Similarity Criteria of Triangles: AA, SSS and SAS Explained

To show from the def­i­n­i­tion that two tri­an­gles are sim­i­lar, you would have to check six facts: three pairs of equal angles and three pairs of sides in the same ratio. With con­gru­ent tri­an­gles you learnt short­cuts (SSS, SAS, ASA, RHS) that need only three match­ing parts. Sim­i­lar tri­an­gles have short­cuts too: the AA (or AAA), SSS and SAS sim­i­lar­ity cri­te­ria. This les­son, from NCERT Class 10 Chap­ter 6 (Tri­an­gles), states each cri­te­rion, explains why it is true using the Basic Pro­por­tion­al­ity The­o­rem, and applies the cri­te­ria to angles, lengths, proofs and the shadow prob­lem that often appears in exams.

Nota­tion and cor­re­spon­dence#

We write $\triangle ABC \sim \triangle DEF$ to mean the tri­an­gles are sim­i­lar with $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$. Then

$$\displaystyle \angle A = \angle D,\quad \angle B = \angle E,\quad \angle C = \angle F \quad\text{and}\quad \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}.$$

The order of the let­ters mat­ters. For the same pair of tri­an­gles, writ­ing $\triangle ABC \sim \triangle EDF$ would claim that $A$ cor­re­sponds to $E$ and $B$ to $D$, which is false. Always write the ver­tices in match­ing order.

AAA and AA sim­i­lar­ity#

An activ­ity#

Draw $BC = 3$ cm with angles of $60^\circ$ at $B$ and $40^\circ$ at $C$, and let the arms meet at $A$. Then draw a sep­a­rate seg­ment $EF = 5$ cm with the same angles, $60^\circ$ at $E$ and $40^\circ$ at $F$, meet­ing at $D$. The third angles are both $180^\circ - 60^\circ - 40^\circ = 80^\circ$, so the tri­an­gles are equian­gu­lar.

Now $\displaystyle \frac{BC}{EF} = \frac{3}{5} = 0.6$. Mea­sure $AB$, $DE$, $AC$ and $DF$: the ratios $\displaystyle \frac{AB}{DE}$ and $\displaystyle \frac{AC}{DF}$ also come out as $0.6$. The sides take care of them­selves.

State­ment#

The­o­rem 6.3 (AAA sim­i­lar­ity cri­te­rion). If in two tri­an­gles the cor­re­spond­ing angles are equal, then their cor­re­spond­ing sides are in the same ratio (pro­por­tion), and hence the two tri­an­gles are sim­i­lar.

Because the angles of a tri­an­gle add up to $180^\circ$, two pairs of equal angles force the third pair to be equal. So in prac­tice we only need two:

AA sim­i­lar­ity cri­te­rion. If two angles of one tri­an­gle are respec­tively equal to two angles of another tri­an­gle, then the two tri­an­gles are sim­i­lar.

Why it is true#

Given: $\triangle ABC$ and $\triangle DEF$ with $\angle A = \angle D$, $\angle B = \angle E$, $\angle C = \angle F$.

To prove: $\displaystyle \frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF}$.

Con­struc­tion: On $DE$ mark $P$ with $DP = AB$, and on $DF$ mark $Q$ with $DQ = AC$. Join $PQ$. (If $AB = DE$ the tri­an­gles are con­gru­ent by ASA and there is noth­ing more to prove.)

Proof.

1. In $\triangle ABC$ and $\triangle DPQ$: $AB = DP$, $\angle A = \angle D$, $AC = DQ$. So $\triangle ABC \cong \triangle DPQ$ (SAS con­gru­ence).
2. Hence $\angle DPQ = \angle B = \angle E$. These are cor­re­spond­ing angles for the lines $PQ$ and $EF$ with trans­ver­sal $DE$, so $PQ \parallel EF$.
3. In $\triangle DEF$, $PQ \parallel EF$, so by the Basic Pro­por­tion­al­ity The­o­rem (whole-side form) $\displaystyle \frac{DP}{DE} = \frac{DQ}{DF}$, that is, $\displaystyle \frac{AB}{DE} = \frac{AC}{DF}$.
4. Repeat­ing the argu­ment at ver­tex $E$ (mark­ing lengths $BA$ and $BC$ from $E$) gives $\displaystyle \frac{AB}{DE} = \frac{BC}{EF}$. $\blacksquare$

This is what makes tri­an­gles spe­cial. For a gen­eral poly­gon, equal angles are not enough (a square and a rec­tan­gle have equal angles). For a tri­an­gle, fix­ing the angles fixes the shape com­pletely, and only the size can change.

