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# 4.3: Triangle Congruence using SSS and SAS

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use the distance formula to analyze triangles on the xy\begin{align*}x-y\end{align*} plane.
• Apply the SSS and SAS Postulate to show two triangles are congruent.

## Review Queue

1. Use the distance formula, (x2x1)2+(y2y1)2\begin{align*}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\end{align*} to find the distance between the two points.
1. (-1, 5) and (4, 12)
2. (-6, -15) and (-3, 8)
1. If we know that AB¯¯¯¯¯¯¯¯||CD¯¯¯¯¯¯¯¯, AD¯¯¯¯¯¯¯¯||BC¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}||\overline{CD}, \ \overline{AD}||\overline{BC}\end{align*}, what angles are congruent? By which theorem?
2. Which side is congruent by the Reflexive Property?
3. Is this enough to say ADCCBA\begin{align*}\triangle ADC \cong \triangle CBA\end{align*}?
1. If we know that B\begin{align*}B\end{align*} is the midpoint of AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} and DE¯¯¯¯¯¯¯¯\begin{align*}\overline{DE}\end{align*}, what segments are congruent?
2. Are there any angles that are congruent by looking at the picture? Which ones and why?
3. Is this enough to say ABECBD\begin{align*}\triangle ABE \cong \triangle CBD\end{align*}?

## Know What?

The “ideal” measurements in a kitchen from the sink, refrigerator and oven are as close to an equilateral triangle as possible. Your parents are remodeling theirs to be as close to this as possible and the measurements are in the picture at the left, below. Your neighbor’s kitchen has the measurements on the right. Are the two triangles congruent? Why or why not?

## SSS Postulate of Triangle Congruence

Consider the question: If I have three lengths: 3 in, 4 in, and 5 in, can I construct more than one triangle?

Investigation 4-2: Constructing a Triangle Given Three Sides

Tools Needed: compass, pencil, ruler, and paper

1. Draw the longest side (5 in) horizontally, halfway down the page.

The drawings in this investigation are to scale.

2. Take the compass and, using the ruler, widen the compass to measure 4 in, the second side.

3. Using the measurement from Step 2, place the pointer of the compass on the left endpoint of the side drawn in Step 1. Draw an arc mark above the line segment.

4. Repeat Step 2 with the third measurement, 3 in. Then, like Step 3, place the pointer of the compass on the right endpoint of the side drawn in Step 1. Draw an arc mark above the line segment. Make sure it intersects the arc mark drawn in Step 3.

5. Draw lines from each endpoint to the arc intersections. These segments are the other two sides of the triangle.

An animation of this construction can be found at: http://www.mathsisfun.com/geometry/construct-ruler-compass-1.html

Can another triangle be drawn with these measurements that look different? NO. Only one triangle can be created from any given three lengths. You can rotate, flip, or move this triangle but it will still be the same size.

Side-Side-Side (SSS) Triangle Congruence Postulate: If 3 sides in one triangle are congruent to 3 sides in another triangle, then the triangles are congruent.

BC¯¯¯¯¯¯¯¯YZ¯¯¯¯¯¯¯, AB¯¯¯¯¯¯¯¯XY¯¯¯¯¯¯¯¯\begin{align*}\overline{BC} \cong \overline{YZ}, \ \overline{AB} \cong \overline{XY}\end{align*}, and AC¯¯¯¯¯¯¯¯XZ¯¯¯¯¯¯¯¯\begin{align*}\overline{AC} \cong \overline{XZ}\end{align*} then ABCXYZ\begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}.

The SSS Postulate is a shortcut. Before, you had to show 3 sides and 3 angles in one triangle were congruent to 3 sides and 3 angles in another triangle. Now you only have to show 3 sides in one triangle are congruent to 3 sides in another.

