4.3: Triangle Congruence using SSS and SAS
Learning Objectives
 Use the distance formula to analyze triangles on the \begin{align*}xy\end{align*} plane.
 Apply the SSS Postulate to prove two triangles are congruent.
 Apply the SAS Postulate to prove two triangles are congruent.
Review Queue
 Determine the distance between the two points.
 (1, 5) and (4, 12)
 (6, 15) and (3, 8)
 Are the two triangles congruent? Explain why or why not.
 \begin{align*}\overline{AB} \  \ \overline{CD}, \overline{AD} \  \ \overline{BC}\!\\ \overline{AB} \cong \overline{CD}, \overline{AD} \cong \overline{BC}\end{align*}
 \begin{align*}B\end{align*} is the midpoint of \begin{align*}\overline{AC}\end{align*} and \begin{align*}\overline{DE}\end{align*}
 At this point in time, how many angles and sides do we have to know are congruent in order to say that two triangles are congruent?
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 with these measurements? In other words, can I construct two different triangles with these same three lengths?
Investigation 42: Constructing a Triangle Given Three Sides
Tools Needed: compass, pencil, ruler, and paper
 Draw the longest side (5 in) horizontally, halfway down the page. The drawings in this investigation are to scale.
 Take the compass and, using the ruler, widen the compass to measure 4 in, the next side.
 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.
 Repeat Step 2 with the last measurement, 3 in. Then, 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.
 Draw lines from each endpoint to the arc intersections. These lines will be the other two sides of the triangle.
Can you draw another triangle, with these measurements that looks different? The answer is NO. Only one triangle can be created from any given three lengths.
An animation of this investigation can be found at: http://www.mathsisfun.com/geometry/constructrulercompass1.html
SideSideSide (SSS) Triangle Congruence Postulate: If three sides in one triangle are congruent to three sides in another triangle, then the triangles are congruent.
Now, we only need to show that all three sides in a triangle are congruent to the three sides in another triangle. This is a postulate so we accept it as true without proof.
Think of the SSS Postulate as a shortcut. You no longer have to show 3 sets of angles are congruent and 3 sets of sides are congruent in order to say that the two triangles are congruent.
Example 1: Write a triangle congruence statement based on the diagram below:
Solution: From the tic marks, we know \begin{align*}\overline{AB} \cong \overline{LM}, \overline{AC} \cong \overline{LK}, \overline{BC} \cong \overline{MK}\end{align*}. Using the SSS Postulate we know the two triangles are congruent. Lining up the corresponding sides, we have \begin{align*}\triangle ABC \cong \triangle LMK\end{align*}.
Don’t forget ORDER MATTERS when writing triangle congruence statements. Here, we lined up the sides with one tic mark, then the sides with two tic marks, and finally the sides with three tic marks.
Example 2: Write a twocolumn proof to show that the two triangles are congruent.
Given: \begin{align*}\overline{AB} \cong \overline{DE}\end{align*}
\begin{align*}C\end{align*} is the midpoint of \begin{align*}\overline{AE}\end{align*} and \begin{align*}\overline{DB}\end{align*}.
Prove: \begin{align*}\triangle ACB \cong \triangle ECD\end{align*}
Solution:
Statement  Reason 

1. \begin{align*}\overline{AB} \cong \overline{DE}\end{align*} \begin{align*}C\end{align*} is the midpoint of \begin{align*}\overline{AE}\end{align*} and \begin{align*}\overline{DB}\end{align*} 
Given 
2. \begin{align*}\overline{AC} \cong \overline{CE}, \overline{BC} \cong \overline{CD}\end{align*}  Definition of a midpoint 
3. \begin{align*}\triangle ACB \cong \triangle ECD\end{align*}  SSS Postulate 
Make sure that you clearly state the three sets of congruent sides BEFORE stating that the triangles are congruent.
Prove Move: Feel free to mark the picture with the information you are given as well as information that you can infer (vertical angles, information from parallel lines, midpoints, angle bisectors, right angles).
SAS Triangle Congruence Postulate
First, it is important to note that SAS refers to SideAngleSide. The placement of the word Angle is important because it indicates that the angle that you are given is between the two sides.
Included Angle: When an angle is between two given sides of a triangle (or polygon).
