Proving two triangles similar
Two triangles look like the same shape at different sizes. What is the least you have to prove to be sure, and how do you write the proof?
Where you are 4 in Proofs about similar figures
When you reach for it
A congruence proof ends with equal sides. Many of the results that matter in geometry are about sides that are not equal but in proportion: a short post and a tall tree casting shadows, the pieces of two crossing chords, the two legs and the hypotenuse of a right triangle. Those results are proved by showing that two triangles are similar, and then reading a proportion off the similarity.
So reach for this whenever the Prove says \triangle ABC \sim \triangle DEF, and also whenever it asks for a ratio of lengths, a product of lengths, or a length you cannot measure directly. In the second case the similarity is a step in the middle of the proof rather than its last line.
You already have the tools for the proof itself from two-column proofs and planning a congruence proof: unpack the Given into pairs of parts, look for the pairs the figure hands you, then cite a criterion. The criteria change, and the pairs of sides now need to be in proportion instead of equal. Everything else carries over.
The procedure
What “similar” asks for. Two triangles are similar when there is a correspondence of their vertices under which every pair of corresponding angles is congruent and every pair of corresponding sides is in the same ratio. Both halves are part of the definition. Writing \triangle ABC \sim \triangle DEF claims the correspondence A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow F, and with it all six of these facts:
Here AB is the length of \overline{AB}. Each side of \triangle DEF is the same number k times its partner in \triangle ABC, and \displaystyle k = \frac{DE}{AB} is the scale factor from \triangle ABC to \triangle DEF. The ratio in the other direction, \displaystyle \frac{AB}{DE}, is the scale factor back again.
This agrees with the definition by motions: two figures are similar when a dilation followed by rigid motions carries one onto the other (similarity transformations). A dilation by k multiplies every length by k and keeps every angle, and rigid motions change neither, so the image has equal angles and sides in the ratio k.
Reading the statement. The order of the letters is the correspondence, exactly as in a congruence statement: first letter with first, second with second, third with third. A side is named by two positions, so it corresponds to the side in the same two positions. For \triangle KLM \sim \triangle TRS:
| Positions | In \triangle KLM | In \triangle TRS |
|---|---|---|
| 1 | \angle K | \angle T |
| 2 | \angle L | \angle R |
| 3 | \angle M | \angle S |
| 1 and 2 | \overline{KL} | \overline{TR} |
| 2 and 3 | \overline{LM} | \overline{RS} |
| 1 and 3 | \overline{KM} | \overline{TS} |
Put the first triangle’s sides on top of every ratio, or on the bottom of every ratio, and never mix the two. A proportion is only true under the correspondence it was read from.
The three criteria. For polygons in general, both halves of the definition have to be checked: every angle of a 2 by 2 square matches every angle of a 2 by 4 rectangle, and the shapes differ. Triangles are special. For them, much less than six facts forces all six, and each of these criteria names a set that is enough.
Prove the facts in a row below, and that criterion lets you conclude \triangle ABC \sim \triangle DEF.
| Criterion | What you prove |
|---|---|
| AA similarity | {\angle A \cong \angle D} and {\angle B \cong \angle E} |
| SAS similarity | \displaystyle {\frac{AB}{DE} = \frac{AC}{DF}} and {\angle A \cong \angle D} |
| SSS similarity | \displaystyle {\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}} |
Each criterion has hypotheses that are easy to pass over, and a proof must meet all of them.
- AA needs two pairs of angles, both matched under the one correspondence you are claiming. Any two of the three pairs will do, because the third pair then follows from the triangle angle sum. It needs no length at all. Full treatment: angle angle similarity.
- SAS needs two ratios of sides that are equal, and the angle must be included: in each triangle it is the angle formed by the two sides in the ratios. The ratios must also pair the sides the correspondence pairs: the two sides at A with the two sides at D, \overline{AB} with \overline{DE} and \overline{AC} with \overline{DF}. Full treatment: side angle side similarity.
- SSS needs all three ratios, and they must be one and the same number. Two equal ratios and a third that differs mean the triangles are not similar under that correspondence. Full treatment: side side side similarity.
