Reference

Triangle Congruence Theorems

Rule

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.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ \overline{A B} ≅ \overline{D E} ∠ A ≅ ∠ D \overline{A C} ≅ \overline{D F} \Rightarrow △ A B C ≅ △ D E F

Proof

Side-Angle-Side Congruence Theorem

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

The primary purpose of the proof 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

△ D E F

So That Two Corresponding Vertices Match

Translate

△ D E F

so that

D

is mapped onto

A .

If this translation maps

△ D E F

onto

△ A B C,

the proof is complete.

Since the image of the translation does not match

△ A B C,

at least one more transformation is needed.

Rotate

△ A E^{'} F^{'}

So That Two Corresponding Sides Match

Rotate

△ A E^{'} F^{'}

counterclockwise about

A

so that a pair of corresponding sides match. If the image of this transformation is

△ A B C,

the proof is complete. Note that this rotation maps

E^{'}

onto

B .

Therefore, the rotation maps

\overline{A E^{'}}

onto

\overline{A B} .

As before, the image does not match

△ A B C .

Therefore, a third rigid motion is required.

Reflect

△ A B F^{''}

So That Two More Corresponding Sides Match

Reflect

△ A B F^{''}

across

A B .

Because reflections preserve angles,

\overline{A F^{''}}

is mapped onto

\overline{A C} .

Additionally, it is given that

A C = A F^{''} .

Therefore,

F^{''}

is mapped onto

C,

which gives that

B F^{''}

is mapped onto

B C .

This time the image matches

△ A B C .

Consequently, after a sequence of rigid motions,

△ D E F

can be mapped onto

△ A B C .

This means that

△ D E F

and

△ A B C

are congruent triangles. The proof is complete.

Rule

Angle-Side-Angle Congruence Theorem

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.

Triangle ABC is congruent to triangle DEF

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ ∠ A ≅ ∠ D \overline{A B} ≅ \overline{D E} ∠ B ≅ ∠ E \Rightarrow △ A B C ≅ △ D E F

Proof

Angle-Side-Angle Congruent Theorem

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

△ D E F

So That Two Corresponding Vertices Match

Translate

△ D E F

so that

D

is mapped onto

A .

If this translation maps

△ D E F

onto

△ A B C,

the proof is complete.

Translation that maps vertex D of triangle DEF onto vertex A of triangle ABC

Since the image of the translation does not match

△ A B C,

at least one more transformation is needed.

Rotate

△ A E^{'} F^{'}

So That Two Corresponding Sides Match

Rotate

△ A E^{'} F^{'}

counterclockwise about

A

so that a pair of corresponding sides match. If the image of this transformation is

△ A B C,

the proof is complete. Note that this rotation maps

E^{'}

onto

B .

Therefore, the rotation maps

\overline{A E^{'}}

onto

\overline{A B} .

As before, the image does not match

△ A B C .

Therefore, a third rigid motion is required.

Reflect

△ A B F^{''}

So That All Corresponding Sides Match

Reflect

△ A B F^{''}

across

A B .

Because reflections preserve angles,

A F^{''}

and

B F^{''}

are mapped onto

A C

and

B C,

respectively. Then, the point of intersection of the original rays

F^{''}

is mapped onto the point of intersection of the image rays

C .

This time the image matches

△ A B C .

Consequently, after a sequence of rigid motions

△ D E F

can be mapped onto

△ A B C .

This means that

△ D E F

and

△ A B C

are congruent triangles.

Rule

Side-Side-Side Congruence Theorem

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.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ \overline{A B} ≅ \overline{D E} \overline{B C} ≅ \overline{E F} \overline{A C} ≅ \overline{D F} \Rightarrow △ A B C ≅ △ D E F

Proof

Side-Side-Side Congruence Theorem

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

The primary purpose of this proof 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

△ D E F

So That Two Corresponding Vertices Match

Translate

△ D E F

so that

D

is mapped onto

A .

If this translation maps

△ D E F

onto

△ A B C,

the proof is complete.

Since the image of the translation does not match

△ A B C,

at least one more transformation is needed.

Rotate

△ A E^{'} F^{'}

So That Two Corresponding Sides Match

Rotate

△ A E^{'} F^{'}

counterclockwise about

A

so that a pair of corresponding sides matches. If the image of this transformation is

△ A B C,

the proof is complete. Note that this rotation maps

E^{'}

onto

B .