SSS sim­i­lar­ity#

An activ­ity#

Draw $\triangle ABC$ with $AB = 3$ cm, $BC = 6$ cm, $CA = 8$ cm and $\triangle DEF$ with $DE = 4.5$ cm, $EF = 9$ cm, $FD = 12$ cm. Then

$$\displaystyle \frac{AB}{DE} = \frac{3}{4.5} = \frac{2}{3}, \qquad \frac{BC}{EF} = \frac{6}{9} = \frac{2}{3}, \qquad \frac{CA}{FD} = \frac{8}{12} = \frac{2}{3}.$$

Mea­sure the angles: $\angle A = \angle D$, $\angle B = \angle E$ and $\angle C = \angle F$ (about $39.6^\circ$, $121.9^\circ$ and $18.6^\circ$ respec­tively).

State­ment#

The­o­rem 6.4 (SSS sim­i­lar­ity cri­te­rion). If in two tri­an­gles the sides of one tri­an­gle are pro­por­tional to (in the same ratio as) the sides of the other tri­an­gle, then their cor­re­spond­ing angles are equal, and hence the two tri­an­gles are sim­i­lar.

Why it is true#

Sup­pose $\displaystyle \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}$ with this com­mon ratio less than $1$ (the case greater than $1$ is the same with the tri­an­gles swapped, and the case equal to $1$ is SSS con­gru­ence). Mark $P$ on $DE$ and $Q$ on $DF$ with $DP = AB$ and $DQ = AC$.

1. Then $\displaystyle \frac{DP}{DE} = \frac{DQ}{DF}$, which gives $\displaystyle \frac{DP}{PE} = \frac{DQ}{QF}$. By the con­verse of the Basic Pro­por­tion­al­ity The­o­rem, $PQ \parallel EF$.
2. So $\angle DPQ = \angle E$ and $\angle DQP = \angle F$ (cor­re­spond­ing angles), and $\triangle DPQ \sim \triangle DEF$ by AA.
3. Hence $\displaystyle \frac{PQ}{EF} = \frac{DP}{DE} = \frac{AB}{DE} = \frac{BC}{EF}$, so $PQ = BC$.
4. Now $\triangle ABC \cong \triangle DPQ$ by SSS con­gru­ence, so $\angle A = \angle D$, $\angle B = \angle DPQ = \angle E$ and $\angle C = \angle DQP = \angle F$. $\blacksquare$

SAS sim­i­lar­ity#

An activ­ity#

Draw $\triangle ABC$ with $AB = 2$ cm, $AC = 4$ cm and $\angle A = 50^\circ$, and $\triangle DEF$ with $DE = 3$ cm, $DF = 6$ cm and $\angle D = 50^\circ$. The sides that include the equal angles are in the same ratio: $\displaystyle \frac{AB}{DE} = \frac{2}{3}$ and $\displaystyle \frac{AC}{DF} = \frac{4}{6} = \frac{2}{3}$. Mea­sur­ing gives $\angle B = \angle E$ (about $100.6^\circ$) and $\angle C = \angle F$ (about $29.4^\circ$), so the tri­an­gles are sim­i­lar.

State­ment#

The­o­rem 6.5 (SAS sim­i­lar­ity cri­te­rion). If one angle of a tri­an­gle is equal to one angle of the other tri­an­gle and the sides includ­ing these angles are pro­por­tional, then the two tri­an­gles are sim­i­lar.

The equal angle must be the one between the two pro­por­tional sides, just as in SAS con­gru­ence.

