SAS Congruence

The **SAS (Side-Angle-Side) criterion** says: if two sides of one triangle and the angle **between** them are congruent to two sides and the angle between them in another triangle, then the two triangles are congruent. $$\overline{AB} \cong \overline{DE}, \quad \angle B \cong \angle E, \quad \overline{BC} \cong \overline{EF} \quad \Longrightarrow \quad \triangle ABC \cong \triangle DEF$$ The angle must be the **included angle**: the angle formed by the two given sides. Its vertex is the point the two sides share. For sides $AB$ and $BC$ the shared point is $B$, so the included angle is $\angle B$. **Why it works:** think of two sticks joined by a hinge. Once the lengths of the sticks and the opening of the hinge are fixed, the distance between the free ends is fixed too. The third side has only one possible length, so there is only one possible triangle. **Careful:** two sides and an angle that is **not** between them (SSA) do **not** prove congruence. The side opposite the angle can sometimes swing into two different positions, giving two different triangles.

Checking SAS step by step

In $\triangle ABC$: $AB = 6$ cm, $\angle B = 50°$, $BC = 8$ cm. In $\triangle DEF$: $DE = 6$ cm, $\angle E = 50°$, $EF = 8$ cm. Are the triangles congruent? If $AC \approx 6.2$ cm, how long is $DF$?

List the matching parts: $AB = DE = 6$ cm, $BC = EF = 8$ cm, $\angle B = \angle E = 50°$ = Two pairs of sides and one pair of angles

Check that the angle is included: Sides $AB$ and $BC$ meet at $B$; sides $DE$ and $EF$ meet at $E$ = $\angle B$ and $\angle E$ are the included angles

Apply SAS: Two sides and the included angle match = $\triangle ABC \cong \triangle DEF$

Use corresponding parts: $AC$ corresponds to $DF$ (first and third letters) = $DF = AC \approx 6.2$ cm

Answer: Yes, $\triangle ABC \cong \triangle DEF$ by SAS, so $DF \approx 6.2$ cm.

Is the angle really included?

In $\triangle PQR$: $PQ = 5$, $QR = 7$, $\angle R = 40°$. In $\triangle XYZ$: $XY = 5$, $YZ = 7$, $\angle Z = 40°$. Can you use SAS?

Find the shared point of the two sides: $PQ$ and $QR$ share the point $Q$ = The included angle would be $\angle Q$

Compare with the given angle: The given angle is $\angle R$, which is not between $PQ$ and $QR$ = The angle is not included

Name the pattern: Side, side, then an angle that is not between them = This is SSA

Conclude: SSA is not a congruence criterion = No conclusion is possible

Answer: No. The given angle is not between the two sides (SSA), so SAS does not apply and the triangles might not be congruent.

Hidden angle: vertical angles

Segments $AE$ and $BD$ cross at $C$. $C$ is the midpoint of both segments: $AC = CE = 4$ cm and $BC = CD = 5$ cm. If $AB = 7$ cm, how long is $DE$?

Use the midpoints: $AC = CE = 4$ cm and $BC = CD = 5$ cm = Two pairs of equal sides

Find the hidden angle: $\angle ACB$ and $\angle ECD$ are vertical angles = $\angle ACB = \angle ECD$

Check it is included: $\angle ACB$ lies between $AC$ and $BC$; $\angle ECD$ lies between $EC$ and $DC$ = $\triangle ACB \cong \triangle ECD$ by SAS

Apply CPCTC: $AB$ corresponds to $ED$ = $DE = AB = 7$ cm

Answer: $DE = 7$ cm, because $\triangle ACB \cong \triangle ECD$ by SAS.

Hidden side: a shared side

In quadrilateral $ABCD$, $AB = CB$ and the diagonal $BD$ bisects $\angle ABC$, so $\angle ABD = \angle CBD = 35°$. If $\angle BAD = 80°$, find $\angle BCD$ and $\angle ADB$.

Write the given parts: $AB = CB$ and $\angle ABD = \angle CBD$ = One pair of sides, one pair of angles

Find the hidden side: Both triangles contain $BD$, and $BD = BD$ = Shared side

Apply SAS: The angle at $B$ lies between $AB$ and $BD$ (and between $CB$ and $BD$) = $\triangle ABD \cong \triangle CBD$

Use CPCTC and the angle sum: $\angle BCD = \angle BAD = 80°$; $\angle ADB = 180° - 35° - 80°$ = $\angle BCD = 80°$, $\angle ADB = 65°$

Answer: $\angle BCD = 80°$ and $\angle ADB = 65°$.

Why SSA fails: two triangles from the same data

Draw a triangle with $\angle A = 30°$, $AB = 10$ and $BC = 6$. How many different triangles fit this description?

