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{{ printedBook.courseTrack.name }} {{ printedBook.name }} Two triangles are congruent if their corresponding sides and angles are congruent. However, there could be cases where not all side lengths or angle measures are known. The good news is that congruence can still be verified depending on which parts are known.
### Catch-Up and Review

The primary goal of this lesson is to investigate exactly how much information about a pair of triangles has to be known in order to conclude that the triangles are congruent.

**Here are a few recommended readings before getting started with this lesson.**

Consider the following statement.

After exploring different cases, what can be said about the given statement?

In the applet, rigid motions can be applied only on $△ABC.$

- To translate $△ABC,$ select its interior region and slide.
- Point $P$ acts as the center of rotation and can be moved by dragging it.
- To rotate $△ABC$ about $P,$ click on any vertex of $△ABC$ and drag it.
- The given line acts as a line of reflection which can be moved by dragging it. Similarly, its inclination can be changed by dragging either of its two points.
- To reflect $△ABC$ across the line of reflection, push the
Reflect

button.

In the previous exploration, it was seen that a pair of triangles can have corresponding congruent angles but not be congruent triangles. Therefore, relying only on the relationship of only angles is not a valid criterion.

Angle-Angle-Angle is $not$ a valid criterion for proving triangle congruence.

Once the two triangles have been drawn, find the side lengths and angle measures of each triangle. Can any relationship between the triangles be found? Repeat the process a few times to see if the relationship remains true.

The previous exploration suggests that two triangles are congruent whenever they have two pairs of corresponding congruent sides and the corresponding included angles are congruent. In fact, this conclusion is formalized in the Side-Angle-Side Congruence Theorem

If two sides and the included angle of a triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent.

Based on the diagram above, the theorem can be written as follows.

$⎩⎪⎪⎨⎪⎪⎧ AB≅DE∠A≅∠DAC≅DF ⇒△ABC≅△DEF $

This proof will be developed based on the given diagram, but it is valid for any pair of triangles.

The primary purpose is finding a rigid motion or sequence of rigid motions that maps one triangle onto the other. This can be done in several ways. One of the ways will be shown here.Translate $△DEF$ So That Two Corresponding Vertices Match

Since the image of the translation does not match $△ABC,$ at least one more transformation is needed.

Rotate $△AE_{′}F_{′}$ So That Two Corresponding Sides Match

As before, the image does not match $△ABC.$ Therefore, a third rigid motion is required.

Reflect $△ABF_{′′}$ So That Two More Corresponding Sides Match

This time the image matches $△ABC.$

In the following diagram, triangles $ADE$ and $BCE$ are congruent, and $∠ADC$ is congruent to $∠BCD.$

How many more pairs of congruent triangles are there in the diagram? Name each congruent triangle pair.

There are two more pairs of congruent triangles. $△ADC△ABD ≅△BCD≅△BAC $

Remember, if two triangles are congruent, then their corresponding sides and angles are congruent.

Start by highlighting the given pair of congruent triangles, $△ADE$ and $△BCE.$

Since these triangles are congruent, their corresponding parts are congruent. This implies that $AD$ is congruent to $BC.$

Notice that $CD$ is a common side for triangles $ADC$ and $BCD.$ Because of the Reflexive Property of Congruence, $CD$ is congruent to itself. Next, list the corresponding congruent parts between these two triangles.
$AD≅BC∠ADC≅∠BCDDC≅CD SideAngleSide $
By the Side-Angle-Side (SAS) Congruence Theorem, it can be concluded that $△ADC$ and $△BCD$ are congruent.

$△ADC≅△BCD$

Below, the corresponding congruent parts between $△ABD$ and $△BAC$ are listed.
$AD≅BC∠ADB≅∠BCADB≅CA SideAngleSide $
One more time, the Side-Angle-Side (SAS) Congruence Theorem can be used to conclude that triangles $ABD$ and $BAC$ are congruent.

$△ABD≅△BAC$

The last two triangles to consider are triangles $ABE$ and $DEC.$ Unlike the first two pairs, these dimensions seem to be quite different. Therefore, it can be concluded that they are not congruent.

Consequently, in the initial diagram, there are two more pairs of congruent triangles in addition to the given one.

Use segment $AB$ and the rays $AX$ and $BY$ to construct two different triangles, one at a time, in such a way that the following conditions are met.

- The angle formed at $A$ has the same measure in both triangles.
- The angle formed at $B$ has the same measure in both triangles.

Once the two triangles are drawn, find the side lengths and angle measures of each triangle. Is there any relationship between the triangles? Repeat the process a few times to see if the relationship remains true.

The following statement could be seen in the previous applet. When two triangles have two pairs of corresponding congruent angles, and the included corresponding sides are congruent, the triangles are then congruent. That leads to the second criteria for triangle congruence.