What each criterion guarantees is the whole definition: after the line \triangle ABC \sim \triangle DEF, every one of the six facts above may be cited. A proof cites them as “corresponding angles of similar triangles are congruent” and “corresponding sides of similar triangles are proportional”, which is the definition of similar triangles read forwards.
Why a criterion is enough. Each one reduces to a congruence criterion. Dilate \triangle ABC by the factor
to get \triangle A'B'C'. It has the same angles as \triangle ABC, and its sides are k times as long, so A'B' = DE.
- Under AA, \triangle A'B'C' and \triangle DEF have \angle A' \cong \angle D, \overline{A'B'} \cong \overline{DE} and \angle B' \cong \angle E, so they are congruent by ASA.
- Under SAS, A'C' = k \cdot AC = DF as well, and the included angles match, so they are congruent by SAS.
- Under SSS, all three sides of \triangle A'B'C' equal the sides of \triangle DEF, so they are congruent by SSS.
In every case a rigid motion carries \triangle A'B'C' onto \triangle DEF, so the dilation followed by that motion carries \triangle ABC onto \triangle DEF, and the triangles are similar. Some courses take AA as a postulate and prove the other two from it; either way, a proof cites all three the same way.
The pattern: name the correspondence, then a criterion. Here it is on Figure 2.
Given. \overline{AB} \perp \overline{BD}; \overline{ED} \perp \overline{BD}; C lies on \overline{BD}; \angle ACB \cong \angle ECD.
Prove. \triangle ABC \sim \triangle EDC.
Name the correspondence. The Prove fixes it: A \leftrightarrow E, B \leftrightarrow D, C \leftrightarrow C. So the angle pairs worth finding are \angle A with \angle E, \angle ABC with \angle EDC, and \angle ACB with \angle ECD, and the side pairs are \overline{AB} with \overline{ED}, \overline{BC} with \overline{DC}, and \overline{AC} with \overline{EC}. When a problem only asks whether two triangles are similar, you choose the correspondence yourself, from the pairs you can prove, and then write the statement with matched vertices in matched positions.
Find the pairs. Unpack each given. The perpendiculars make right angles at B and D, and any two right angles are congruent. The angles at C are given congruent. Then look for pairs the figure supplies, as in a congruence proof: a shared angle, vertical angles at a crossing, alternate interior angles between given parallels.
Choose the criterion. Two pairs of angles under the named correspondence: AA. No lengths are needed, and none are given.
Write it out.
| Statements | Reasons |
|---|---|
| 1. \overline{AB} \perp \overline{BD} and \overline{ED} \perp \overline{BD} | given |
| 2. \angle ABC and \angle EDC are right angles | definition of perpendicular lines: line 1 |
| 3. \angle ABC \cong \angle EDC | Right Angle Congruence Theorem: line 2 |
| 4. \angle ACB \cong \angle ECD | given |
| 5. \triangle ABC \sim \triangle EDC | AA similarity: lines 3 and 4 |
Line 5 cites two lines, and each is an angle pair in matching positions: \angle ABC and \angle EDC have their vertices in position 2, and \angle ACB and \angle ECD in position 3.
Use it. Most proofs go one line further and read a proportion off the similarity. Sides in positions 1 and 2, then 2 and 3:
| Statements | Reasons |
|---|---|
| 6. \displaystyle {\frac{AB}{ED} = \frac{BC}{DC}} | corresponding sides of similar triangles are proportional: line 5 |
Solving line 6 for ED:
With a mirror lying flat at C, a person whose eye is at A sees the top E of a tree. The model takes the law of reflection as its assumption: light leaves a mirror at the same angle to the mirror as it arrives, which is what makes \angle ACB \cong \angle ECD a given. Three lengths along the ground and up to an eye then give the height of the tree, which nobody climbed.
Worked examples
Naming the correspondence from the angles
In \triangle PQR, m\angle P = 48^\circ and m\angle Q = 77^\circ. In \triangle STU, m\angle T = 48^\circ and m\angle U = 55^\circ. Show that the triangles are similar, write the similarity statement beginning with \triangle PQR, and write the proportion of corresponding sides. Then, given PQ = 10 and TS = 15, find the scale factor from \triangle PQR to its partner and say what it tells you about SU and TU.