In the picture to the left, the markings indicate that \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{BC}\end{align*} are the given sides, so \begin{align*}\angle B\end{align*} would be the included angle.
Consider the question: If I have two sides of length 2 in and 5 in and the angle between them is \begin{align*}45^\circ\end{align*}, can I construct only one triangle?
Investigation 43: Constructing a Triangle Given Two Sides and Included Angle Tools Needed: protractor, pencil, ruler, and paper
 Draw the longest side (5 in) horizontally, halfway down the page. The drawings in this investigation are to scale.
 At the left endpoint of your line segment, use the protractor to measure a \begin{align*}45^\circ\end{align*} angle. Mark this measurement.
 Connect your mark from Step 2 with the left endpoint. Make your line 2 in long, the length of the second side.
 Connect the two endpoints by drawing the third side.
Can you draw another triangle, with these measurements that looks different? The answer is NO. Only one triangle can be created from any given two lengths and the INCLUDED angle.
SideAngleSide (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.
The markings in the picture are enough to say that \begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}.
So, in addition to SSS congruence, we now have SAS. Both of these postulates can be used to say that two triangles are congruent. When doing proofs, you might be able to use either SSS or SAS to prove that two triangles are congruent. There is no set way to complete a proof, so when faced with the choice to use SSS or SAS, it does not matter. Either would be correct.
Example 3: What additional piece of information would you need to prove that these two triangles are congruent using the SAS Postulate?
a) \begin{align*}\angle ABC \cong \angle LKM\end{align*}
b) \begin{align*}\overline{AB} \cong \overline{LK}\end{align*}
c) \begin{align*}\overline{BC} \cong \overline{KM}\end{align*}
d) \begin{align*}\angle BAC \cong \angle KLM\end{align*}
Solution: For the SAS Postulate, you need two sides and the included angle in both triangles. So, you need the side on the other side of the angle. In \begin{align*}\triangle ABC\end{align*}, that is \begin{align*}\overline{BC}\end{align*} and in \begin{align*}\triangle LKM\end{align*} that is \begin{align*}\overline{KM}\end{align*}. The correct answer is c.
Example 4: Write a twocolumn proof to show that the two triangles are congruent.
Given: \begin{align*}C\end{align*} is the midpoint of \begin{align*}\overline{AE}\end{align*} and \begin{align*}\overline{DB}\end{align*}
Prove: \begin{align*}\triangle ACB \cong \triangle ECD\end{align*}
Solution:
Statement  Reason 

1. \begin{align*}C\end{align*} is the midpoint of \begin{align*}\overline{AE}\end{align*} and \begin{align*}\overline{DB}\end{align*}  Given 
2. \begin{align*}\overline{AC} \cong \overline{CE}, \overline{BC} \cong \overline{CD}\end{align*}  Definition of a midpoint 
3. \begin{align*}\angle ACB \cong \angle DCE\end{align*}  Vertical Angles Postulate 
4. \begin{align*}\triangle ACB \cong \triangle ECD\end{align*}  SAS Postulate 
In Example 4, we could have only proven the two triangles congruent by SAS. If we were given that \begin{align*}\overline{AB} \cong \overline{DE}\end{align*}, then we could have also proven the two triangles congruent by SSS.
SSS in the Coordinate Plane
In the coordinate plane, the easiest way to show two triangles are congruent is to find the lengths of the 3 sides in each triangle. Finding the measure of an angle in the coordinate plane can be a little tricky, so we will avoid it in this text. Therefore, you will only need to apply SSS in the coordinate plane. To find the lengths of the sides, you will need to use the distance formula, \begin{align*}\sqrt{(x_2x_1)^2+(y_2y_1)^2}\end{align*}.
