Before any theorems will be introduced, try to discover some properties of angles using the interactive applet. While exploring, think about how those properties could be proven. The applet allows for translations and rotations of the angles. Consider a pair of vertical angles.
Vertical angles can be mapped onto each other by using a rotation. Since rotations are rigid motions, the angle measures are preserved. This leads to the following theorem.
Vertical angles are always congruent.
Based on the characteristics of the diagram, the following relations hold true.
$∠1≅∠3$
$∠2≅∠4$
Analyzing the diagram, it can be observed that $∠1$ and $∠2$ form a straight angle, so these are supplementary angles. Similarly, $∠2$ and $∠3$ are also supplementary angles.
Therefore, by the Angle Addition Postulate, the sum of $m∠1$ and $m∠2$ is $180_{∘},$ and the sum of $m∠2$ and $m∠3$ is also $180_{∘}.$ These facts can be used to express $m∠2$ in terms of $m∠1$ and also in terms of $m∠3.$
Angle Addition Postulate | Isolate $m∠2$ |
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$m∠1+m∠2$ $=$ $180_{∘}$ | $m∠2$ $=$ $180_{∘}−m∠1$ |
$m∠2+m∠3$ $=$ $180_{∘}$ | $m∠2$ $=$ $180_{∘}−m∠3$ |
$LHS−180_{∘}=RHS−180_{∘}$
$LHS⋅(-1)=RHS⋅(-1)$
The previous proof can be summarized in the following two-column table.
Statements | Reasons |
$ℓ_{1}$ and $ℓ_{2}$ lines | Given |
$∠1$ and $∠2$ supplementary | Definition of straight angle |
$m∠1+m∠2=180_{∘}$ | Definition of supplementary angles |
$m∠2=180_{∘}−m∠1$ | Subtraction Property of Equality |
$∠2$ and $∠3$ supplementary | Definition of straight angle |
$m∠2+m∠3=180_{∘}$ | Definition of supplementary angles |
$m∠2=180_{∘}−m∠3$ | Subtraction Property of Equality |
$180_{∘}−m∠1=180_{∘}−m∠3$ | Transitive Property of Equality |
$m∠1=m∠3$ | Subtraction and Multiplication Properties of Equality |
Consider the points $A,$ $B,$ $C,$ and $D$ on each ray that starts at the point of intersection $E$ of the two lines.
Suppose that the points $A$ and $B$ are rotated $180º$ about point $E.$Use the given expressions to form an equation for $x.$ Identify the relationship between $∠1$ and $∠3,$ as well as $∠2$ and $∠4$ by analyzing their positions.
$x=19$ | |
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$m∠1=3x+14$ | $m∠2=6x−5$ |
$m∠1=3(19)+14$ | $m∠2=6(19)−5$ |
$m∠1=57+14$ | $m∠2=114−5$ |
$m∠1=71_{∘}$ | $m∠2=109_{∘}$ |
Next, by analyzing the position of $∠1$ and $∠3,$ as well as $∠2$ and $∠4,$ it can be noted that these are vertical angles. Therefore, by the Vertical Angles Theorem, they are two pairs of congruent angles. $m∠1=m∠3m∠2=m∠4 ⇒m∠3=71_{∘}m∠4=109_{∘} $ In this way, it was obtained that $∠1$ and $∠3$ are each $71_{∘},$ and $∠2$ and $∠4$ are each $109_{∘}.$
Consider two parallel lines cut by a transversal. The applet shows a pair of corresponding angles, $∠A$ and $∠B.$ Is it possible to translate one line so that one of these angles maps onto another?
The observed relation between corresponding angles is presented and proven in the following theorem.
If parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.
If $ℓ_{1}∥ℓ_{2},$ then $∠1≅∠5,$ $∠2≅∠6,$ $∠3≅∠7,$ and $∠4≅∠8.$
Note that the converse statement is also true.