SolutionEach triangle is missing one angle, and the triangle angle sum gives it:
Now pair the equal angles: P \leftrightarrow T at 48^\circ, Q \leftrightarrow S at 77^\circ, and R \leftrightarrow U at 55^\circ. The first two pairs are enough for AA, so
The letters are placed by the angles, not by the alphabet. \triangle PQR \sim \triangle STU would claim P \leftrightarrow S, a 48^\circ angle with a 77^\circ one, and is false.
Reading sides by positions 1 and 2, 2 and 3, 1 and 3:
\overline{PQ} and \overline{TS} are partners, so the scale factor from \triangle PQR to \triangle TSU is
Every side of \triangle TSU is 1.5 times its partner:
Parallel sides of a trapezoid
Given. ABCD is a trapezoid with \overline{AB} \parallel \overline{DC}; its diagonals \overline{AC} and \overline{BD} cross at E.
Prove. \triangle ABE \sim \triangle CDE. Then, given AB = 18, DC = 12 and AC = 20, find AE and EC.
Plan. The Prove names A \leftrightarrow C, B \leftrightarrow D, E \leftrightarrow E. The parallel sides supply angles. With \overline{AC} as the transversal, \angle BAE and \angle DCE are alternate interior angles, since B and D lie on opposite sides of \overline{AC}. With \overline{BD} as the transversal, \angle ABE and \angle CDE are too. Their vertices sit in positions 1 and 2 of both triangles, which is what AA needs. (The vertical angles at E would serve as either pair.)
| Statements | Reasons |
|---|---|
| 1. \overline{AB} \parallel \overline{DC} | given |
| 2. \angle BAE \cong \angle DCE | Alternate Interior Angles Theorem: line 1, transversal \overline{AC} |
| 3. \angle ABE \cong \angle CDE | Alternate Interior Angles Theorem: line 1, transversal \overline{BD} |
| 4. \triangle ABE \sim \triangle CDE | AA similarity: lines 2 and 3 |
The lengths. Corresponding sides are proportional, and \overline{AB} and \overline{CD} are partners, so
So AE = 3t and EC = 2t for some length t. E lies between A and C, so by the segment addition postulate AE + EC = AC:
AE = 12 and EC = 8. Check. \displaystyle \frac{12}{8} = \frac{3}{2} and 12 + 8 = 20.
Three sides and nothing else
\triangle KLM has KL = 8, LM = 14 and KM = 10. \triangle XYZ has XY = 15, YZ = 12 and XZ = 21. Are the triangles similar? If so, write the statement and the three pairs of congruent angles. Is \triangle GHI, with GH = 12, HI = 15 and GI = 20, similar to \triangle KLM?
SolutionNo angle is known, so the only criterion that can apply is SSS, and the first job is the correspondence. If the triangles are similar, every side of one is k times its partner for one positive number k, and multiplying by the same positive number keeps lengths in order. So the shortest side must go with the shortest, the middle with the middle, the longest with the longest.
In order, the sides are 8, 10, 14 and 12, 15, 21, and
All three ratios are the one number \displaystyle \frac{3}{2}, so the triangles are similar by SSS.
The side pairs are \overline{KL} with \overline{YZ}, \overline{KM} with \overline{XY}, and \overline{LM} with \overline{XZ}. A vertex is where two sides meet, so partners of vertices come from partners of sides. K is where \overline{KL} and \overline{KM} meet, and their partners \overline{YZ} and \overline{XY} meet at Y, so K \leftrightarrow Y. In the same way L \leftrightarrow Z and M \leftrightarrow X:
with \angle K \cong \angle Y, \angle L \cong \angle Z and \angle M \cong \angle X. Reading the statement back gives \overline{KL} with \overline{YZ}, \overline{LM} with \overline{ZX} and \overline{KM} with \overline{YX}, the pairs we found.
For \triangle GHI the sides in order are 12, 15, 20, and
The third ratio differs. Since any similarity would have to pair the sides in order, and this pairing fails, no correspondence works: \triangle GHI is not similar to \triangle KLM.
A shared angle, with the sides paired across
In Figure 4, AD = 4, DB = 8, AE = 6, EC = 2 and BC = 10. Prove that \triangle ADE \sim \triangle ACB, and find DE.
