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Reference

Rule

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

If $∠A≅∠D$ and $∠B≅∠E,$ then $△ABC∼△DEF.$

Consider two triangles $△ABC$ and $△DEF,$ 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, $△DEF$ can be dilated with the scale factor $k=DEAB $ about $D,$ forming the new triangle $△DE_{′}F_{′}.$

Since a dilation is a similarity transformation, it can be concluded that $△DE_{′}F_{′}$ and $△DEF$ are similar triangles. Next, it has to be proven that a rigid motion that maps $△DE_{′}F_{′}$ onto $△ABC$ 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.
$DEDE_{′} =k $

In this case, the scale factor $k$ is $DEAB .$ Applying the Transitive Property of Equality, an equation can be formed and simplified. $DEDE_{′} =DEAB ⇕DE_{′}=AB $

It has been obtained that the two angles and the included side of $△DE_{′}F_{′}$ are congruent to the corresponding two angles and the included side of $△ABC.$
Therefore, by the Angle-Side-Angle (ASA) Congruence Theorem, the two triangles are congruent.
$△ABC≅△DE_{′}F_{′} $

Since congruent figures can be transformed into each other using rigid motions, and $△ABC$ and $△DE_{′}F_{′}$ are congruent triangles, there is a rigid motion placing $△DE_{′}F_{′}$ onto $△ABC.$
The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps $△DEF$ onto $△ABC.$

Therefore, it can be concluded that $△ABC$ and $△DEF$ are similar triangles.

$△ABC∼△DEF$

The proof is now complete.

Rule

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

If $DEAB =EFBC =FDCA ,$ then $△ABC∼△DEF.$

Consider two triangles $△ABC$ and $△DEF,$ whose corresponding sides are proportional.

These triangles can be proven to be similar by identifying a similarity transformation that maps one triangle onto the other. First, $△DEF$ can be dilated with the scale factor $k=DEAB $ about $D,$ forming the new triangle $△DE_{′}F_{′}.$

Because dilation is a similarity transformation, it can be concluded that $△DE_{′}F_{′}$ and $△DEF$ are similar triangles. Now, it has to be proven that a rigid motion that maps $△DE_{′}F_{′}$ onto $△ABC$ exists. The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor.$DEDE_{′} =DFDF_{′} =EFE_{′}F_{′} =k $

In this case, the scale factor $k$ is $DEAB .$ Since all of the sides of $△ABC$ and $△DEF$ are proportional, the scale factor can be expressed by any of the following ratios.
$k=DEAB =EFBC =DFCA $

Applying the Transitive Property of Equality, three equations can be formed and simplified. $DEDE_{′} DFDF_{′} EFE_{′}F_{′} =DEAB =DFCA =EFBC ⇒DE_{′}DF_{′}E_{′}F_{′} =AB=CA=BC $

These relations imply that the three sides of $△DE_{′}F_{′}$ are congruent to the three sides of $△ABC.$ Therefore, by the Side-Side-Side (SSS) Congruence Theorem, the two triangles are congruent.
$△ABC≅△DE_{′}F_{′} $

Since congruent figures can be transformed into each other using rigid motions, and $△ABC$ and $△DE_{′}F_{′}$ are congruent triangles, there is a rigid motion placing $△DE_{′}F_{′}$ onto $△ABC.$
The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps $△DEF$ onto $△ABC.$

Therefore, it can be concluded that $△ABC$ and $△DEF$ are similar triangles.

$△ABC∼△DEF$

The proof is now complete.

Rule

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.

If $DEAB =DFAC $ and $∠A≅∠D,$ then $△ABC∼△DEF.$

Consider two triangles $△ABC$ and $△DEF,$ whose two pairs of corresponding sides are proportional and the included angles are congruent.

These triangles can be proven to be similar by identifying a similarity transformation that maps one triangle onto the other. First, $△DEF$ can be dilated with the scale factor $k=DEAB $ about $D,$ forming the new triangle $△DE_{′}F_{′}.$

Because dilation is a similarity transformation, it can be concluded that $△DE_{′}F_{′}$ and $△DEF$ are similar triangles. Now, it has to be proven that a rigid motion that maps $△DE_{′}F_{′}$ onto $△ABC$ exists. The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor.$DEDE_{′} =DFDF_{′} =k $

In this case, the scale factor $k$ is $DEAB .$ Since $AB$ and $AC$ are proportional to $DE$ and $DF$ respectively, the scale factor can be expressed by any of the following ratios.
$k=DEAB =DFAC $

Applying the Transitive Property of Equality, three equations can be formed and simplified. $DEDE_{′} =DEAB DFDF_{′} =DFAC ⇒⇒ DE_{′}=ABDF_{′}=AC $

These relations imply that the two sides of $△DE_{′}F_{′}$ are congruent to the corresponding two sides of $△ABC.$ Moreover, the included angles $∠A$ and $∠D$ are also congruent.
Therefore, by the Side-Angle-Side (SAS) Congruence Theorem, the two triangles are congruent.
$△ABC≅△DE_{′}F_{′} $

Since congruent figures can be transformed into each other using rigid motions, and $△ABC$ and $△DE_{′}F_{′}$ are congruent triangles, there is a rigid motion placing $△DE_{′}F_{′}$ onto $△ABC.$
The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps $△DEF$ onto $△ABC.$

Therefore, it can be concluded that $△ABC$ and $△DEF$ are similar triangles.

$△ABC∼△DEF$

The proof is now complete.

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