If $∠1≅∠5,$ $∠2≅∠6,$ $∠3≅∠7,$ or $∠4≅∠8,$ then $ℓ_{1}∥ℓ_{2}.$
This theorem can be proven by an indirect proof. Let $ℓ_{1}$ and $ℓ_{2}$ be two lines intersected by a transversal line $ℓ_{3}$ forming corresponding congruent angles $∠1$ and $∠2.$
Since the goal is to prove that $ℓ_{1}$ is parallel to $ℓ_{2}$, it will be temporarily assumed that $ℓ_{1}$ and $ℓ_{2}$ are not parallel. $Temporary Assumptionℓ_{1}∦ℓ_{2} $ By the Parallel Postulate, there exists a line $n$ parallel to $ℓ_{2}$ that passes through the point of intersection between $ℓ_{1}$ and $ℓ_{3}.$ This line forms $∠3$ and $∠4.$
By the Angle Addition Postulate, $m∠1$ is equal to the sum of $m∠3$ and $m∠4.$ $m∠1=m∠3+m∠4 $ Since $n$ and $ℓ_{2}$ are parallel lines that are cut by a transversal, by the Corresponding Angles Theorem, $∠3$ and $∠2$ are congruent. By the definition of congruence, these angles have the same measure. $∠3≅∠2⇕m∠3=m∠2 $ By the Substitution Property of Equality, $m∠2$ can be substituted for $m∠3$ into the equation for $m∠1.$ $m∠1=m∠3+m∠4↓m∠1=m∠2+m∠4 $ From the above equation and since $m∠4$ is a positive number, it can be concluded that $m∠1$ is greater than $m∠2.$ $m∠1>m∠2 $ This contradicts the given fact that $∠1$ and $∠2$ are congruent. The contradiction came from assuming that $ℓ_{1}$ and $ℓ_{2}$ are not parallel lines. Therefore, $ℓ_{1}$ and $ℓ_{2}$ must be parallel lines.
In a Flowerland Village house, there are stairs with hand railings like shown in the diagram. The measures of $∠1$ and $∠2$ are expressed as $5t−2$ and $4t+12,$ respectively.
What are the measures of $∠1$ and $∠2?$How do measures of $∠1$ and $∠2$ relate to each other? Use the given expressions to form an equation for $t.$
Like corresponding angles, alternate interior angles are also formed by two parallel lines cut by a transversal.
If $ℓ_{1}∥ℓ_{2},$ then $∠1≅∠2$ and $∠3≅∠4.$
To prove that alternate interior angles are congruent, it will be shown that $∠1$ and $∠2$ are congruent.
Notice that by definition $∠2$ and $∠5$ are vertical angles. By the Vertical Angles Theorem, they are therefore congruent angles. $∠2≅∠5 $ Furthermore, by definition $∠5$ and $∠1$ are corresponding angles. Hence, by the Corresponding Angles Theorem, $∠5$ and $∠1$ are also congruent angles. $∠5≅∠1 $ Applying the Transitive Property of Congruence, $∠2$ and $∠1$ can be concluded to be congruent angles as well. ${∠2≅∠5∠5≅∠1 ⇒∠2≅∠1 $ The same reasoning applies to the other pair of alternate interior angles. Therefore, when a pair of parallel lines is cut by a transversal, the pairs of alternate interior angles are congruent.
The previous proof can be summarized in the following two-column table.
Statements | Reasons |
$∠2$ and $∠5$ are vertical angles | Def. of vertical angles |
$∠2≅∠5$ | Vertical Angles Theorem |
$∠5$ and $∠1$ are corresponding angles | Def. of corresponding angles |
$∠5≅∠1$ | Corresponding Angles Theorem |
$∠2≅∠1$ | Transitive Property of Congruence |
Apart from the points of intersection, consider two more points on each line.
Next, $A,$ $B,$ $C,$ and $D$ will be translated parallel to the transversal until the points $A,$ $C,$ and $D$ lie on $ℓ_{2}.$ Then, $A,$ $B,$ and $C$ will be rotated $180_{∘}$ about $F.$ It should be noted that since point $D$ lies on the transversal, when translating it to $ℓ_{2}$ the point will fall into the same position as $F.$ Therefore, $D$ will not be affected by the rotation around $F.$The converse statement is also true.
If $∠1≅∠2$ or $∠3≅∠4,$ then $ℓ_{1}∥ℓ_{2}.$
The proof will be based on the given diagram, but it holds true for any pair of lines cut by a transversal. Consider only one pair of congruent alternate interior angles and one more angle.
It needs to be proven that $ℓ_{1}$ and $ℓ_{2}$ are parallel lines. It is already given that $∠1$ is congruent to $∠2.$ $∠1≅∠2 $ From the diagram, it can be noted that $∠2$ and $∠α$ are vertical angles. By the Vertical Angles Theorem, these angles are congruent. $∠2≅∠α $ Notice the common angle of $∠2$ in both relationships. By the Transitive Property of Congruence, because $∠1$ is congruent to $∠2$ and $∠2$ is congruent to $∠α,$ $∠1$ is congruent $∠α.$ ${∠1≅∠2∠2≅∠α ⇓∠1≅∠α $ It can also be noted that $∠1$ and $∠α$ are corresponding angles. Hence, the Converse Corresponding Angles Theorem will be applied.