Why it is true#

Sup­pose $\angle A = \angle D$ and $\displaystyle \frac{AB}{DE} = \frac{AC}{DF}$ (less than $1$). Mark $P$ on $DE$ and $Q$ on $DF$ with $DP = AB$ and $DQ = AC$. Then $\displaystyle \frac{DP}{DE} = \frac{DQ}{DF}$, so $PQ \parallel EF$ by the con­verse of the Basic Pro­por­tion­al­ity The­o­rem, and $\triangle DPQ \sim \triangle DEF$ by AA. Also $\triangle ABC \cong \triangle DPQ$ by SAS con­gru­ence. So $\triangle ABC$ has the same angles as $\triangle DPQ$, which has the same angles as $\triangle DEF$, and $\triangle ABC \sim \triangle DEF$. $\blacksquare$

Method: prov­ing tri­an­gles sim­i­lar#

1. Name the two tri­an­gles and look for infor­ma­tion: par­al­lel lines (alter­nate or cor­re­spond­ing angles), ver­ti­cally oppo­site angles, a com­mon angle, right angles, or given lengths.
2. Choose the cri­te­rion: two angles known, use AA; three sides known, use SSS; one angle and its two sides known, use SAS.
3. Write the cor­re­spon­dence care­fully, with equal angles in match­ing posi­tions.
4. State the cri­te­rion and con­clude, for exam­ple "$\triangle POQ \sim \triangle SOR$ (AA)".
5. Use the sim­i­lar­ity: equal angles, or equal ratios of cor­re­spond­ing sides, to fin­ish the ques­tion.

Worked exam­ples#

Exam­ple 1: par­al­lel lines give AA#

$PQ \parallel RS$, and the seg­ments $PS$ and $QR$ inter­sect at $O$ (so $P, O, S$ are collinear and $Q, O, R$ are collinear). Prove that $\triangle POQ \sim \triangle SOR$.

Solu­tion. With $PS$ as trans­ver­sal, $\angle P = \angle S$ (alter­nate angles, since $PQ \parallel RS$). With $QR$ as trans­ver­sal, $\angle Q = \angle R$ (alter­nate angles). Two pairs of equal angles, so $\triangle POQ \sim \triangle SOR$ (AA). The third pair also matches: $\angle POQ = \angle SOR$ as ver­ti­cally oppo­site angles. $\blacksquare$

Exam­ple 2: SSS to find an angle#

In $\triangle ABC$, $AB = 3.8$ cm, $BC = 6$ cm, $CA = 3\sqrt{3}$ cm, $\angle A = 80^\circ$ and $\angle B = 60^\circ$. In $\triangle PQR$, $RQ = 7.6$ cm, $QP = 12$ cm and $PR = 6\sqrt{3}$ cm. Find $\angle P$.

Solu­tion. Com­pare the sides in order of size:

$$\displaystyle \frac{AB}{RQ} = \frac{3.8}{7.6} = \frac{1}{2}, \qquad \frac{BC}{QP} = \frac{6}{12} = \frac{1}{2}, \qquad \frac{CA}{PR} = \frac{3\sqrt{3}}{6\sqrt{3}} = \frac{1}{2}.$$

All three ratios are equal, so $\triangle ABC \sim \triangle RQP$ (SSS). This gives $A \leftrightarrow R$, $B \leftrightarrow Q$, $C \leftrightarrow P$, so $\angle P = \angle C$. By the angle sum prop­erty,

$$\angle C = 180^\circ - 80^\circ - 60^\circ = 40^\circ.$$

Answer: $\angle P = 40^\circ$.

Exam­ple 3: SAS with a prod­uct of lengths#

The seg­ments $AC$ and $BD$ inter­sect at $O$, and $OA \cdot OB = OC \cdot OD$. Show that $\angle A = \angle C$ and $\angle B = \angle D$.