Plan. Both triangles contain the angle at A, so one pair of angles comes free. No second pair of angles is known, so the criterion has to be SAS, with \angle A as the included angle. That needs the two sides at A in one triangle to be in proportion with the two sides at A in the other.
The whole sides are AB = 4 + 8 = 12 and AC = 6 + 2 = 8. Pairing each short piece with the whole side it lies on fails:
Pairing them across works:
So \overline{AD} goes with \overline{AC} and \overline{AE} with \overline{AB}. That makes D \leftrightarrow C and E \leftrightarrow B, with A \leftrightarrow A, which is the order \triangle ADE \sim \triangle ACB in the Prove.
| Statements | Reasons |
|---|---|
| 1. {AD = 4}, {DB = 8}, {AE = 6}, {EC = 2} | given |
| 2. {AB = 12} and {AC = 8} | segment addition postulate: line 1 |
| 3. \displaystyle {\frac{AD}{AC} = \frac{1}{2}} | arithmetic: lines 1 and 2 |
| 4. \displaystyle {\frac{AE}{AB} = \frac{1}{2}} | arithmetic: lines 1 and 2 |
| 5. \angle DAE \cong \angle CAB | reflexive property of congruence: both are \angle A |
| 6. \triangle ADE \sim \triangle ACB | SAS similarity: lines 3, 4 and 5 |
In each triangle the angle at A lies between the two sides in the ratios, so it is included.
Finding DE. \overline{DE} is in positions 2 and 3, so its partner is \overline{CB}, and the scale factor from \triangle ACB down to \triangle ADE is \displaystyle \frac{1}{2}:
The similarity also gives \angle ADE \cong \angle ACB: the angle at D matches the angle at C, not the angle at B. Since AB \ne AC, the angles at B and C differ, so \overline{DE} is not parallel to \overline{BC}. Similar triangles sharing an angle do not always come from a parallel cut; the correspondence tells you which kind you have.
Two altitudes and a product of lengths
Given. In \triangle ABC, \overline{AD} \perp \overline{CB} with D on \overline{CB}, and \overline{BE} \perp \overline{CA} with E on \overline{CA}.
Prove. \triangle ADC \sim \triangle BEC, and CA \cdot CE = CB \cdot CD. Then, given CD = 4, CB = 10 and CA = 8, find CE.
Plan. The correspondence is A \leftrightarrow B, D \leftrightarrow E, C \leftrightarrow C. The perpendiculars give right angles at D and E, in position 2 of each triangle. The angle at C belongs to both triangles: D lies on \overrightarrow{CB} and E on \overrightarrow{CA}, so \angle ACD and \angle BCE are both \angle ACB. Two pairs of angles: AA.
A product of lengths comes from a proportion by clearing fractions, so the plan continues past the similarity: read off the proportion that contains CA, CE, CB and CD, then multiply.
| Statements | Reasons |
|---|---|
| 1. \overline{AD} \perp \overline{CB} and \overline{BE} \perp \overline{CA} | given |
| 2. \angle ADC and \angle BEC are right angles | definition of perpendicular lines: line 1 |
| 3. \angle ADC \cong \angle BEC | Right Angle Congruence Theorem: line 2 |
| 4. \angle ACD \cong \angle BCE | reflexive property of congruence: both are \angle ACB |
| 5. \triangle ADC \sim \triangle BEC | AA similarity: lines 3 and 4 |
| 6. \displaystyle {\frac{CA}{CB} = \frac{CD}{CE}} | corresponding sides of similar triangles are proportional: line 5 |
| 7. {CA \cdot CE = CB \cdot CD} | multiplication property of equality: line 6 |
Line 6 pairs \overline{AC} (positions 1 and 3) with \overline{BC} (positions 1 and 3), and \overline{DC} (positions 2 and 3) with \overline{EC} (positions 2 and 3). Line 7 multiplies both sides by CB \cdot CE.
The number. 8 \cdot CE = 10 \cdot 4 = 40, so CE = 5. Check. \displaystyle \frac{8}{10} = \frac{4}{5}, as line 6 requires.
Using a similarity to prove lines parallel
Given. \overline{AC} and \overline{BD} cross at E; AE = 6, EC = 9, BE = 4, ED = 6.
Prove. \overline{AB} \parallel \overline{CD}.