Example 1: Write a triangle congruence statement based on the picture below:

Solution: From the tic marks, we know AB¯¯¯¯¯¯¯¯LM¯¯¯¯¯¯¯¯¯, AC¯¯¯¯¯¯¯¯LK¯¯¯¯¯¯¯¯, BC¯¯¯¯¯¯¯¯MK¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{LM}, \ \overline{AC} \cong \overline{LK}, \ \overline{BC} \cong \overline{MK}\end{align*}. From the SSS Postulate, the triangles are congruent. Lining up the corresponding sides, we have ABCLMK\begin{align*}\triangle ABC \cong \triangle LMK\end{align*}.

Don’t forget ORDER MATTERS when writing congruence statements. Line up the sides with the same number of tic marks.

Example 2: Write a two-column proof to show that the two triangles are congruent.

Given: AB¯¯¯¯¯¯¯¯DE¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{DE}\end{align*}

C\begin{align*}C\end{align*} is the midpoint of AE¯¯¯¯¯¯¯¯\begin{align*}\overline{AE}\end{align*} and DB¯¯¯¯¯¯¯¯\begin{align*}\overline{DB}\end{align*}.

Prove: ACBECD\begin{align*}\triangle ACB \cong \triangle ECD\end{align*}

Solution:

Statement Reason

1. AB¯¯¯¯¯¯¯¯DE¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{DE}\end{align*}

C\begin{align*}C\end{align*} is the midpoint of AE¯¯¯¯¯¯¯¯\begin{align*}\overline{AE}\end{align*} and DB¯¯¯¯¯¯¯¯\begin{align*}\overline{DB}\end{align*}

Given
2. AC¯¯¯¯¯¯¯¯CE¯¯¯¯¯¯¯¯, BC¯¯¯¯¯¯¯¯CD¯¯¯¯¯¯¯¯\begin{align*}\overline{AC} \cong \overline{CE}, \ \overline{BC} \cong \overline{CD}\end{align*} Definition of a midpoint
3. ACBECD\begin{align*}\triangle ACB \cong \triangle ECD\end{align*} SSS Postulate

Prove Move: You must clearly state the three sets of sides are congruent BEFORE stating the triangles are congruent.

Prove Move: Mark the picture with the information you are given as well as information that you see in the picture (vertical angles, information from parallel lines, midpoints, angle bisectors, right angles). This information may be used in a proof.

## SAS Triangle Congruence Postulate

SAS refers to Side-Angle-Side. The placement of the word Angle is important because it indicates that the angle you are given is between the two sides.

Included Angle: When an angle is between two given sides of a polygon.

B\begin{align*}\angle{B}\end{align*} would be the included angle for sides AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} and BC¯¯¯¯¯¯¯¯\begin{align*}\overline{BC}\end{align*}.

Consider the question: If I have two sides of length 2 in and 5 in and the angle between them is 45\begin{align*}45^\circ\end{align*}, can I construct one triangle?

Investigation 4-3: Constructing a Triangle Given Two Sides and Included Angle

Tools Needed: protractor, pencil, ruler, and paper

1. Draw the longest side (5 in) horizontally, halfway down the page.

The drawings in this investigation are to scale.

2. At the left endpoint of your line segment, use the protractor to measure a 45\begin{align*}45^\circ\end{align*} angle. Mark this measurement.

3. Connect your mark from Step 2 with the left endpoint. Make your line 2 in long, the length of the second side.

4. Connect the two endpoints to draw the third side.

Can you draw another triangle, with these measurements that looks different? NO. Only one triangle can be created from any two lengths and the INCLUDED angle.

Side-Angle-Side (SAS) Triangle Congruence Postulate: If two sides and the included angle in one triangle are congruent to two sides and the included angle in another triangle, then the two triangles are congruent.

AC¯¯¯¯¯¯¯¯XZ¯¯¯¯¯¯¯¯, BC¯¯¯¯¯¯¯¯YZ¯¯¯¯¯¯¯\begin{align*}\overline{AC} \cong \overline{XZ}, \ \overline{BC} \cong \overline{YZ}\end{align*}, and CZ\begin{align*}\angle{C} \cong \angle{Z}\end{align*}, then ABCXYZ\begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}.