Solu­tion. Divid­ing $OA \cdot OB = OC \cdot OD$ by $OC \cdot OB$ gives

$$\displaystyle \frac{OA}{OC} = \frac{OD}{OB}.$$

Con­sider $\triangle AOD$ and $\triangle COB$. The sides $OA, OD$ of the first and $OC, OB$ of the sec­ond are pro­por­tional, and the angles they include are equal: $\angle AOD = \angle COB$ (ver­ti­cally oppo­site angles). So $\triangle AOD \sim \triangle COB$ (SAS), with $A \leftrightarrow C$ and $D \leftrightarrow B$. Hence $\angle A = \angle C$ and $\angle D = \angle B$. $\blacksquare$

Exam­ple 4: the girl and the lamp-post#

A girl of height $90$ cm is walk­ing away from the base of a lamp-post at a speed of $1.2$ m/s. The lamp is $3.6$ m above the ground. Find the length of her shadow after $4$ sec­onds.

Solu­tion. Let $AB$ be the lamp-post, $CD$ the girl after $4$ sec­onds, and $DE = x$ m her shadow. In $4$ sec­onds she walks

$$BD = 1.2 \times 4 = 4.8 \text{ m}.$$

In $\triangle ABE$ and $\triangle CDE$: $\angle B = \angle D = 90^\circ$ (the lamp-post and the girl both stand ver­ti­cally), and $\angle E$ is com­mon. So $\triangle ABE \sim \triangle CDE$ (AA), and

$$\displaystyle \frac{BE}{DE} = \frac{AB}{CD}.$$

Here $BE = 4.8 + x$, $AB = 3.6$ m and $CD = 90$ cm $= 0.9$ m. So

$$\displaystyle \begin{aligned} \frac{4.8 + x}{x} &= \frac{3.6}{0.9} = 4 \\ 4.8 + x &= 4x \\ 3x &= 4.8 \\ x &= 1.6 \end{aligned}$$

Answer: The shadow is $1.6$ m long after $4$ sec­onds.

Exam­ple 5: a length from AA#

In $\triangle ABC$, $D$ is on $AB$ and $E$ is on $AC$ with $DE \parallel BC$. If $AD = 4$ cm, $AB = 10$ cm and $BC = 15$ cm, find $DE$.

Solu­tion. $\angle ADE = \angle ABC$ and $\angle AED = \angle ACB$ (cor­re­spond­ing angles, $DE \parallel BC$). So $\triangle ADE \sim \triangle ABC$ (AA), and

$$\displaystyle \frac{DE}{BC} = \frac{AD}{AB} \implies DE = \frac{4}{10} \times 15 = 6.$$

Answer: $DE = 6$ cm.

Exam­ple 6: a tri­an­gle sim­i­lar to part of itself#

$D$ is a point on side $BC$ of $\triangle ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.

Solu­tion. Com­pare $\triangle CDA$ and $\triangle CAB$. Since $D$ lies on $BC$, $\angle DCA$ and $\angle ACB$ are the same angle, so $\angle C$ is com­mon. Also $\angle CDA = \angle CAB$ (given). So $\triangle CDA \sim \triangle CAB$ (AA), with $C \leftrightarrow C$, $D \leftrightarrow A$, $A \leftrightarrow B$. Cor­re­spond­ing sides give

$$\displaystyle \frac{CD}{CA} = \frac{CA}{CB} \implies CA^2 = CB \cdot CD. \qquad \blacksquare$$

For exam­ple, if $CB = 9$ cm and $CD = 4$ cm, then $CA^2 = 36$ and $CA = 6$ cm. Spot­ting a tri­an­gle sim­i­lar to a piece of itself is a trick that returns again and again in geom­e­try.

A fourth cri­te­rion for right tri­an­gles#

For right tri­an­gles there is one more test: if the hypotenuse and one side of one right tri­an­gle are pro­por­tional to the hypotenuse and one side of another right tri­an­gle, the two tri­an­gles are sim­i­lar. This is the RHS sim­i­lar­ity cri­te­rion. It becomes use­ful once you work with Pythago­ras' the­o­rem. So the full list is AA, SSS, SAS and, for right tri­an­gles only, RHS.