Find the height of $B$ above line $AC$: Opposite a $30°$ angle in a right triangle the side is half the hypotenuse: $h = 10 \div 2$ = $h = 5$

Compare $BC$ with the height: $5 < 6 < 10$: side $BC$ is longer than the height but shorter than $AB$ = $BC$ can reach line $AC$ at two points

Find how far each point is from the foot of the height: $\sqrt{6^2 - 5^2} = \sqrt{11} \approx 3.32$ = One point on each side of the foot

Find both possible lengths of $AC$: The foot is $\approx 8.66$ from $A$: $8.66 - 3.32$ and $8.66 + 3.32$ = $AC \approx 5.34$ or $AC \approx 11.98$

Answer: Two different triangles, with $AC \approx 5.34$ or $AC \approx 11.98$. The same side, side, angle data gives two shapes, so SSA cannot prove congruence.

SAS is one of the most used tools in geometry proofs, because the information it needs often comes for free: - **Vertical angles**: when two segments cross, the opposite angles are equal, so you already have the angle - **Shared sides**: when two triangles share a side, that side is congruent to itself - **Midpoints and bisectors**: a midpoint gives two equal segments, an angle bisector gives two equal angles Once SAS proves two triangles congruent, every other pair of corresponding parts is equal too (**CPCTC**). That is how surveyors measure across a pond they cannot walk through, and how builders know two hinged frames opened to the same angle are identical.

Mistake: Using SAS when the angle is not between the two sides

Why: Two sides and a non-included angle (SSA) can describe two different triangles, as the $30°$, $10$, $6$ example shows.

Correct: Find the point the two sides share. The angle at that vertex is the only one SAS may use.

Mistake: Matching the vertices in the wrong order

Why: Writing $\triangle ABC \cong \triangle FED$ when $B$ matches $E$ and $A$ matches $D$ pairs the wrong parts and leads to wrong CPCTC conclusions.

Correct: Write the vertices in matching order. If $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$, write $\triangle ABC \cong \triangle DEF$.

Mistake: Thinking there is not enough information because a part is not written in the problem

Why: Vertical angles and shared sides are equal even when no one says so. Students miss them and give up on a proof that works.

Correct: Before you decide, look for crossing segments (vertical angles) and a side that belongs to both triangles (shared side).

Mistake: Using CPCTC before the triangles are proven congruent

Why: Corresponding parts are equal only because the triangles are congruent. Without that, you are assuming what you want to prove.

Correct: First prove $\triangle ABC \cong \triangle DEF$ (for example by SAS), then use CPCTC for the remaining parts.

Folding A-frames

Sawhorses, ladders and easels are two legs joined by a hinge. The leg lengths and the opening angle decide the whole shape.

Two easels each have legs of 150 cm opened to $25°$ at the hinge. By SAS the triangles they form are congruent, so their feet are the same distance apart.

Measuring across a pond

Surveyors use SAS to find a distance they cannot walk, by building a congruent copy of the triangle on dry land.

To measure $AB$ across a pond, pick a point $C$ on land. Extend $AC$ past $C$ to $D$ with $CD = AC$, and $BC$ past $C$ to $E$ with $CE = BC$. The angles at $C$ are vertical angles, so $\triangle ACB \cong \triangle DCE$ by SAS and $DE = AB$.

Clock hands

The two hands of a clock are two sides with a fixed length, and the time sets the angle between them.

At 2:00 and at 10:00 the angle between the hands is $60°$ both times, so by SAS the tips are the same distance apart at both times.

**SAS**: two sides and the **included angle** of one triangle congruent to those of another means the triangles are congruent

The included angle is at the vertex the two sides share: for $AB$ and $BC$ it is $\angle B$

**SSA is not a criterion**: two sides and a non-included angle can give two different triangles

Look for hidden congruent parts: **vertical angles** where segments cross, a **shared side**, midpoints and angle bisectors

After proving congruence, use **CPCTC** to get the remaining sides and angles

Criteria that work: SSS, SAS, ASA, AAS. Patterns that do not: SSA and AAA

Q: Why is SSA not a congruence criterion?

A: Because the side opposite the given angle can swing like a compass and meet the third side at two points. With $\angle A = 30°$, $AB = 10$ and $BC = 6$ you get two triangles, one with $AC \approx 5.34$ and one with $AC \approx 11.98$.

Q: Does SSA ever work?

A: Yes, in one special case: when the given angle is opposite the **longer** of the two given sides, only one triangle is possible. The best known example is a right angle opposite the hypotenuse (the HL criterion). In general, though, do not use SSA.

Q: Is SAS a postulate or a theorem?

A: Many textbooks accept SAS as a postulate (a basic rule without proof) and use it to prove ASA and SSS. Euclid proved it as his Proposition 4 by placing one triangle on top of the other. Either way, you may use it in proofs.

Q: How do I choose between SSS, SAS, ASA and AAS?

A: Count what you know. Three sides: SSS. Two sides: check whether the known angle is between them (SAS) or not (SSA, no conclusion). Two angles: check whether the known side is between them (ASA) or not (AAS).

Exit Presentation

SAS Congruence

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Geometry / Congruence & Similarity

SAS Congruence

Prove triangles congruent with two sides and the included angle, spot hidden equal parts, and see why SSA is not enough.

Intermediate25 min

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