Example 3: What additional piece of information do you need to show that these two triangles are congruent using the SAS Postulate?

a) ABCLKM\begin{align*}\angle{ABC} \cong \angle{LKM}\end{align*}

b) AB¯¯¯¯¯¯¯¯LK¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{LK}\end{align*}

c) BC¯¯¯¯¯¯¯¯KM¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{BC} \cong \overline{KM}\end{align*}

d) BACKLM\begin{align*}\angle{BAC} \cong \angle{KLM}\end{align*}

Solution: For the SAS Postulate, you need the side on the other side of the angle. In ABC\begin{align*}\triangle ABC\end{align*}, that is BC¯¯¯¯¯¯¯¯\begin{align*}\overline{BC}\end{align*} and in LKM\begin{align*}\triangle LKM\end{align*} that is KM¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{KM}\end{align*}. The answer is c.

Example 4: Write a two-column proof to show that the two triangles are congruent.

Given: C\begin{align*}C\end{align*} is the midpoint of AE¯¯¯¯¯¯¯¯\begin{align*}\overline{AE}\end{align*} and DB¯¯¯¯¯¯¯¯\begin{align*}\overline{DB}\end{align*}

Prove: ACBECD\begin{align*}\triangle ACB \cong \triangle ECD\end{align*}

Solution:

Statement Reason
1. C\begin{align*}C\end{align*} is the midpoint of AE¯¯¯¯¯¯¯¯\begin{align*}\overline{AE}\end{align*} and DB¯¯¯¯¯¯¯¯\begin{align*}\overline{DB}\end{align*} Given
2. AC¯¯¯¯¯¯¯¯CE¯¯¯¯¯¯¯¯, BC¯¯¯¯¯¯¯¯CD¯¯¯¯¯¯¯¯\begin{align*}\overline{AC} \cong \overline{CE}, \ \overline{BC} \cong \overline{CD}\end{align*} Definition of a midpoint
3. ACBDCE\begin{align*}\angle{ACB} \cong \angle{DCE}\end{align*} Vertical Angles Postulate
4. ACBECD\begin{align*}\triangle ACB \cong \triangle ECD\end{align*} SAS Postulate

## SSS in the Coordinate Plane

The only way we will show two triangles are congruent in an \begin{align*}x-y\end{align*} plane is using SSS. To do this, you need to use the distance formula.

Example 5: Find the distances of all the line segments from both triangles to see if the two triangles are congruent.

Solution: Begin with \begin{align*}\triangle ABC\end{align*} and its sides.

\begin{align*}AB & = \sqrt{(-6-(-2))^2+(5-10)^2}\\ & = \sqrt{(-4)^2+(-5)^2}\\ & = \sqrt{16+25}\\ & = \sqrt{41}\end{align*}

\begin{align*}BC & = \sqrt{(-2-(-3))^2+(10-3)^2}\\ & = \sqrt{(1)^2+(7)^2}\\ & = \sqrt{1+49}\\ & = \sqrt{50} = 5 \sqrt{2}\end{align*}

\begin{align*}AC & = \sqrt{(-6-(-3))^2 + (5-3)^2}\\ & = \sqrt{(-3)^2+(2)^2}\\ & = \sqrt{9+4}\\ & = \sqrt{13}\end{align*}

Now, find the distances of all the sides in \begin{align*}\triangle DEF\end{align*}.

\begin{align*}DE & = \sqrt{(1-5)^2 + (-3-2)^2}\\ & = \sqrt{(-4)^2 + (-5)^2}\\ & = \sqrt{16+25}\\ & = \sqrt{41}\end{align*}

\begin{align*}EF & = \sqrt{(5-4)^2+(2-(-5))^2}\\ & = \sqrt{(1)^2+(7)^2}\\ & = \sqrt{1+49}\\ & = \sqrt{50} = 5\sqrt{2}\end{align*}

\begin{align*}DF & = \sqrt{(1-4)^2+(-3-(-5))^2}\\ & = \sqrt{(-3)^2+(2)^2}\\ & = \sqrt{9+4}\\ & = \sqrt{13}\end{align*}

\begin{align*}AB = DE, \ BC = EF\end{align*}, and \begin{align*}AC = DF\end{align*}, so two triangles are congruent by SSS.

