In this section, we will take a look at **ASA Theorem** and understand how to prove congruence and similarity between triangles without all sides and angles of triangles.

## ASA Theorem geometry

In geometry, two triangles are congruent when either all the sides of one triangle are equal to all sides of another triangle respectively. Or all the three angles of both the triangles should be equal respectively. But with the ASA criterion, we can show congruent triangles with the help of two angles and one side of each triangle.

ASA theorem, as the name suggests, considers two angles and one side of one triangle equal to another triangle respectively. Here two adjacent angles and the included side between these angles are taken. But one should remember that ASA is not the same as **AAS**. As ASA has the included side of the two triangles, but in AAS the selected side is the unincluded side of both angles.

## ASA similarity and congruence theorem

We can easily find similar triangles and congruent triangles with the help of the ASA similarity and congruence theorem.

### ASA similarity theorem

We know that if two triangles are similar then all the corresponding sides are in proportionality and all the corresponding pairs are congruent from the definition of similar triangles. However, in order to ensure the similarity of two triangles we only need information about two angles with the ASA similarity theorem.

**ASA similarity theorem** : Two triangles are similar if two corresponding angles of one triangle are congruent to the two corresponding angles of another triangle. Also, the corresponding sides are proportional.

Mathematically we represent as, if $\angle A=\angle X,\angle B=\angle Y,$ then $\u25b3ABC~\u25b3XYZ.$ And$\frac{AB}{XY}=\frac{BC}{YZ}=\frac{AC}{XZ}.$

Generally ASA similarity is more well known as the **AA similarity theorem**, as there is nothing further to check because of only one ratio of sides. Also when two angle measures are given, we can easily find the third angle as the total angle measure isid="2696588" role="math" $180\xb0$. So we can easily check the equality of corresponding angles of two triangles and determine the similarity of both triangles.

### ASA congruence theorem

ASA congruence theorem stands for Angle-Side-Angle and gives the congruent relation between two triangles.

**ASA congruence theorem**: Two triangles are congruent if two adjacent angles and the included side on one triangle are congruent to the two angles and included side of another triangle.

Mathematically we say that, if$\angle B\cong \angle M,BC\cong MN,\angle C\cong \angle N,$ then id="2696597" role="math" $\u25b3ABC\cong \u25b3LMN.$

As the angles and sides are congruent they will also be equal. So$\angle B=\angle M,BC=MN,\angle C=\angle N,$then$\u25b3ABC\cong \u25b3LMN.$

## ASA theorem proof

Now let us take a look at ASA theorem proof for similarity and congruence.

### ASA similarity theorem proof

For two triangles $\u25b3ABC$ and $\u25b3XYZ,$ it is given from the statement of ASA similarity theorem that $\angle A=\angle X,\angle B=\angle Y.$

To prove: $\u25b3ABC~\u25b3XYZ.$ And $\frac{AB}{XY}=\frac{BC}{YZ}=\frac{AC}{XZ}.$

Now as two angles $\angle A$ and $\angle B$ are already given in $\u25b3ABC,$ we can easily find $\angle C$ by taking $180\xb0-(\angle A+\angle B).$And the same will be the case for the triangle$\u25b3XYZ.$

We will construct a line PQ in triangle $\u25b3XYZ$ such that $XP=AB$ and $XQ=AC.$ Also it is given that $\angle A=\angle X.$ Then by using SAS congruence theorem we get that $\u25b3ABC~\u25b3XPQ.$

Since $\u25b3ABC~\u25b3XPQ$ then the corresponding parts of congruent triangles are congruent.

$\Rightarrow \angle B=\angle P\left(1\right)$

Also, it is given that

$\angle B=\angle Y\left(2\right)$

From equation (1) and equation (2) $\angle P=\angle Y.$

Since $\angle P$ and $\angle Y$ forms corresponding angles and XY works as transversal $PQ\parallel YZ.$

Using Basic Proportionality Theorem in $\u25b3XYZ,$

$\frac{XP}{XY}=\frac{XQ}{XZ}$

**Basic Proportionality theorem** states that if a line is parallel to one side of a triangle and it intersects the other two sides at two different points then those sides are in proportion.

From the construction of PQ we replace $XP=AB$and $XQ=AC$ in the above equation.

$\Rightarrow \frac{AB}{XY}=\frac{AC}{XZ}$

Similarly, $\frac{AB}{XY}=\frac{BC}{YZ}$

$\Rightarrow \frac{AB}{XY}=\frac{BC}{YZ}=\frac{AC}{XZ}$

Hence $\u25b3ABC~\u25b3XYZ.$

### ASA congruence theorem proof

We are given from the statement of ASA congruence theorem that $\angle B=\angle M,\angle C=\angle N,$ and $BC=MN.$

To prove: $\u25b3ABC\cong \u25b3LMN$

To prove the above statement we will consider different cases.