If two angles and the included side of a triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.

Based on the diagram above, the theorem can be written as follows.

$⎩⎪⎪⎨⎪⎪⎧ ∠A≅∠DAB≅DE∠B≅∠E ⇒△ABC≅△DEF $

This proof will be developed based on the given diagram, but it is valid for any pair of triangles.

The goal of the proof is to find a rigid motion or sequence of rigid motions that maps one triangle onto the other. This can be done in several ways. One of the ways will be shown here.Translate $△DEF$ So That Two Corresponding Vertices Match

Since the image of the translation does not match $△ABC,$ at least one more transformation is needed.

Rotate $△AE_{′}F_{′}$ So That Two Corresponding Sides Match

As before, the image does not match $△ABC.$ Therefore, a third rigid motion is required.

Reflect $△ABF_{′′}$ So That All Corresponding Sides Match

This time the image matches $△ABC.$

Consider the following diagram.

What is the value of $x+y+z?${"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["9.75"]}}

Take note that $QS $ is a common side for two triangles. Use the fact that if two triangles are congruent, their corresponding sides and angles are congruent.

Notice that $QS $ is a common side for triangles $PQS$ and $RSQ.$
Then, the value of $y$ can be found by substituting $x=1.5$ into the Equation (II) and solving the resulting equation for $y.$
Finally, the required sum can be calculated by substituting the values found for $x,$ $y,$ and $z.$

By the Reflexive Property of Congruence, $QS $ is congruent to itself. Additionally, $∠PQS$ and $∠RSQ$ are congruent, as are $∠PSQ$ and $∠RQS.$
$∠PQS≅∠RSQQS ≅QS ∠PSQ≅∠RQS AngleSideAngle $
Consequently, $△PQS$ and $△RSQ$ are congruent because of the Angle-Side-Angle (ASA) Congruence Theorem. Therefore, the corresponding sides and angles are congruent.
$△PQS≅△RSQ⇒⎩⎪⎪⎨⎪⎪⎧ ∠PPQ PS ≅∠R≅RS≅RQ $
By definition, congruent angles have the same measure, and congruent segments have the same length. Therefore, the congruence statements on the right-hand side support the formation of the following three equations.
$⎩⎪⎪⎨⎪⎪⎧ 8z+2=506=2y+x5x−3=4.5 (I)(II)(III) $
By solving Equation (I), the value of $z$ can be found.
Next, solve Equation (III) to find the value of $x.$

$5x−3=4.5$

$x=1.5$

$6=2y+x$

Substitute

$x=1.5$

$6=2y+1.5$

Solve for $y$

SubEqn

$LHS−1.5=RHS−1.5$

$4.5=2y$

DivEqn

$LHS/2=RHS/2$

$24.5 =y$

UseCalc

Use a calculator

$2.25=y$

RearrangeEqn

Rearrange equation

$y=2.25$

At the beginning of the lesson, it was shown that the Angle-Angle-Angle is not a valid criterion for determining triangle congruence. Next, using the following applet, it will be investigated if the Side-Side-Side is a valid criterion.

Use segments $AB,$ $AC,$ and $BC$ to construct two different triangles. Construct the triangles one at a time. Once the two triangles are drawn, find the angle measures of each triangle. Is there any relationship between the triangles? Repeat the process a couple of times to see if the relationship holds true.

As seen in the previous exploration, the Side-Side-Side is a valid criterion for checking triangle congruence.

If the three sides of a triangle are congruent to the three sides of another triangle, then the triangles are congruent.

Based on the diagram above, the theorem can be written as follows.

$⎩⎪⎪⎨⎪⎪⎧ AB≅DEBC≅EFAC≅DF ⇒△ABC≅△DEF $

This proof will be developed based on the given diagram, but it is valid for any pair of triangles.

The primary purpose is finding a rigid motion or sequence of rigid motions that maps one triangle onto the other. This can be done in several ways. One of them will be shown here.Translate $△DEF$ So That Two Corresponding Vertices Match

Since the image of the translation does not match $△ABC,$ at least one more transformation is needed.

Rotate $△AE_{′}F_{′}$ So That Two Corresponding Sides Match

As before, the image does not match $△ABC.$ Therefore, a third rigid motion is required.

Reflect $△ABF_{′′}$ So That Two More Corresponding Sides Match

The points $C$ and $F_{′′}$ are on opposite sides of $AB.$ Now, consider $CF_{′}.$ Let $G$ denote the point of intersection between $AB$ and $CF_{′′}.$

It can be noted that $AC=AF_{′′}$ and $BC=BF_{′′}.$ By the Converse Perpendicular Bisector Theorem, $AB$ is a perpendicular bisector of $CF_{′′}.$ Points along the perpendicular bisector are equidistant from the endpoints of the segment, so $CG=GF_{′′}.$

Finally, $F_{′′}$ can be mapped onto $C$ by a reflection across $AB$ by reflecting $△ABF_{′′}$ across $AB.$ Because reflections preserve angles, $AF_{′′}$ and $BF_{′′}$ are mapped onto $AC$ and $BC,$ respectively.
This time the image matches $△ABC.$

Given three random segments, it is not always possible to construct a triangle. But, when possible, this triangle will be unique. This fact implies that the angle measures of that triangle are also unique.