Com­mon mis­takes#

• Writ­ing the ver­tices in the wrong order, for exam­ple $\triangle ABC \sim \triangle EDF$ when $A$ matches $D$.
• Using SAS with an angle that is not between the two pro­por­tional sides.
• Pair­ing sides at ran­dom in SSS. Pair short­est with short­est, mid­dle with mid­dle and longest with longest.
• Mix­ing units, such as $90$ cm and $3.6$ m in the same ratio. Con­vert first.
• Con­clud­ing "sim­i­lar" from one pair of equal angles. AA needs two pairs.
• For­get­ting that the shadow prob­lem uses the whole base $BE = BD + DE$, not just $BD$.

Try these#

1. One tri­an­gle has angles $50^\circ$ and $70^\circ$; another has angles $70^\circ$ and $60^\circ$. Are they sim­i­lar? Answer: Yes (AA); both have angles $50^\circ, 60^\circ, 70^\circ$.
2. Are tri­an­gles with sides $4, 6, 8$ cm and $6, 9, 12$ cm sim­i­lar? Answer: Yes (SSS), ratio $\displaystyle \tfrac{2}{3}$.
3. Are tri­an­gles with sides $5, 7, 9$ cm and $10, 14, 17$ cm sim­i­lar? Answer: No, since $\displaystyle \tfrac{9}{17} \ne \tfrac{1}{2}$.
4. A $1.2$ m stick casts a $0.8$ m shadow at the same time as a tree casts a $6$ m shadow. Find the height of the tree. Answer: $9$ m.
5. $\triangle ABC \sim \triangle PQR$ with $AB = 6$ cm, $PQ = 9$ cm and $QR = 12$ cm. Find $BC$. Answer: $8$ cm.
6. In $\triangle ABC$, $AB = 3$ cm, $AC = 5$ cm, $\angle A = 60^\circ$; in $\triangle PQR$, $PQ = 6$ cm, $PR = 10$ cm, $\angle P = 60^\circ$. Are they sim­i­lar? Answer: Yes (SAS), scale fac­tor $2$.

Key terms#

Equian­gu­lar tri­an­gles

Tri­an­gles whose cor­re­spond­ing angles are all equal.

AAA / AA cri­te­rion

Equal cor­re­spond­ing angles (in prac­tice, two pairs) make two tri­an­gles sim­i­lar.

SSS cri­te­rion

Three pairs of sides in the same ratio make two tri­an­gles sim­i­lar.

SAS cri­te­rion

One equal angle with the sides includ­ing it in the same ratio makes two tri­an­gles sim­i­lar.

RHS cri­te­rion

For right tri­an­gles, hypotenuse and one side in the same ratio make the tri­an­gles sim­i­lar.

Cor­re­spon­dence

The match­ing of ver­tices recorded by the order of let­ters in $\triangle ABC \sim \triangle DEF$.

Included angle

The angle formed between two given sides of a tri­an­gle.

Ver­ti­cally oppo­site angles

The equal, oppo­site angles formed where two straight lines cross.

Com­mon ques­tions#

Why is there no "SSA" sim­i­lar­ity cri­te­rion?#

An angle that is not between the two pro­por­tional sides does not fix the tri­an­gle's shape: two dif­fer­ently shaped tri­an­gles can share such data. So, as with con­gru­ence, the angle must be the included one.

Is AA really enough, or do I need AAA?#

AA is enough. The third angle is $180^\circ$ minus the other two, so it is auto­mat­i­cally equal.

How do I find the cor­rect cor­re­spon­dence?#

Match equal angles first. For SSS, match sides by size: small­est to small­est, and so on. The ver­tex oppo­site a side cor­re­sponds to the ver­tex oppo­site its part­ner.

Are con­gru­ent tri­an­gles sim­i­lar?#

Yes. They sat­isfy every cri­te­rion with ratio $1$.

Where are these cri­te­ria used later?#

In proofs about areas and right tri­an­gles, in coor­di­nate geom­e­try, and in heights-and-dis­tances prob­lems like the lamp-post exam­ple.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Kise­lev, A. P. Kise­lev's Geom­e­try, Book I: Planime­try (adapted by A. Given­tal). Sum­iz­dat.
3. Cox­eter, H. S. M. and Gre­itzer, S. L. Geom­e­try Revis­ited. Math­e­mat­i­cal Asso­ci­a­tion of Amer­ica.
4. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.