Example 6: Determine if the two triangles are congruent.

Solution: Start with \begin{align*}\triangle ABC\end{align*}.

\begin{align*}AB & = \sqrt{(-2-(-8))^2+(-2-(-6))^2}\\ & = \sqrt{(6)^2+(4)^2}\\ & = \sqrt{36+16}\\ & = \sqrt{52} = 2\sqrt{13}\end{align*}

\begin{align*}BC & = \sqrt{(-8-(-6))^2+(-6-(-9))^2}\\ & = \sqrt{(-2)^2+(3)^2}\\ & = \sqrt{4+9}\\ & = \sqrt{13}\end{align*}

\begin{align*}AC & = \sqrt{(-2-(-6))^2+(-2-(-9))^2}\\ & = \sqrt{(4)^2+(7)^2}\\ & = \sqrt{16+49}\\ & = \sqrt{65}\end{align*}

Now find the sides of \begin{align*}\triangle DEF\end{align*}.

\begin{align*}DE & = \sqrt{(3-6)^2 + (9-4)^2}\\ & = \sqrt{(-3)^2 + (5)^2}\\ & = \sqrt{9+25}\\ & = \sqrt{34}\end{align*}

\begin{align*}EF & = \sqrt{(6-10)^2+(4-7)^2}\\ & = \sqrt{(-4)^2+(-3)^2}\\ & = \sqrt{16+9}\\ & = \sqrt{25} = 5\end{align*}

\begin{align*}DF & = \sqrt{(3-10)^2+(9-7)^2}\\ & = \sqrt{(-7)^2+(2)^2}\\ & = \sqrt{49+4}\\ & = \sqrt{53}\end{align*}No sides have equal measures, so the triangles are not congruent.

Know What? Revisited From what we have learned in this section, the two triangles are not congruent because the distance from the fridge to the stove in your house is 4 feet and in your neighbor’s it is 4.5 ft. The SSS Postulate tells us that all three sides have to be congruent in order for the triangles to be congruent.

## Review Questions

• Questions 1-10 are similar to Example 1.
• Questions 11-16 are similar to Example 3.
• Questions 17-23 are similar to Examples 2 and 4.
• Questions 24-27 are similar to Examples 5 and 6.

Are the pairs of triangles congruent? If so, write the congruence statement and why.

State the additional piece of information needed to show that each pair of triangles is congruent.

1. Use SAS
2. Use SSS
3. Use SAS
4. Use SAS
5. Use SSS
6. Use SAS

Fill in the blanks in the proofs below.