Consequently,

\overline{A E^{'}}

is mapped onto

\overline{A B} .

As before, the image does not match

△ A B C .

Therefore, a third rigid motion is required.

Reflect

△ A B F^{''}

So That Two More Corresponding Sides Match

The points $C$ and $F^{''}$ are on opposite sides of $A B .$ Now, consider $\overline{C F^{'}} .$ Let $G$ denote the point of intersection between $A B$ and $\overline{C F^{''}} .$

It can be noted that $A C = A F^{''}$ and $B C = B F^{''} .$ By the Converse Perpendicular Bisector Theorem, $A B$ is a perpendicular bisector of $\overline{C F^{''}} .$ Points along the perpendicular bisector are equidistant from the endpoints of the segment, so $C G = G F^{''} .$

Finally,

F^{''}

can be mapped onto

C

by a reflection across

A B

by reflecting

△ A B F^{''}

across

A B .

Because reflections preserve angles,

A F^{''}

and

B F^{''}

are mapped onto

A C

and

B C,

respectively.

This time the image matches

△ A B C .

Consequently, the application of a sequence of rigid motions allows

△ D E F

to be mapped onto

△ A B C .

This means that

△ D E F

and

△ A B C

are congruent triangles. The proof is complete.

Rule

Angle-Angle-Side Congruence Theorem

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 ≅ ∠ E \overline{B C} ≅ \overline{E F} \Rightarrow △ A B C ≅ △ D E F

Proof

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

The primary purpose of the proof 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

△ D E F

So That Two Corresponding Vertices Match

Translate

△ D E F

so that

F

is mapped onto

C .

If this translation maps

△ D E F

onto

△ A B C,

the proof is complete.

Since the image of the translation does not match

△ A B C,

at least one more transformation is needed.

Rotate

△ C D^{'} E^{'}

So That Two Corresponding Sides Match

Rotate

△ C D^{'} E^{'}

clockwise about

C

so that a pair of corresponding sides match. If the image of this transformation is

△ A B C,

the proof is complete. Note that this rotation maps

E^{'}

onto

B .

Therefore, the rotation maps

\overline{C E^{'}}

onto

\overline{C B} .

As before, the image does not match

△ A B C .

Therefore, a third rigid motion is required.

Reflect

△ A B F^{''}

So That Two More Corresponding Sides Match

It is given that two angles of $△ A B C$ are congruent to two angles of $△ B C D^{''} .$ Hence, by the Third Angle Theorem, $∠ B C D^{''}$ is congruent to $∠ B C A .$

Triangles ABC and CBD'' with a common side CB

Reflect

△ C B D^{''}

across

B C .

Because reflections preserve angles,

\overline{B D^{''}}

and

\overline{C D^{''}}

are mapped onto

\overline{B A}

and

\overline{C A},

respectively. Then, the point of intersection of the original segments

D^{''}

is mapped onto the point of intersection of the image segments

A .

This time the image matches

△ A B C .

Consequently, after a sequence of rigid motions,

△ D E F

can be mapped onto

△ A B C .

This means that

△ D E F

and

△ A B C

are congruent triangles. The proof is complete.

Rule

Isosceles Triangle Theorem

If two sides of a triangle are congruent, then the angles opposite them are congruent.

Based on this diagram, the following relation holds true.

$\overline{A B} ≅ \overline{A C}$ $\Rightarrow$ $∠ B ≅ ∠ C$

The Isosceles Triangle Theorem is also known as the Base Angles Theorem.

Proof

Geometric Approach

Consider a triangle $A B C$ with two congruent sides, or an isosceles triangle.

In this triangle, let

P

be the point of intersection of

\overline{B C}

and the angle bisector of

∠ A .

From the diagram, the following facts about

△ B A P

and

△ C A P

can be observed.

Statement	Reason
$∠ B A P ≅ ∠ C A P$	Definition of an angle bisector
$\overline{B A} ≅ \overline{C A}$	Given
$\overline{A P} ≅ \overline{A P}$	Reflexive Property of Congruence

Therefore,

△ B A P

and

△ C A P

have two pairs of corresponding congruent sides and one pair of congruent included angles. By the Side-Angle-Side Congruence Theorem,

△ B A P

and

△ C A P

are congruent triangles.