Plan, working backwards. Lines are proved parallel by a pair of angles (proving lines parallel). With \overline{AC} as a transversal, the pair would be \angle EAB and \angle ECD, alternate interior angles, since B and D lie on opposite sides of \overline{AC}. Those two angles sit in \triangle AEB and \triangle CED, at the vertices A and C. So the goal becomes \triangle AEB \sim \triangle CED, with A \leftrightarrow C, E \leftrightarrow E, B \leftrightarrow D.
The crossing gives the vertical angles \angle AEB \cong \angle CED, each included between the two sides at E. The sides at E compare as
Equal ratios and the included angle: SAS.
| Statements | Reasons |
|---|---|
| 1. \overline{AC} and \overline{BD} cross at E | given |
| 2. \angle AEB \cong \angle CED | Vertical Angles Theorem: line 1 |
| 3. {AE = 6}, {EC = 9}, {BE = 4}, {ED = 6} | given |
| 4. \displaystyle {\frac{AE}{CE} = \frac{2}{3}} | arithmetic: line 3 |
| 5. \displaystyle {\frac{BE}{DE} = \frac{2}{3}} | arithmetic: line 3 |
| 6. \triangle AEB \sim \triangle CED | SAS similarity: lines 4, 5 and 2 |
| 7. \angle EAB \cong \angle ECD | corresponding angles of similar triangles are congruent: line 6 |
| 8. \overline{AB} \parallel \overline{CD} | converse of the Alternate Interior Angles Theorem: line 7, transversal \overline{AC} |
Here the similarity is neither the Given nor the Prove. It is the bridge between lengths, which the Given talks about, and angles, which parallel lines need.
Boundaries
Similar is not congruent. A similarity gives sides in proportion, not equal sides. Congruent triangles are the similar triangles with scale factor 1, so every congruence criterion is a similarity criterion as well, but not the other way round.
Every pair must have a reason. As in congruence proofs, a figure may be trusted for positions (which lines cross, which angle is shared, which point lies between two others) and never for sizes (what a diagram may assume). Two angles that look equal, or two sides that look to be in the same ratio, are not a pair until a given, a definition or a theorem makes them one.
The SAS angle must be included. Two sides in proportion and an angle that is not between them is the side, side, angle pattern, and for similarity, as for congruence, it does not force the triangles to match (why side side angle fails). Exercise 9 is a case.
SSS needs one ratio, three times. If two ratios agree and the third does not, as for \triangle GHI in Example 3, the triangles are not similar under that correspondence.
The criteria are for triangles only. For four or more sides, equal angles do not force proportional sides (a square and a long rectangle) and proportional sides do not force equal angles (a square and a thin rhombus with the same side). Proving two polygons similar means checking both halves of the definition, or cutting them into triangles.
The correspondence decides the proportion. Two triangles can be similar under one correspondence and not under another, and a proportion read from the wrong one is false even when the triangles are similar. In Example 4, \displaystyle \frac{AD}{AB} = \frac{AE}{AC} is false while \displaystyle \frac{AD}{AC} = \frac{AE}{AB} is true.
Where this goes next
Almost every proportion in geometry is proved by this pattern. A line parallel to one side of a triangle cuts off a smaller triangle that shares an angle with the whole and has corresponding angles at the parallel line, so AA applies, and the proportion it gives is the triangle proportionality theorem. The altitude to the hypotenuse of a right triangle makes two smaller right triangles, each sharing an acute angle with the whole, and the proportions from those two similarities add up to the Pythagorean theorem. Two chords or secants of a circle form triangles whose angles match, and clearing the fractions in the proportion turns it into equal products of lengths, as in Example 5; that is the power of a point.
The same step, run with numbers instead of letters, is solving proportions in similar triangles, and it measures heights from shadows and mirrors. In a right triangle, AA says that one acute angle fixes the shape, so the ratio of any two sides depends on that angle alone, which is why trigonometric ratios depend only on the angle.
Exercises
Reading a similarity statement
- \triangle RST \sim \triangle XYZ. List the three pairs of congruent angles, and write the ratios of corresponding sides as one proportion.
- \triangle GHJ \sim \triangle MKL. Name the side that corresponds to \overline{HJ}, the side that corresponds to \overline{GJ}, and the angle that corresponds to \angle J.