1. Given: \begin{align*}\overline{AB} \cong \overline{DC}, \ \overline{BE} \cong \overline{CE}\end{align*} Prove: \begin{align*}\triangle ABE \cong \triangle ACE\end{align*}
Statement Reason
1. 1.
2. \begin{align*}\angle{AEB} \cong \angle{DEC}\end{align*} 2.
3. \begin{align*}\triangle ABE \cong \triangle ACE\end{align*} 3.
1. Given: \begin{align*}\overline{AB} \cong \overline{DC}, \ \overline{AC} \cong \overline{DB}\end{align*} Prove: \begin{align*}\triangle ABC \cong \triangle DCB\end{align*}
Statement Reason
1. 1.
2. 2. Reflexive PoC
3. \begin{align*}\triangle ABC \cong \triangle DCB\end{align*} 3.
1. Given: \begin{align*}B\end{align*} is a midpoint of \begin{align*}\overline{DC}\end{align*} \begin{align*}\overline{AB} \perp \overline{DC}\end{align*} Prove: \begin{align*}\triangle ABD \cong \triangle ABC\end{align*}
Statement Reason
1. \begin{align*}B\end{align*} is a midpoint of \begin{align*}\overline{DC}, \overline{AB} \perp \overline{DC}\end{align*} 1.
2. 2. Definition of a midpoint
3. \begin{align*}\angle{ABD}\end{align*} and \begin{align*}\angle{ABC}\end{align*} are right angles 3.
4. 4. All right angles are \begin{align*}\cong\end{align*}
5. 5.
6. \begin{align*}\triangle{ABD} \cong \triangle{ABC}\end{align*} 6.
1. Given: \begin{align*}\overline{AB}\end{align*} is an angle bisector of \begin{align*}\angle{DAC}\end{align*} \begin{align*}\overline{AD} \cong \overline{AC}\end{align*} Prove: \begin{align*}\triangle ABD \cong \triangle ABC\end{align*}
Statement Reason
1.
2. \begin{align*}\angle{DAB} \cong \angle{BAC}\end{align*}
3. Reflexive PoC
4. \begin{align*}\triangle ABD \cong \triangle ABC\end{align*}
1. Given: \begin{align*}B\end{align*} is the midpoint of \begin{align*}\overline{DC}\end{align*} \begin{align*}\overline{AD} \cong \overline{AC}\end{align*} Prove: \begin{align*}\triangle ABD \cong \triangle ABC\end{align*}
Statement Reason
1.
2. Definition of a Midpoint
3. Reflexive PoC
4. \begin{align*}\triangle ABD \cong \triangle ABC\end{align*}
1. Given: \begin{align*}B\end{align*} is the midpoint of \begin{align*}\overline{DE}\end{align*} and \begin{align*}\overline{AC}\end{align*} \begin{align*}\angle{ABE}\end{align*} is a right angle Prove: \begin{align*}\triangle ABE \cong \triangle CBD\end{align*}
Statement Reason
1. Given
2. \begin{align*}\overline{DB} \cong \overline{BE}, \ \overline{AB} \cong \overline{BC}\end{align*}
3. Definition of a Right Angle
4. Vertical Angle Theorem
5. \begin{align*}\triangle ABE \cong \triangle CBD\end{align*}
1. Given: \begin{align*}\overline{DB}\end{align*} is the angle bisector of \begin{align*}\angle{ADC}\end{align*} \begin{align*}\overline{AD} \cong \overline{DC}\end{align*} Prove: \begin{align*}\triangle ABD \cong \triangle CBD\end{align*}
Statement Reason
1.
2. \begin{align*}\angle{ADB} \cong \angle{BDC}\end{align*}
3.
4. \begin{align*}\triangle ABD \cong \triangle CBD\end{align*}

Find the lengths of the sides of each triangle to see if the two triangles are congruent. Leave your answers under the radical.

1. \begin{align*}\triangle ABC: \ A(-1, 5), \ B(-4, 2), \ C(2, -2)\end{align*} and \begin{align*}\triangle DEF: \ D(7, -5), \ E(4, 2), \ F(8, -9)\end{align*}
2. \begin{align*}\triangle ABC: \ A(-8, -3), \ B(-2, -4), \ C(-5, -9)\end{align*} and \begin{align*}\triangle DEF: \ D(-7, 2), \ E(-1, 3), \ F(-4, 8)\end{align*}

1. \begin{align*}\sqrt{74}\end{align*}
2. \begin{align*}\sqrt{538}\end{align*}
1. \begin{align*}\angle BAC \cong \angle DCA, \angle DAC \cong \angle BCA\end{align*} by the Alternate Interior Angles Theorem.
2. \begin{align*}\overline{AC} \cong \overline{AC}\end{align*}
3. Not yet, this would be ASA.
1. \begin{align*}\overline{DB} \cong \overline{BE}, \overline{AB} \cong \overline{BC}\end{align*}
2. \begin{align*}\angle DBC \cong \angle ABE\end{align*} by the Vertical Angles Theorem.
3. By the end of this section, yes, we will be able to show that these two triangles are congruent by SAS.

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Subjects:
8 , 9 , 10
Date Created:
Feb 22, 2012