SSS and SAS Similarity

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Learning goals

Establish triangle similarity by checking three corresponding side ratios for SSS
Apply SAS similarity using two proportional sides and their included angle
Justify SSS and SAS similarity with dilation followed by congruence
Distinguish insufficient evidence from a demonstrated failure of triangle similarity

Three side ratios establish SSS similarity

Three matching lengths can establish similarity without any angle measurement. Triangles with side lengths $5, 7, 9$ and $7.5, 10.5, 13.5$ have the same factor $1.5$ for every matched pair. Scaling the first triangle by $1.5$ gives exactly the three lengths of the second.

Every corresponding side uses the same positive factor. Equal tick marks would mean equal lengths, so these proportional sides are labeled with their lengths instead.

SSS similarity says that two nondegenerate triangles are similar if all three pairs of corresponding sides have the same positive length ratio. This is different from SSS congruence, where the corresponding lengths are equal. Congruence is the factor- $1$ case.

Why SSS similarity follows from SSS congruence#

For the numerical triangles above, dilate the $5, 7, 9$ triangle by $1.5$ . Its new lengths are $7.5, 10.5, 13.5$ . By the earlier SSS congruence criterion, the resized triangle is congruent to the target. Rigid motions can therefore complete the match, giving similarity.

The same argument works with any positive side lengths forming nondegenerate triangles. Suppose the chosen correspondence gives

\begin{aligned}\frac{DE}{AB}&=k,\\\frac{EF}{BC}&=k,\\\frac{FD}{CA}&=k.\end{aligned}

Dilate $ABC$ by $k$ , calling the image $A'B'C'$ . The accepted length-scaling property gives

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

All three matching lengths are now equal, so $\triangle A'B'C'\cong\triangle DEF$ by SSS congruence. Its rigid-motion match, together with the dilation, proves $\triangle ABC\sim\triangle DEF$ . The proof uses congruence to establish similarity, not similarity to justify itself.

Worked example 1 Check all three matched pairs

Problem. The sides of $ABC$ are

\begin{aligned}AB&=8\text{ cm}\\BC&=11\text{ cm}\\CA&=13\text{ cm}.\end{aligned}

The sides of $DEF$ are

\begin{aligned}DE&=12\text{ cm}\\EF&=16.5\text{ cm}\\FD&=19.5\text{ cm}.\end{aligned}

Decide whether the ordered statement $\triangle ABC\sim\triangle DEF$ is justified, and state the factor from $ABC$ to $DEF$ .

Solution. Keep target lengths over original lengths in every ratio.

\begin{aligned}\frac{DE}{AB}&=\frac{12}{8},\\&=1.5,\\\frac{EF}{BC}&=\frac{16.5}{11},\\&=1.5,\\\frac{FD}{CA}&=\frac{19.5}{13},\\&=1.5.\end{aligned}

All three ratios agree, so SSS proves the ordered similarity, with scale factor $1.5$ . The positive triples satisfy the triangle inequality, so neither describes a flat or impossible triangle.

Check your understanding

The corresponding side ratios for two nondegenerate triangles are $1.6$ , $1.6$ , and $1.6$ . What is justified?

A Similarity by SSS B Congruence because the ratios are equal C Similarity only if an angle is also given D No conclusion because the factor is not an integer

Preserve one matching and one ratio direction

When a correspondence is given, test that correspondence. When no matching is specified, a positive factor preserves the order of side lengths, so compare shortest with shortest, middle with middle, and longest with longest. Then recover the vertex matching from the side endpoints.

Worked example 2 Two matching ratios do not settle the third

Problem. One triangle has side lengths $6, 9, 12$ and another has side lengths $10, 15, 21$ . No correspondence is specified. Are the triangles similar? Explain why another ordering cannot repair a failed test.

Solution. Both triples are positive and satisfy the triangle inequality. Under any similarity, the shortest sides must match, as must the middle and longest sides.

\begin{aligned}\frac{10}{6}&=\frac53,\\\frac{15}{9}&=\frac53,\\\frac{21}{12}&=\frac74.\end{aligned}

The last ratio differs, so no one scale factor matches all three sides. The triangles are not similar. Matching the sides in a different order would pair unequal size ranks, which a positive scale factor cannot do.

If only a proposed named correspondence fails, say that it fails; another correspondence may work. Here the sorted-side argument rules out all of them.

Two sides need their included angle for SAS

Two lengths and the opening between them can also fix a scaled shape. Suppose two sides of one triangle are $4$ and $7$ , and their included angle is $62^\circ$ . A second triangle has corresponding sides $6$ and $10.5$ with the same included angle. Both sides scale by $1.5$ , and the opening stays fixed.

The marked angle is formed by the two selected sides in each triangle.

An included angle is the angle formed by the two sides being compared. For $AB$ and $AC$ , it is $\angle BAC$ , at $A$ . For $AB$ and $BC$ , it is $\angle ABC$ , at $B$ .

SAS similarity says that two nondegenerate triangles are similar if two corresponding side ratios are equal and the angles between those side pairs are equal. The common ratio is positive, and the matched angle must be the included angle in both triangles.

Why SAS similarity follows from SAS congruence#

Dilate the numerical $4, 7$ triangle by $1.5$ . The two selected lengths become $6$ and $10.5$ , while their included angle remains $62^\circ$ . The resized triangle is congruent to the second by earlier SAS congruence. Thus the original triangles are similar.