Therefore, any other triangle with the same side lengths will also have the same angle measures. Consequently, the two triangles are congruent. Notice that, in contrast, having the same angle measures does notIn the following diagram, $R_{1}$ is a rectangle, $S_{1}$ and $S_{2}$ are squares, $T_{1},$ $T_{2},$ and $△MKL$ are isosceles triangles, $DK$ is congruent to $CL,$ and $GM$ is congruent to $JM.$

If $m∠JCK=3z+8,$ what is the value of $x?${"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["70"]}}

Using the Segment Addition Postulate and the Side-Side-Side (SSS) Congruence Theorem, prove that $△DGL$ is congruent to $△CJK.$ Then, find the measure of $∠JCK.$ Use the fact that $m∠JCK+m∠DCB+m∠JCH+x=360_{∘}.$

To find the value of $x,$ notice that $x,$ $m∠DCB,$ $m∠JCK,$ and $m∠JCH$ add up $360_{∘}.$
$x+m∠DCB+m∠JCK+m∠JCH=360_{∘} $
Since $R_{1}$ is a rectangle and $S_{2}$ is a square, $∠DCB$ and $∠JCH$ are right angles. Therefore, these angles have a measure of $90_{∘}$ each. Also, it is given that $m∠JCK=3z+8.$ Use this information to solve for $x.$
An expression of $x$ was found in terms of $z.$ Therefore, to find the value of $x,$ the value of $z$ should first be known. ### Finding the Value of $z$

#### Proving that $DL≅CK$

Notice that $KL$ is common to both triangles. By using the Segment Addition Postulate, the following pair of equations can be written.
${DL=DK+KLCK=CL+LK (I)(II) $
Since $DK$ is congruent to $CL,$ these segments have the same length, that is, $DK=CL.$ Simlarly, $KL=LK.$ By substituting these expressions into Equation (I), a relation between $DL$ and $CK$ will be obtained.
This equation implies that $DL$ and $CK$ are congruent.
#### Proving that $GL≅JK$

Once more, the Segment Addition Postulate can be used to rewrite $GL$ and $JK.$
${GL=GM+MLJK=JM+MK (I)(II) $
Because $△MKL$ is isosceles, $MK$ is congruent to $ML.$ Therefore, $ML=MK.$ Keep in mind that it is given that $GM$ and $JM$ are congruent. That means $GM=JM.$ These two expressions can be substituted into Equation (I).
Based on the equation just obtained, it can be concluded that $GL$ and $JK$ are congruent.
#### Proving that $DG≅CJ$

#### Proving that $△DGL≅△CJK$

Previously, the following three congruence statements were obtained.
$DL≅CKGL≅JKDG≅CJ SideSideSide $
The Side-Side-Side (SSS) Congruence Theorem allows to conclude that $△DGL$ is congruent to $△CJK.$
### Finding the Value of $x$

Finally, to find the value of $x,$ substitute $z=34$ into the equation $x=172−3z.$ Then solve for $x.$

If $△DGL$ and $△CJK$ can be proven to be congruent, that would provide the needed information to find the value of $z.$ Therefore, focus on those two triangles.

$DL=DK+KL$

Substitute values and simplify

$DL=CK$

$DL≅CK$

$GL=GM+ML$

Substitute values and simplify

$GL=JK$

$GL≅JK$

Since $R_{1}$ is a rectangle, $S_{1}$ and $S_{2}$ are squares, and $T_{1}$ and $T_{2}$ are isosceles triangles, the following consequences can be drawn.

Given | Consequence |
---|---|

$R_{1}$ is a rectangle | $DA≅CB$ |

$S_{1}$ is a square | $DG≅DE$ |

$S_{2}$ is a square | $CH≅CJ$ |

$T_{1}$ is an isosceles triangle | $DE≅DA$ |

$T_{2}$ is an isosceles triangle | $CB≅CH$ |

Next, organize the information in the right-hand column in a flow chart and use the Transitive Property of Congruence to prove that $DG≅CJ.$

Since corresponding parts of congruent triangles are congruent, it can be concluded that $∠DGL$ is congruent to $∠CJK.$ Therefore, $z=34.$

Notice that the ASA criterion requires the congruent sides to be included between the two pairs of corresponding congruent angles. Using the following applet, investigate what happens when the congruent sides are not the included sides.

