three ratios are \displaystyle \frac{b}{a}, so the triangles are similar by SSS similarity.
Call the equal sides of one a and a, and of the other b and b, with the angle between them \theta in both. The two ratios are both \displaystyle \frac{b}{a}, and \theta is the included angle in each, so the triangles are similar by SAS similarity.
\displaystyle \frac{AD}{AB} = \frac{1}{2} = \frac{AE}{AC} because D and E are midpoints, and the angle at A is shared and included, so \triangle ADE \sim \triangle ABC by SAS with scale factor \displaystyle \frac{1}{2}. DE matches BC, so \displaystyle DE = \frac{1}{2} BC. Also \angle ADE \cong \angle ABC, and these are corresponding angles along \overline{AB}, so \overline{DE} \parallel \overline{BC}. This is the triangle midsegment theorem.
\displaystyle AC_1 = 5\sqrt{3} - \sqrt{11} (about 5.34) and \displaystyle AC_2 = 5\sqrt{3} + \sqrt{11} (about 11.98). B is 10 \sin 30^\circ = 5 above the ray, at a foot \displaystyle 10 \cos 30^\circ = 5\sqrt{3} along it, and each C is 6 from B, so it lies \displaystyle \sqrt{6^2 - 5^2} = \sqrt{11} to either side of the foot. In order of size the sides are 5.34, 6, 10 and 6, 10, 11.98, and the ratios \displaystyle \frac{6}{5.34} and \displaystyle \frac{10}{6} already differ, so the triangles are not similar. Two sides in proportion and an angle that is not between them do not fix a triangle’s shape, so there is no SSA similarity test.
A square with side 1 and a rhombus with side 2 and a 60^\circ angle. Every side ratio is 2, but the square’s angles are all 90^\circ and the rhombus has a 60^\circ angle, so they are not similar. SSS works for triangles because three lengths fix a triangle’s angles; four lengths do not fix a quadrilateral’s angles, since a four-sided frame can be pushed out of square.