△ B A P ≅ △ C A P

Corresponding parts of congruent figures are congruent. Therefore,

∠ B

and

∠ C

are congruent.

∠ B ≅ ∠ C

It has been proven that if two sides of a triangle are congruent, then the angles opposite them are congruent.

Proof

Using Transformations

Consider an isosceles triangle $△ A B C .$

A line passing through $A$ and the midpoint of $\overline{B C}$ will be drawn. Let $P$ be the midpoint.

Since $\overline{B P}$ and $\overline{P C}$ are congruent, the distance between $B$ and $P$ is equal to the distance between $C$ and $P .$ Therefore, $B$ is the image of $C$ after a reflection across $A P .$ Also, because $A$ lies on $A P,$ a reflection across $A P$ maps $A$ onto itself. The same is true for $P .$

Reflection Across $A P$
Preimage	Image
$C$	$B$
$A$	$A$
$P$	$P$

The table shows that the images of the vertices of

△ C A P

are the vertices of

△ B A P .

It can be concluded that

△ B A P

is the image of

△ C A P

after a reflection across

A P .

Since a reflection is a rigid motion, this proves that the triangles are congruent.

Corresponding parts of congruent figures are congruent, so

∠ B

and

∠ C

are congruent.

∠ B ≅ ∠ C

Rule

Converse Isosceles Triangle Theorem

If two angles of a triangle are congruent, then the sides opposite them are congruent.

Isosceles triangle with a movable point that can be used to build triangles with different pairs of congruent angles.

Based on this diagram, the following relation holds true.

$∠ B ≅ ∠ C$ $\Rightarrow$ $\overline{A B} ≅ \overline{A C}$

This theorem is the converse theorem to the Isosceles Triangle Theorem. It is also known as the Converse Base Angles Theorem.

Proof

Consider a triangle $A B C$ with two congruent angles.

Let $P$ be the point of intersection of $\overline{B C}$ and the angle bisector of $∠ A .$ Since $\overline{A P}$ is the angle bisector of $∠ A,$ then $∠ B A P ≅ ∠ C A P .$

By the Reflexive Property of Congruence,

\overline{A P}

△ A B P

is congruent to

\overline{A P}

△ A C P .

Because of the Angle-Angle-Side Congruence Theorem, both triangles are congruent.

△ A B P ≅ △ A C P

Since corresponding parts of congruent triangles are congruent, it follows that

A B

is congruent to

A C .

\overline{A B} ≅ \overline{A C}

It has been proven that if two angles of a triangle are congruent, then the sides opposite them are congruent.

Rule

Hinge Theorem

If two sides of a triangle are congruent to two sides of another triangle, the triangle with the larger included angle has the larger third side.

Two triangles, ABC and A'B'C', that have two pairs of congruent sides and have the same included angle.

Based on the diagram above, the following relation holds true.

$m ∠ B A C > m ∠ B^{'} A^{'} C^{'} \Rightarrow \overline{B C} > \overline{B^{'} C^{'}}$

The Hinge Theorem receives its name because the included angle of the congruent sides acts like a hinge. The more the sides are open, the further away their ends are from each other.

Proof

Consider $△ A B C$ and $△ A^{'} B^{'} C$ such that $\overline{A B} = \overline{A^{'} B^{'}}$ and $\overline{A C} = \overline{A^{'} C^{'}},$ where $m ∠ B A C > m ∠ B^{'} A^{'} C .$

Place a point $D^{'}$ on $△ A^{'} B^{'} C^{'}$ so that $m ∠ D^{'} A^{'} C^{'} = m ∠ B A C$ and $\overline{A^{'} D^{'}} = \overline{A B} .$

Draw $△ A^{'} C^{'} D^{'}$ and $△ B^{'} C^{'} D^{'} .$

Note that two sides of $△ A B C$ and their included angle are congruent to two sides of $△ A^{'} D^{'} C^{'}$ and their included angle. Because of the Side-Angle-Side Congruence Theorem, $\overline{B C} = \overline{D^{'} C^{'}} .$ It now suffices to prove that $\overline{D^{'} C^{'}} > \overline{B^{'} C^{'}} .$ To do so, note that $△ A^{'} B^{'} D$ is isosceles.