In general, suppose

\begin{aligned}\frac{DE}{AB}&=k,\\\frac{DF}{AC}&=k,\end{aligned}

and $\angle BAC\cong\angle EDF$ . These are the included angles at $A$ and $D$ . Dilate $ABC$ by $k$ . The resulting triangle satisfies

\begin{aligned}A'B'&=DE,\\A'C'&=DF,\\\angle B'A'C'&\cong\angle EDF.\end{aligned}

The first two equalities come from length scaling; the angle equality comes from angle preservation. SAS congruence gives a rigid-motion match between $A'B'C'$ and $DEF$ . Together with the dilation, that proves $\triangle ABC\sim\triangle DEF$ .

Worked example 3 Establish similarity before using an unknown side

Problem. In $\triangle JKL$ ,

\begin{aligned}JK&=3\text{ cm}\\JL&=8\text{ cm}\\KL&=7\text{ cm}\\\angle KJL&=60^\circ.\end{aligned}

In $\triangle PQR$ ,

\begin{aligned}QP&=7.5\text{ cm}\\QR&=20\text{ cm}\\\angle PQR&=60^\circ.\end{aligned}

Prove a correct ordered similarity statement and find $PR$ .

Solution. Compare the two sides meeting at the given angle.

\begin{aligned}\frac{QP}{JK}&=\frac{7.5}{3},\\&=2.5,\\\frac{QR}{JL}&=\frac{20}{8},\\&=2.5.\end{aligned}

The equal $60^\circ$ angles are included, so SAS proves $\triangle JKL\sim\triangle QPR$ . In this order $KL$ matches $PR$ , giving

\begin{aligned}PR&=2.5\cdot7,\\&=17.5\text{ cm}.\end{aligned}

After SAS establishes the matching, the completed triangles show the factor 2.5 and the derived PR 17.5.

Check your understanding

The sides being compared are $AB,BC$ and $DE,EF$ . Which angle pair must match to apply SAS similarity directly?

\angle A

and

\angle D

\angle C

and

\angle F

\angle B

and

\angle E

\angle A

and

\angle F

Check your understanding

In the SAS similarity proof, a dilation first matches the two given sides while preserving their included angle. Which earlier result completes the match?

A Angle sum alone B SAS congruence followed by rigid motions C The unknown third-side ratio assumed in advance D A second dilation with an unrelated factor

Insufficient evidence does not mean dissimilar

Worked example 4 Locate the angle before naming the criterion

Problem. The known lengths are

\begin{aligned}AB&=10\\BC&=7\\DE&=15\\EF&=10.5.\end{aligned}

Also,

\begin{aligned}\angle A&=\angle D,\\&=30^\circ.\end{aligned}

Does this information establish $\triangle ABC\sim\triangle DEF$ by SAS? State exactly what is missing.

The schematic marks locate the given angles at A and D; the compared side pairs meet at B and E.

Solution. Both listed side ratios are $1.5$ , but the given equal angles are not between those side pairs. SAS would need $\angle ABC\cong\angle DEF$ .

The stated information does not establish similarity. It also does not establish that the triangles are dissimilar. The missing included-angle equality might hold; it has simply not been supplied. The next example shows why a nonincluded angle cannot replace it as a general criterion.

Worked example 5 An actual counterexample to the nonincluded-angle rule

Problem. Construct a triangle $ABD$ with angles

\begin{aligned}A&=26^\circ\\D&=68^\circ\\B&=86^\circ.\end{aligned}

Draw a ray inside angle $ABD$ making $44^\circ$ with $BD$ , and let it meet $AD$ at $C$ . Show that $ABC$ and $ABD$ have two pairs of sides in the same ratio and an equal nonincluded angle, yet are not similar.

Solution. The starting triangle exists because the positive angles sum to $180^\circ$ ; its two base angles can be copied at the endpoints of a segment. The $44^\circ$ ray lies inside the $86^\circ$ angle, so it meets the opposite side between $A$ and $D$ . Thus $A,C,D$ are in that order.

In triangle $BCD$ , the angles at $B$ and $D$ are $44^\circ$ and $68^\circ$ . Its angle at $C$ is therefore $68^\circ$ . The equal base angles give $BC=BD$ by the isosceles-triangle converse.

Triangles $ABC$ and $ABD$ share side $AB$ and the angle at $A$ . Hence their two matched side ratios, $AB/AB$ and $BD/BC$ , both equal $1$ , and their equal angle at $A$ is nonincluded for these side pairs.

Their angles at $B$ are different.

\begin{aligned}m\angle ABC&=86^\circ-44^\circ,\\&=42^\circ,\\m\angle ACB&=180^\circ-26^\circ-42^\circ,\\&=112^\circ.\end{aligned}

The angle sets $26^\circ,42^\circ,112^\circ$ and $26^\circ,86^\circ,68^\circ$ cannot be paired equally in any order. These particular triangles are not similar.

The equal side pair BC and BD is marked. The shared angle at A is not the angle between AB and those sides.

Check your understanding

Two proportional side pairs and an equal nonincluded angle are given. Which statement is always justified from that information alone?

A The triangles are necessarily similar B That information alone does not guarantee similarity C The triangles are necessarily dissimilar D The triangles are necessarily congruent

Choose a criterion from the actual evidence

The earlier AA criterion uses two corresponding angle pairs. SSS uses three proportional side pairs; SAS uses two proportional side pairs and their included angle.

For each problem, fix the correspondence, choose the criterion its givens support, and keep one ratio direction. An unknown side is a consequence of established similarity, not evidence that may be assumed to establish it.

Check your understanding

Which evidence establishes similarity using only side lengths?

A Two proportional side pairs B One proportional side pair and one equal angle C Two equal angles D Three proportional corresponding side pairs

SSS and SAS Similarity

Learning goals

Three side ratios establish SSS similarity

Preserve one matching and one ratio direction

Two sides need their included angle for SAS

Insufficient evidence does not mean dissimilar

Choose a criterion from the actual evidence

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