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+(510)^2}\\ &= \sqrt{(4)^2+(5)^2}\\ &= \sqrt{16+25}\\ &= \sqrt{41}\end{align*}
\begin{align*}BC &= \sqrt{(2(3))^2+(103)^2}\\ &= \sqrt{(1)^2+(7)^2}\\ &= \sqrt{1+49}\\ &= \sqrt{50}=5\sqrt{2}\end{align*}
\begin{align*}AC &= \sqrt{(6(3))^2+(53)^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{(15)^2+(32)^2}\\ &= \sqrt{(4)^2+(5)^2}\\ &= \sqrt{16+25}\\ &= \sqrt{41}\end{align*}
\begin{align*}EF &= \sqrt{(54)^2+(2(5))^2}\\ &= \sqrt{(1)^2+(7)^2}\\ &= \sqrt{1+49}\\ &= \sqrt{50}=5\sqrt{2}\end{align*}
\begin{align*}DF &= \sqrt{(14)^2+(3(5))^2}\\ &= \sqrt{(3)^2+(2)^2}\\ &= \sqrt{9+4}\\ &= \sqrt{13}\end{align*}
We see that \begin{align*}AB = DE, BC = EF\end{align*}, and \begin{align*}AC = DF\end{align*}. Recall that if two lengths are equal, then they are also congruent. Therefore, \begin{align*}\overline{AB} \cong \overline{DE}, \overline{BC} \cong \overline{EF}\end{align*}, and \begin{align*}\overline{AC} \cong \overline{DF}\end{align*}. Because the corresponding sides are congruent, we can say that \begin{align*}\triangle ABC \cong \triangle DEF\end{align*} by SSS.
Example 6: Determine if the two triangles are congruent.
Solution: Use the distance formula to find all the lengths. 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{(36)^2+(94)^2}\\ &= \sqrt{(3)^2+(5)^2}\\ &= \sqrt{9+25}\\ &= \sqrt{34}\end{align*}
\begin{align*}EF &= \sqrt{(610)^2+(47)^2}\\ &= \sqrt{(4)^2+(3)^2}\\ &= \sqrt{16+9}\\ &= \sqrt{25}=5\end{align*}
\begin{align*}DF &= \sqrt{(310)^2+(97)^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.
Review Questions
Are the pairs of triangles congruent? If so, write the congruence statement.
State the additional piece of information needed to show that each pair of triangles are congruent.
 Use SAS
 Use SSS
 Use SAS
 Use SAS
 Use SSS
 Use SAS
Fill in the blanks in the proofs below.
 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. 
 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. 
 Given: \begin{align*}B\end{align*} is a midpoint of \begin{align*}\overline{DC}\end{align*} \begin{align*}\overline {AB} \bot \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} \bot \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. 
Write a twocolumn proof for the given information below.
 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*}
 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*}
 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*}
 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*}
Determine if the two triangles are congruent, using the distance formula. Leave all of your answers in simplest radical form (simplify all radicals, no decimals).
 \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*}
 \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*}
Constructions
 Construct a triangle with sides of length 5cm, 3cm, 2cm.
 Copy the triangle below using a straightedge and compass.
 Use the two sides and the given angle to construct \begin{align*}\triangle ABC\end{align*}.
 Use the two sides and the given angle to construct \begin{align*}\triangle ABC\end{align*}.
 Was the information given in problem 29 in SAS order? If not, your triangle may not be the only triangle that you can construct using the given information. Construct the second possible triangle.
Review Queue Answers

 \begin{align*}\sqrt{74}\end{align*}
 \begin{align*}\sqrt{538}\end{align*}
 Yes, \begin{align*}\triangle CAD \cong \triangle ACB\end{align*} because \begin{align*}\angle CAD \cong \angle ACB\end{align*} and \begin{align*}\angle BAC \cong \angle ACD\end{align*} by Alternate Interior Angles. \begin{align*}\overline{AC} \cong \overline{AC}\end{align*} by the Reflexive PoC and \begin{align*}\angle ADC \cong \angle ABC\end{align*} by the \begin{align*}3^{rd}\end{align*} Angle Theorem.
 At this point in time, we do not have enough information to show that the two triangles are congruent. We know that \begin{align*}\overline{AB} \cong \overline{BC}\end{align*} and \begin{align*}\overline{DB} \cong \overline{BE}\end{align*} from the definition of a midpoint. By vertical angles, we know that \begin{align*}\angle DBC \cong \angle ABE\end{align*}. This is only two sides and one pair of angles; not enough info, yet.
 We need to know three pairs of congruent sides and two pairs of congruent angles. From this, we can assume the third pair of angles are congruent from the \begin{align*}3^{rd}\end{align*} Angle Theorem.
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