Use segment $AB$ and the rays $AX$ and $BY$ to construct two different triangles, one at a time, in such a way that these conditions are met:

- The angle formed at $A$ has the same measure in both triangles.
- The angle formed at $C,$ the intersection of the rays, has the same measure in both triangles.

Once the two triangles are drawn, find the side lengths and angle measures of each triangle. Is there any relationship between the triangles? Repeat the process a couple of times to see if the relationship holds true.

As seen in the previous exploration, the Angle-Angle-Side condition is a valid criterion for triangle congruence.

If two angles and a non-included side of a triangle are congruent to two angles and the corresponding non-included side of another triangle, then the triangles are congruent.

Based on the diagram above, the theorem can be written as follows.

$⎩⎪⎪⎨⎪⎪⎧ ∠A≅∠D∠B≅∠EBC≅EF ⇒△ABC≅△DEF $

This proof will be developed based on the given diagram, but it is valid for any pair of triangles.

The primary purpose is finding a rigid motion or sequence of rigid motions that maps one triangle onto the other. This can be done in several ways. One of the ways will be shown here.Translate $△DEF$ so that two corresponding vertices match

Rotate $△CD_{′}E_{′}$ so that two corresponding sides match

As before, the image does not match $△ABC.$ Therefore, a third rigid motion is required.

Reflect $△ABF_{′′}$ so that two more corresponding sides match

It is given that two angles of $△ABC$ are congruent to two angles of $△BCD_{′′}.$ Hence, by the Third Angle Theorem, $∠BCD_{′′}$ is congruent to $∠BCA.$

Reflect $△CBD_{′′}$ across $BC.$ Because reflections preserve angles, $BD_{′′}$ and $CD_{′′}$ are mapped onto $BA$ and $CA,$ respectively. Then, the point of intersection of the original rays $D_{′′}$ is mapped onto the point of intersection of the image rays $A.$
This time the image matches $△ABC.$

Dylan bought a new boomerang to play with his friends next summer. In the drawing printed on the boomerang, $∠A$ and $∠C$ are congruent, and $BF$ and $BE$ are congruent.

Show that $AE$ is congruent to $CF.$

See solution.

Separate triangles $ABE$ and $CBF$ and notice they have a common angle. Then, use the Angle-Angle-Side (AAS) Congruence Theorem.

Start by separating $△ABE$ and $△CBF$ from the design.

Notice that $∠B$ is common to both triangles. By the Reflexive Property of Congruence, $∠B$ is congruent to itself. Also, it is given that $∠A$ is congruent to $∠C,$ and $BF$ is congruent to $BE.$
$∠A≅∠C∠B≅∠BBF≅BE AngleAngleSide $
Applying the Angle-Angle-Side (AAS) Congruence Theorem, it is obtained that $△ABE$ is congruent to $△CBF.$ Consequently, their corresponding parts are congruent, which means that $AE$ is congruent to $CF.$

With the help of the following applet, investigate if the Side-Side-Angle is a valid criterion for determining triangle congruence.

Use segments $AB$ and $AC$ to construct two different triangles in such a way that the angle formed at $B$ has the same measure in both triangles. Once the two triangles are drawn, find the side lengths and angle measures of each triangle. Are the triangles congruent in all cases?

With the previous applet, it can be checked that, in general, the Side-Side-Angle is not a valid criterion to determine triangle congruence. For instance, the following triangles meet the conditions of this criterion, and they are not congruent.

However, this criteria is valid in the particular case that both triangles are right triangles.

If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the two triangles are congruent.

Based on the diagram, the following relations hold true.

$⎩⎪⎪⎪⎨⎪⎪⎪⎧ m∠A=90_{∘}m∠D=90_{∘}AB≅DEBC≅EF ⇒△ABC≅△DEF $

Consider $△ABC$ and $△DEF$ shown below.

By applying the Pythagorean Theorem in each triangle, the following equations can be written. ${c_{2}=a_{2}+b_{1}c_{2}=a_{2}+b_{2} (I)(II) $ The expression on the right hand-side of the first equation can be substituted into the second equation. Then, a relation between $b_{1}$ and $b_{2}$ can be found.${c_{2}=a_{2}+b_{1}c_{2}=a_{2}+b_{2} (I)(II) $

Substitute

$(II):$ $c_{2}=a_{2}+b_{1}$

${c_{2}=a_{2}+b_{1}a_{2}+b_{1}=a_{2}+b_{2} $

${c_{2}=a_{2}+b_{1}∣b_{1}∣=∣b_{2}∣ $

Therefore, by the Side-Side-Side Congruence Theorem, the triangles are congruent.

$△ABC≅△DEF$

In the following chart, all the criteria for triangle congruence seen in the lesson are listed.

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