- \triangle ABC \sim \triangle DEF, with AB = 6, BC = 8, CA = 10 and DE = 9. Find the scale factor from \triangle ABC to \triangle DEF, then EF and FD.
- In \triangle PQR and \triangle XYZ, \angle P \cong \angle Y and \angle Q \cong \angle X. Write a similarity statement that begins with \triangle PQR, and name the criterion.
Choosing the criterion
Decide whether the two triangles must be similar. If they must, write the similarity statement and name the criterion. If not, say why not.
- \triangle ABC has m\angle A = 40^\circ and m\angle B = 65^\circ; \triangle DEF has m\angle D = 65^\circ and m\angle E = 75^\circ.
- \triangle ABC has AB = 5, BC = 7, CA = 9; \triangle PQR has PQ = 21, QR = 27, RP = 15.
- \triangle KLM has KL = 4, LM = 6, MK = 8; \triangle STU has ST = 6, TU = 9, US = 13.
- \triangle JKL has JK = 6, JL = 9 and m\angle J = 50^\circ; \triangle MNP has MN = 10, MP = 15 and m\angle M = 50^\circ.
- \triangle JKL has JK = 9, KL = 7 and m\angle J = 40^\circ; \triangle MNP has MN = 18, NP = 14 and m\angle M = 40^\circ.
- \triangle ABC has a right angle at C, AC = 5 and BC = 12; \triangle DEF has a right angle at F, {DF = 7.5} and EF = 18. Then, given AB = 13, find DE.
Finding lengths
- In Figure 2, Leo’s eye is at A, {1.5\ \text{m}} above his feet at B. A mirror lies flat on the ground at C, {2\ \text{m}} from B, and a flagpole \overline{DE} stands {14\ \text{m}} beyond the mirror. Leo sees the top E in the mirror. Taking the law of reflection as the model (light leaves the mirror at the same angle to it as it arrives), \angle ACB \cong \angle ECD. How tall is the flagpole?
- In Figure 3, AB = 15, DC = 10 and BD = 20. Find BE and DE.
- In \triangle PQR, point S lies on \overline{PR} and \angle PQS \cong \angle R, so \triangle PQS \sim \triangle PRQ (Exercise 17 asks for the proof). If PS = 4, PR = 9 and QR = 7.5, find PQ and QS.
- In \triangle ABC and \triangle DEF, \angle A \cong \angle D, AB = x, AC = 12, DE = 6 and DF = 8. Find the value of x for which SAS similarity gives \triangle ABC \sim \triangle DEF.
- A triangle with sides 3, 5 and 6 is similar to a triangle with sides 9, x + 7 and 2x + 2, with the sides listed in corresponding order. Find x.
Writing the proof
Write each as a two-column proof.
- Given. \overline{JK} \parallel \overline{LM}; \overline{JM} and \overline{KL} cross at N. Prove. \triangle JKN \sim \triangle MLN.
- Given. In \triangle PQR, S lies on \overline{PR}; \angle PQS \cong \angle R. Prove. \triangle PQS \sim \triangle PRQ.
- Given. In \triangle XYZ, W lies on \overline{XY} and V on \overline{XZ}; XW = 2, WY = 10, XV = 3, VZ = 5. Prove. \triangle XWV \sim \triangle XZY. Then, given YZ = 14, find WV.
- Given. Quadrilateral ABCD has diagonal \overline{AC}, with AB = 4, BC = 6, AC = 8, CD = 12 and DA = 16. Prove. \overrightarrow{AC} bisects \angle BAD.
Reasoning
- AA asks for two pairs of angles and no sides. Explain why two pairs are enough, and why the sides then have to be in proportion.
- Prove that any two equilateral triangles are similar, and name the criterion. Then decide whether any two isosceles triangles are similar.
- One rhombus has sides of length 3 and another has sides of length 6. Explain why their sides are in proportion but the rhombuses need not be similar, and why the same thing cannot happen with two triangles.
- Prove that if \triangle ABC \sim \triangle DEF and \triangle DEF \sim \triangle GHI, then \triangle ABC \sim \triangle GHI.
Answers
- \angle R \cong \angle X, \angle S \cong \angle Y, \angle T \cong \angle Z, and \displaystyle \frac{RS}{XY} = \frac{ST}{YZ} = \frac{RT}{XZ}. Letters in the same position correspond.
