ASA Theorem

We know that triangles can be congruent and also similar to each other. And we always consider all its sides and angles to prove it. But not anymore, here we will learn a triangle criterion by which we can easily prove congruent triangles.

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 triangles, StudySmarter Originals

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.

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

ASA Similarity triangles, StudySmarter Originals

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" $180°$. 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.

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

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

ASA congruence triangles, StudySmarter Originals

ASA theorem proof

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

ASA similarity theorem proof

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

To prove: $△ABC~△XYZ.$ And $\frac{AB}{XY}=\frac{BC}{YZ}=\frac{AC}{XZ}.$

ASA triangle with constructed line , StudySmarter Originals

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

We will construct a line PQ in triangle $△XYZ$ 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 $△ABC~△XPQ.$

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

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 $△XYZ,$

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

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 $△ABC~△XYZ.$

ASA congruence theorem proof

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

ASA congruent triangles, StudySmarter Originals

To prove: $△ABC\cong △LMN$

To prove the above statement we will consider different cases.

Case 1: Assume that $AB=LM.$

ASA congruent triangles with AB and LM equal, StudySmarter Originals

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

Case 2: Suppose $AB>LM.$

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

ASA congruent triangle with constructed point X, StudySmarter Originals

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

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

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

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

$$\begin{gathered} \angle ACB=\angle LNM=\angle XCB \\ \Rightarrow \angle ACB=\angle XCB \end{gathered}$$

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 $△ABC$ and $△XCB$ are equal. Hence we can consider only one triangle $△ABC$such that $AB=LM.$ So this is the same as case 1, and from that we get that $△ABC\cong △LMN.$

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.

ASA congruent triangles with constructed point Y, StudySmarter Originals

Hence we get that $△ABC\cong △LMN.$

ASA Theorem example

Let us see some examples related to ASA theorems.