Because of the Isosceles Triangle Theorem,

m ∠ A^{'} D^{'} B^{'} = m ∠ A^{'} B^{'} D^{'} .

Furthermore,

∠ A^{'} D^{'} B^{'}

must measure more than

∠ C^{'} D^{'} B^{'}

because of the Angle Addition Postulate.

m ∠ A^{'} D^{'} B^{'} > m ∠ C^{'} D^{'} B^{'}

The same holds true for

∠ C^{'} B^{'} D^{'}

and

∠ A^{'} B^{'} D^{'}

m ∠ C^{'} B^{'} D^{'} > m ∠ A^{'} B^{'} D^{'}

Since

m ∠ A^{'} D^{'} B^{'} = m ∠ A^{'} B^{'} D^{'},

the Substitution Property of Equality allows to substitute

m ∠ A^{'} B^{'} D^{'}

for

m ∠ A^{'} D^{'} B

in the above inequality, and because of the Transitive Property of Inequality, the following can be concluded.

m ∠ C^{'} B^{'} D^{'} > m ∠ C^{'} D^{'} B^{'}

This inequality can be used along with the Triangle Larger Angle Theorem to prove that

\overline{C^{'} D^{'}} > \overline{B^{'} C^{'}} .

Finally, since

\overline{C^{'} D^{'}} = \overline{B C},

use the Substitution Property of Equality one more time to obtain the desired inequality.

$\overline{B C} > \overline{B^{'} C^{'}}$

Rule

Converse Hinge Theorem

If two sides of a triangle are congruent to two sides of another triangle, the triangle with the larger third side also has the larger included angle.

Based on the diagram above, the following relation holds true.

$\overline{B C} > \overline{B^{'} C^{'}} \Rightarrow m ∠ B A C > m ∠ B^{'} A^{'} C^{'}$

This theorem is the converse of the Hinge Theorem.

Proof

Consider $△ A B C$ and $△ A^{'} B^{'} C$ such that $\overline{A B} = \overline{A^{'} B^{'}}$ and $\overline{A C} = \overline{A^{'} C^{'}},$ where $\overline{B C} > \overline{B^{'} C^{'}} .$

This theorem can be proven by contradiction. Since the goal is to prove that $m ∠ B A C > m ∠ B^{'} A^{'} C^{'}$ the opposite statement will be assumed, that is, $m ∠ B A C \leq m ∠ B^{'} A^{'} C^{'} .$ Because this is a non-strict inequality, both $m ∠ B A C = m ∠ B^{'} A^{'} C^{'}$ and $m ∠ B A C < m ∠ B^{'} A^{'} C^{'}$ have to be considered.

$m ∠ B A C = m ∠ B^{'} A^{'} C^{'}$

m ∠ B A C = m ∠ B^{'} A^{'} C^{'},

then,

△ A B C

and

△ A^{'} B^{'} C^{'}

share two congruent sides and their included angles are congruent. Because of the Side-Angle-Side Congruence Theorem, they are congruent.

△ A B C ≅ △ A^{'} B^{'} C

Since the triangles are congruent, their sides are congruent as well. This means that

\overline{B C} = \overline{B^{'} C^{'}},

but this contradicts the fact that the given triangle is such that

\overline{B C} > \overline{B^{'} C^{'}} .

$m ∠ B A C < m ∠ B^{'} A^{'} C^{'}$

If $m ∠ B A C < m ∠ B^{'} A^{'} C^{'}$ then the Hinge Theorem states that $\overline{B C} < \overline{B^{'} C^{'}},$ but this also contradicts the fact that $\overline{B C} > \overline{B^{'} C^{'}} .$

Conclusion

The assumption that $m ∠ B A C$ is less than or equal to $m ∠ B^{'} A^{'} C^{'}$ contradicts the hypothesis. Therefore, this assumption must be false. Consequently, the initial conclusion of the theorem is true.

$m ∠ B A C > m ∠ B^{'} A^{'} C^{'} .$

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Proof

Proof

Proof

Proof

Proof

Proof

Proof

Proof

Proof

m∠BAC=m∠B′A′C′

m∠BAC<m∠B′A′C′

Conclusion

$m ∠ B A C = m ∠ B^{'} A^{'} C^{'}$

$m ∠ B A C < m ∠ B^{'} A^{'} C^{'}$