** Case 1**: Assume that $AB=LM.$

In $\u25b3ABC$ and $\u25b3LMN,AB=LM$ from our assumption. And it is given that $\angle B=\angle M$ and $BC=MN.$ So by SAS congruence theorem $\u25b3ABC\cong \u25b3LMN.$

** Case 2**: Suppose $AB>LM.$

Then we construct point X on AB such that id="2696624" role="math" $XB=LM.$

We have $XB=LM,\angle B=\angle M$ and $BC=MN.$ Using SAS congruence theorem $\u25b3XBC\cong \u25b3LMN.$

Now it is given that $\angle ACB=\angle LNM\left(1\right)$

And from the above congruence,id="2696626" role="math" $\u25b3XBC\cong LMN,$ we get that id="2696647" role="math" $\angle XCB=\angle LNM\left(2\right)$

So from equations (1) and (2), we get

$\angle ACB=\angle LNM=\angle XCB\phantom{\rule{0ex}{0ex}}\Rightarrow \angle ACB=\angle XCB$

But from our assumption of $AB>LM,$ and also by looking at the figure this is not possible. So $\angle ACB=\angle XCB$ can only occur when both the points A and X coincides and $AC=XC.$

So we are again left with the fact that $\u25b3ABC$ and $\u25b3XCB$ are equal. Hence we can consider only one triangle $\u25b3ABC$such that $AB=LM.$ So this is the same as case 1, and from that we get that $\u25b3ABC\cong \u25b3LMN.$

** Case 3**: Suppose $AB<LM.$

Then construct a point Y on LM such that $AB=YM$ and we repeat the same argument as in case 2.

Hence we get that $\u25b3ABC\cong \u25b3LMN.$

## ASA Theorem example

Let us see some examples related to ASA theorems.

Calculate BD and CE in the given figure, if $AC\parallel BD,AE=3cm,AC=6cm,BE=4cm,DE=8cm.$

**Solution****:**

In $\u25b3ACE$ and $\u25b3BDE,$ as $AC\parallel BD$ then $\angle A=\angle B$ because they are alternate interior angles. Also $\angle AEC=\angle BED$ forms vertically opposite angles.

Then by ASA similarity theorem $\u25b3ACE~\u25b3BDE.$

We also get from the ASA similarity theorem that $\frac{CE}{DE}=\frac{AE}{BE}=\frac{AC}{BD}.$

Then by substituting all the given values in the above equation,

$\frac{CE}{8}=\frac{3}{4}=\frac{6}{BD}$

$\Rightarrow BD=\frac{4}{3}\times 6=8cm$

And id="2696649" role="math" $CE=\frac{3}{4}\times 8=6cm$

Hence $\mathit{B}\mathit{D}\mathbf{=}\mathbf{8}\mathbf{}\mathit{c}\mathit{m}\mathbf{,}\mathbf{}\mathit{C}\mathit{E}\mathbf{=}\mathbf{6}\mathbf{}\mathit{c}\mathit{m}.$

Calculate the value of x when $\angle D=55\xb0,\angle F=65\xb0,\angle B=(2x+30)\xb0.$

**Solution:**

From the figure we can see that $\angle A=\angle D,AC=DF,\angle C=\angle F.$ Then by ASA congruence theorem we get that $\u25b3ABC\cong \u25b3DEF.$

$\Rightarrow \angle A+\angle B+\angle C=180\xb0$

Now substituting all the given values we get,

$55\xb0+(2x+30)\xb0+65\xb0=180\xb0\phantom{\rule{0ex}{0ex}}55\xb0+30\xb0+65\xb0+2x=180\xb0\phantom{\rule{0ex}{0ex}}150\xb0+2x=180\xb0\phantom{\rule{0ex}{0ex}}2x=180\xb0-150\xb0\phantom{\rule{0ex}{0ex}}2x=30\xb0\phantom{\rule{0ex}{0ex}}x=\frac{30\xb0}{2}\phantom{\rule{0ex}{0ex}}\mathit{x}\mathbf{=}\mathbf{15}\mathbf{\xb0}$

## ASA Theorem - Key takeaways

- ASA congruence theorem: Two triangles are congruent if two adjacent angles and the included side on one triangle are congruent to the two angles and included side of another triangle.
- ASA similarity theorem: Two triangles are similar if two corresponding angles of one triangle are congruent to the two corresponding angles of another triangle. Also, the corresponding sides are proportional.
- ASA similarity is mostly known as the AA similarity theorem.
- ASA theorem is not the same as the AAS theorem.

###### Learn with 11 ASA Theorem flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about ASA Theorem

What is the ASA theorem?

Two triangles are congruent if two adjacent angles and the included side on one triangle are congruent to the two angles and included side of another triangle.

How do you use ASA theorem?

ASA theorem is used to find congruence between two triangles if two angles and included side are given.

How do you know if a triangle is ASA?

If two angles and the included side are given and they are equal to the corresponding angles and side, then that triangles are ASA.

How do you prove ASA theorem?

ASA theorem is proved with the help of the SAS theorem.

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more