Converse Corresponding Angles Theorem |
If two lines are cut by a transversal so that the corresponding angles are congruent, then the lines are parallel. |
Since $∠1$ and $∠α$ are corresponding congruent angles, $ℓ_{1}$ and $ℓ_{2}$ are parallel lines. Finally, all the steps will be summarized in a two-column proof.
Statement | Reason |
$∠1≅∠2$ | Given |
$∠2≅∠α$ | Vertical Angles Theorem |
$∠1≅∠α$ | Transitive Property of Congruence |
$ℓ_{1}∥ℓ_{2}$ | Converse Corresponding Angles Theorem |
Similar properties can be discovered for alternate exterior angles.
If parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent.
If $ℓ_{1}∥ℓ_{2},$ then $∠1≅∠2$ and $∠3≅∠4.$
In order to prove that alternate exterior angles are congruent, it will be shown that $∠1$ and $∠2$ are congruent.
Notice that by definition, $∠2$ and $∠8$ are corresponding angles. Therefore, by the Corresponding Angles Theorem, they are congruent angles. $∠2≅∠8 $ Furthermore, by definition, $∠8$ and $∠1$ are vertical angles. Therefore, by the Vertical Angles Theorem, $∠8$ and $∠1$ are congruent angles. $∠8≅∠1 $ Then, by applying the Transitive Property of Congruence, $∠2$ and $∠1$ can be concluded to be congruent angles as well. ${∠2≅∠8∠8≅∠1 ⇒∠2≅∠1 $ The same reasoning applies to the other pair of alternate exterior angles. Therefore, when a pair of parallel lines is cut by a transversal, the pairs of alternate exterior angles are congruent.
The previous proof can be summarized in the following two-column table.
Statements | Reasons |
$∠2$ and $∠8$ are corresponding angles | Def. of corresponding angles |
$∠2≅∠8$ | Corresponding Angles Theorem |
$∠8$ and $∠1$ are vertical angles | Def. of vertical angles |
$∠8≅∠1$ | Vertical Angles Theorem |
$∠2≅∠1$ | Transitive Property of Congruence |
Consider the points of intersection as well as two more points on each line.
Next, $A,$ $B,$ $C,$ and $D$ will be translated in the direction of the transversal so that points $A,$ $C,$ and $D$ lie on $ℓ_{2}.$ Then, $A,$ $B,$ and $C$ will be rotated $180_{∘}$ about $F.$$∠ADB≅∠GFH$ and $∠CDB≅∠EFH$
If $∠1≅∠2$ or $∠3≅∠4,$ then $ℓ_{1}∥ℓ_{2}.$
The proof will be based on the given diagram, but it holds true for any pair of lines cut by a transversal. Consider only one pair of congruent alternate exterior angles and one more angle.
It needs to be proven that $ℓ_{1}$ and $ℓ_{2}$ are parallel lines. It is already given that $∠1$ is congruent to $∠2.$ $∠1≅∠2 $ From the diagram, it can also be noted that $∠2$ and $∠α$ are vertical angles. By the Vertical Angles Theorem, these angles are congruent. $∠2≅∠α $ Notice the common angle of $∠2$ in both relationships. By the Transitive Property of Congruence, because $∠1$ is congruent to $∠2$ and $∠2$ is congruent to $∠α,$ $∠1$ is congruent $∠α.$ ${∠1≅∠2∠2≅∠α ⇓∠1≅∠α $ It can also be noted that $∠1$ and $∠α$ are corresponding angles. Hence, the Converse Corresponding Angles Theorem can be applied.
Converse Corresponding Angles Theorem |
If two lines are cut by a transversal so that the corresponding angles are congruent, then the lines are parallel. |
Since $∠1$ and $∠α$ are corresponding congruent angles, $ℓ_{1}$ and $ℓ_{2}$ are parallel lines. Each step of the proof will now be summarized in a two-column proof.
Statement | Reason |
$∠1≅∠2$ | Given |
$∠2≅∠α$ | Vertical Angles Theorem |
$∠1≅∠α$ | Transitive Property of Congruence |
$ℓ_{1}∥ℓ_{2}$ | Converse Corresponding Angles Theorem |
In order to build Tulip Street on the south side of the Lilian river, which goes through Flowerland Village, there is a need to build a bridge. Devontay, an architect, proposed the following plan for the bridge.
It is known that the measure of $∠1$ is equal to $4a+11$ and the measure of $∠2$ is equal to $8a−53.$ What are the measures of $∠1$ and $∠2?$
How do the measures of $∠1$ and $∠2$ relate to each other? Use the given expressions to form an equation for $a.$
By analyzing the diagram it can be noted that $∠1$ and $∠2$ are alternate interior angles.
