Reference

Triangle Similarity Theorems

Rule

Angle-Angle Similarity Theorem

If two angles of a triangle are congruent to two angles of another triangle, then the triangles are similar.

Triangles ABC and DEF with congruent angles marked at A and D and B and E.

If $∠ A ≅ ∠ D$ and $∠ B ≅ ∠ E,$ then $△ A B C \sim △ D E F .$

Proof

Angle-Angle Similarity Theorem

Consider two triangles $△ A B C$ and $△ D E F,$ whose two corresponding angles are congruent.

These triangles can be proven to be similar by identifying a similarity transformation that maps one triangle onto the other. First, $△ D E F$ can be dilated with the scale factor $k = \frac{A B}{D E}$ about $D,$ forming the new triangle $△ D E^{'} F^{'} .$

Since a dilation is a similarity transformation, it can be concluded that

△ D E^{'} F^{'}

and

△ D E F

are similar triangles. Next, it has to be proven that a rigid motion that maps

△ D E^{'} F^{'}

onto

△ A B C

exists. The corresponding angles of similar figures are congruent, so

∠ E^{'}

and

∠ E

are congruent angles.

∠ E^{'} ≅ ∠ E

Additionally, since

∠ E

is congruent to

∠ B,

by the Transitive Property of Congruence,

∠ E^{'}

is congruent to

∠ B .

∠ E^{'} ≅ ∠ B

The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor.

\frac{D E ^{'}}{D E} = k

In this case, the scale factor

k

\frac{A B}{D E} .

Applying the Transitive Property of Equality, an equation can be formed and simplified.

\frac{D E ^{'}}{D E} = \frac{A B}{D E} ⇕ D E^{'} = A B

It has been obtained that the two angles and the included side of

△ D E^{'} F^{'}

are congruent to the corresponding two angles and the included side of

△ A B C .

Therefore, by the Angle-Side-Angle (ASA) Congruence Theorem, the two triangles are congruent.

△ A B C ≅ △ D E^{'} F^{'}

Since congruent figures can be transformed into each other using rigid motions, and

△ A B C

and

△ D E^{'} F^{'}

are congruent triangles, there is a rigid motion placing

△ D E^{'} F^{'}

onto

△ A B C .

The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps

△ D E F

onto

△ A B C .

Therefore, it can be concluded that $△ A B C$ and $△ D E F$ are similar triangles.

$△ A B C \sim △ D E F$

The proof is now complete.

Rule

Side-Side-Side Similarity Theorem

If corresponding sides of two triangles are proportional, then the triangles are similar.

If $\frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D},$ then $△ A B C \sim △ D E F .$

Proof

Consider two triangles $△ A B C$ and $△ D E F,$ whose corresponding sides are proportional.

Because dilation is a similarity transformation, it can be concluded that

△ D E^{'} F^{'}

and

△ D E F

are similar triangles. Now, it has to be proven that a rigid motion that maps

△ D E^{'} F^{'}

onto

△ A B C

exists. The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor.

\frac{D E ^{'}}{D E} = \frac{D F ^{'}}{D F} = \frac{E ^{'} F ^{'}}{E F} = k

In this case, the scale factor

k

\frac{A B}{D E} .

Since all of the sides of

△ A B C

and

△ D E F

are proportional, the scale factor can be expressed by any of the following ratios.

k = \frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{D F}

Applying the Transitive Property of Equality, three equations can be formed and simplified.

\frac{D E ^{'}}{D E} \frac{D F ^{'}}{D F} \frac{E ^{'} F ^{'}}{E F} = \frac{A B}{D E} = \frac{C A}{D F} = \frac{B C}{E F} \Rightarrow D E^{'} D F^{'} E^{'} F^{'} = A B = C A = B C

These relations imply that the three sides of

△ D E^{'} F^{'}

are congruent to the three sides of

△ A B C .

Therefore, by the Side-Side-Side (SSS) Congruence Theorem, the two triangles are congruent.

△ A B C ≅ △ D E^{'} F^{'}

Since congruent figures can be transformed into each other using rigid motions, and

△ A B C

and

△ D E^{'} F^{'}

are congruent triangles, there is a rigid motion placing

△ D E^{'} F^{'}

onto

△ A B C .

The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps $△ D E F$ onto $△ A B C .$

Therefore, it can be concluded that $△ A B C$ and $△ D E F$ are similar triangles.

$△ A B C \sim △ D E F$

The proof is now complete.

Rule

Side-Angle-Side Similarity Theorem

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

Triangles ABC and DEF with congruent angles marked at A and D.

If $\frac{A B}{D E} = \frac{A C}{D F}$ and $∠ A ≅ ∠ D,$ then $△ A B C \sim △ D E F .$

Proof

Consider two triangles $△ A B C$ and $△ D E F,$ whose two pairs of corresponding sides are proportional and the included angles are congruent.

Because dilation is a similarity transformation, it can be concluded that

△ D E^{'} F^{'}

and

△ D E F

are similar triangles. Now, it has to be proven that a rigid motion that maps

△ D E^{'} F^{'}

onto

△ A B C

exists. The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor.

\frac{D E ^{'}}{D E} = \frac{D F ^{'}}{D F} = k

In this case, the scale factor

k

\frac{A B}{D E} .

Since

\overline{A B}

and

\overline{A C}

are proportional to

\overline{D E}

and

\overline{D F}

respectively, the scale factor can be expressed by any of the following ratios.

k = \frac{A B}{D E} = \frac{A C}{D F}

Applying the Transitive Property of Equality, three equations can be formed and simplified.

\frac{D E ^{'}}{D E} = \frac{A B}{D E} \frac{D F ^{'}}{D F} = \frac{A C}{D F} \Rightarrow \Rightarrow D E^{'} = A B D F^{'} = A C

These relations imply that the two sides of

△ D E^{'} F^{'}

are congruent to the corresponding two sides of

△ A B C .

Moreover, the included angles

∠ A

and

∠ D

are also congruent.

Therefore, by the Side-Angle-Side (SAS) Congruence Theorem, the two triangles are congruent.

△ A B C ≅ △ D E^{'} F^{'}

Since congruent figures can be transformed into each other using rigid motions, and

△ A B C

and

△ D E^{'} F^{'}

are congruent triangles, there is a rigid motion placing

△ D E^{'} F^{'}

onto

△ A B C .

The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps $△ D E F$ onto $△ A B C .$

Therefore, it can be concluded that $△ A B C$ and $△ D E F$ are similar triangles.

$△ A B C \sim △ D E F$

The proof is now complete.

Proof

Proof

Proof

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