- \overline{KL}, \overline{ML} and \angle L. The statement pairs G \leftrightarrow M, H \leftrightarrow K, J \leftrightarrow L, so \overline{HJ} (positions 2 and 3) goes with \overline{KL} and \overline{GJ} (positions 1 and 3) with \overline{ML}.
- Scale factor \displaystyle \frac{3}{2}, EF = 12, FD = 15. \overline{DE} is the partner of \overline{AB}, so {k = 9/6 = 3/2}; then {EF = 1.5 \times 8} and {FD = 1.5 \times 10}.
- \triangle PQR \sim \triangle YXZ, by AA. The given pairs put Y under P and X under Q, and the remaining vertex Z goes under R.
- \triangle ABC \sim \triangle FDE, by AA. The third angles are m\angle C = 180^\circ - 40^\circ - 65^\circ = 75^\circ and m\angle F = 180^\circ - 65^\circ - 75^\circ = 40^\circ, so {A \leftrightarrow F} at 40^\circ, {B \leftrightarrow D} at 65^\circ and {C \leftrightarrow E} at 75^\circ.
- \triangle ABC \sim \triangle RPQ, by SSS with scale factor 3. In order the sides are 5, 7, 9 and 15, 21, 27, and \displaystyle \frac{15}{5} = \frac{21}{7} = \frac{27}{9} = 3. \overline{AB} pairs with \overline{RP} and \overline{BC} with \overline{PQ}, so B, where they meet, goes with P; likewise C \leftrightarrow Q and A \leftrightarrow R.
- Not similar. In order the sides are 4, 6, 8 and 6, 9, 13. \displaystyle \frac{6}{4} = \frac{9}{6} = 1.5 but \displaystyle \frac{13}{8} = 1.625. A similarity would have to pair the sides in order, so no correspondence works.
- \triangle JKL \sim \triangle MNP, by SAS. \displaystyle \frac{MN}{JK} = \frac{10}{6} = \frac{5}{3} and \displaystyle \frac{MP}{JL} = \frac{15}{9} = \frac{5}{3}, and the 50^\circ angles are included: \angle J lies between \overline{JK} and \overline{JL}, and \angle M between \overline{MN} and \overline{MP}.
- Not necessarily similar. \displaystyle \frac{MN}{JK} = \frac{NP}{KL} = 2, but the angle between \overline{JK} and \overline{KL} is \angle K, not \angle J, so this is side, side, angle. L lies on the other side of the 40^\circ angle at J, at distance 7 from K. That side passes about 5.8 from K, and 7 is more than that but less than JK = 9, so the circle of radius 7 about K crosses it twice, at JL \approx 2.95 and JL \approx 10.84. Two differently shaped triangles fit, with angles at L of about 124.3^\circ and 55.7^\circ (supplementary). The data for \triangle MNP are the same doubled, so it too can take either shape: similar if both triangles take the same shape, not similar otherwise.
- \triangle ABC \sim \triangle DEF, by SAS, and DE = 19.5. \displaystyle \frac{DF}{AC} = \frac{7.5}{5} = 1.5 and \displaystyle \frac{EF}{BC} = \frac{18}{12} = 1.5, and each right angle is between the two legs. Then DE = 1.5 \times 13.
- 10.5 m. \triangle ABC \sim \triangle EDC by AA, from the right angles at B and D and the given angles at C. So \displaystyle \frac{ED}{AB} = \frac{DC}{BC} = \frac{14}{2} = 7, and ED = 7 \times 1.5.
- BE = 12 and DE = 8. By Example 2, \triangle ABE \sim \triangle CDE, so \displaystyle \frac{BE}{DE} = \frac{AB}{CD} = \frac{15}{10} = \frac{3}{2}. With BE = 3t, DE = 2t and BE + ED = 20, 5t = 20 and t = 4.
- PQ = 6 and QS = 5. From \triangle PQS \sim \triangle PRQ, \displaystyle \frac{PQ}{PR} = \frac{PS}{PQ} = \frac{QS}{RQ}. The first two give PQ^2 = PS \cdot PR = 36, so PQ = 6, and the ratio is \displaystyle \frac{6}{9} = \frac{2}{3}. Then \displaystyle QS = \frac{2}{3}(7.5) = 5.