Therefore, by the Alternate Interior Angles Theorem, these angles are congruent. Hence, the measures of these angles are the same. $∠1m∠1 ≅∠2⇓=m∠2 $ By substituting $m∠1$ with $4a+11$ and $m∠2$ with $8a−53,$ the equation for $a$ can be formed.$m∠1=4a+11$, $m∠2=8a−53$
$LHS−4a=RHS−4a$
$LHS+53=RHS+53$
$LHS/4=RHS/4$
Rearrange equation
Any point on a perpendicular bisector of a line segment is equidistant from the endpoints of the line segment.
Based on the characteristics of the diagram, $CM$ is the perpendicular bisector of $AB.$ Therefore, $C$ is equidistant from $A$ and $B.$
$AC=BC$
Suppose $CM$ is the perpendicular bisector of $AB.$ Then $M$ is the midpoint of $AB.$
Consider a triangle with vertices $A,$ $M,$ and $C,$ and another triangle with vertices and $B,$ $M,$ and $C.$
Both $△ACM$ and $△BCM$ have a right angle and congruent legs $AM$ and $BM.$ Since all right angles are congruent, $∠AMC≅∠BMC.$ Furthermore, by the Reflexive Property of Congruence, $CM$ is congruent to itself.
By the Side-Angle-Side Congruence Theorem, the triangles are congruent. Therefore, since corresponding parts of congruent figures are congruent, their hypotenuses $AC$ and $BC$ are also congruent. By the definition of congruent segments, $AC$ and $BC$ have the same length. This means that $C$ is equidistant from $A$ and $B.$
Using this reasoning it can be proven that any point on a perpendicular bisector is equidistant from the endpoints of the segment.
The proof can be summarized in the following two-column table.
Statements | Reasons |
$AM≅MB$ $∠AMC$ and $∠BMC$ are right angles |
Definition of a perpendicular bisector. |
$∠AMC≅∠BMC$ | All right angles are congruent. |
$CM≅CM$ | Reflexive Property of Congruence. |
$△ACM≅△BCM$ | SAS Congruence Theorem. |
$AC≅BC$ | Corresponding parts of congruent figures are congruent. |
$AC=BC$ | Definition of congruent segments. |
Suppose $CM$ is the perpendicular bisector of $AB.$
Using the given points $A,$ $B,$ $C,$ and $M$ as vertices, two triangles can be formed. The resulting triangles, $△ACM$ and $△BCM,$ can be proven to be congruent by identifying a congruence transformation that maps one triangle onto the other.
Since $AM$ and $BM$ are congruent, the distance between $A$ and $M$ is equal to the distance between $B$ and $M.$ Therefore, $A$ is the image of $B$ after a reflection across $CM.$
Reflection Across $CM$ | |
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Preimage | Image |
$B$ | $A$ |
$C$ | $C$ |
$M$ | $M$ |
$AC=CB⇓CM⊥ABandAM=MB$
Consider $AB$ and a point $C$ equidistant from $A$ and $B.$
To prove that $C$ lies on the perpendicular bisector of $AB,$ it will be shown that the line perpendicular to $AB$ through $C$ bisects $AB.$ If $M$ is the point of intersection between the line and the segment, it must be proven that $AM=MB.$Additionally, it was already known that $CM$ and $AB$ are perpendicular.
$CM⊥ABandAM=MB$
By the definition of a perpendicular bisector, $CM$ is the perpendicular bisector of $AB.$ Therefore, $C$ lies on the perpendicular bisector of $AB.$
In Flowerland Village, there are two related families, Funnystongs and Cleverstongs, who live opposite each other. Mr. Funnystong and Mr. Cleverstong want to pave a road between the houses so that every point of the road is equidistant to their houses.
If the houses are $16$ meters away, how far from the houses and along what line should the road be paved?Distance: $8$ meters from each house.
Direction: Along the perpendicular bisector to the segment with endpoints at the houses.
What does the the Perpendicular Bisector Theorem state?
Recall what the Perpendicular Bisector Theorem states.
Any point on a perpendicular bisector is equidistant from the endpoints of the line segment.
With this theorem in mind, the position of the road can be determined. To do so, draw a segment whose endpoints are located at the houses.
Before drawing the perpendicular bisector of this segment, its midpoint should be found. Since the distance between the houses is $16$ meters, the perpendicular bisector will pass through a point that is $216 =8$ meters away from the houses.
Based on the theorem, it can be said that each point on the bisector is equidistant from the houses. Therefore, the road between the houses should be paved along the segment's perpendicular bisector.