- x = 9. SAS needs {AB/DE = AC/DF}, that is {x/6 = 12/8}, so {x = 6 \times 1.5}.
- x = 8. The first pair gives the ratio \displaystyle \frac{9}{3} = 3. Then x + 7 = 3 \cdot 5 = 15 gives x = 8, and the third pair agrees: 2(8) + 2 = 18 = 3 \cdot 6.
- Line (1): \overline{JK} \parallel \overline{LM} (given). (2): \angle KJN \cong \angle LMN (Alternate Interior Angles Theorem: line 1, transversal \overline{JM}). (3): \angle JKN \cong \angle MLN (Alternate Interior Angles Theorem: line 1, transversal \overline{KL}). (4): \triangle JKN \sim \triangle MLN (AA similarity: lines 2 and 3). The angles at J and M are in position 1 and those at K and L in position 2. The vertical angles at N could replace either pair.
- Line (1): \angle PQS \cong \angle R (given). (2): \angle QPS \cong \angle RPQ (reflexive property of congruence: both are \angle P, since S lies on \overline{PR}). (3): \triangle PQS \sim \triangle PRQ (AA similarity: lines 1 and 2). The angle at Q in \triangle PQS is matched with the angle at R in \triangle PRQ, both in position 2.
- WV = 3.5. Line (1): XW = 2, WY = 10, XV = 3, VZ = 5 (given). (2): XY = 12 and XZ = 8 (segment addition postulate: line 1). (3): \displaystyle \frac{XW}{XZ} = \frac{1}{4} and \displaystyle \frac{XV}{XY} = \frac{1}{4} (arithmetic: lines 1 and 2). (4): \angle WXV \cong \angle ZXY (reflexive property of congruence). (5): \triangle XWV \sim \triangle XZY (SAS similarity: lines 3 and 4). Then \displaystyle \frac{WV}{ZY} = \frac{1}{4}, so \displaystyle WV = \frac{14}{4} = 3.5.
- Line (1): AB = 4, BC = 6, AC = 8, CD = 12, DA = 16 (given). (2): \displaystyle {\frac{AB}{AC} = \frac{BC}{CD} = \frac{AC}{AD} = \frac{1}{2}} (arithmetic: line 1). (3): \triangle ABC \sim \triangle ACD (SSS similarity: line 2). (4): \angle BAC \cong \angle CAD (corresponding angles of similar triangles are congruent: line 3). (5): \overrightarrow{AC} bisects \angle BAD (definition of angle bisector: line 4, with B and D on opposite sides of \overline{AC}). The sides in order are 4, 6, 8 and 8, 12, 16, which fixes A \leftrightarrow A, B \leftrightarrow C, C \leftrightarrow D.
- Two pairs of angles give the third pair, since the angles of each triangle add to 180^\circ. Dilate the first triangle by the ratio of one pair of corresponding sides: the image has that side equal to its partner and the same angles, so it is congruent to the second triangle by ASA. Its other sides are the first triangle’s sides times the same factor, and they equal the second triangle’s sides, so all three ratios are that factor.
- In any equilateral triangle every angle is 180^\circ \div 3 = 60^\circ, so two equilateral triangles have two pairs of 60^\circ angles under any correspondence: similar by AA. (SSS works too: all three ratios are the ratio of the side lengths.) Two isosceles triangles need not be similar: one with angles 40^\circ, 70^\circ, 70^\circ and one with 80^\circ, 50^\circ, 50^\circ share no angle, so no correspondence matches their angles.
- Each side of the second rhombus is twice a side of the first, so all four ratios are 2. But the angles need not match: a rhombus can be a square, or it can lean over into a thin shape with two small angles, while keeping its side. Similar polygons need equal angles too. A triangle cannot lean in this way: its three sides fix its angles (by SSS congruence), so three sides in one ratio already force equal angles, which is SSS similarity.
- From the first similarity, \angle A \cong \angle D and \angle B \cong \angle E; from the second, \angle D \cong \angle G and \angle E \cong \angle H. By the transitive property of congruence, \angle A \cong \angle G and \angle B \cong \angle H, so \triangle ABC \sim \triangle GHI by AA.