SAS Theorem

SAS congruence theorem

SAS congruence theorem gives the congruent relation between two triangles. When all the corresponding angles are congruent to each other and all the corresponding sides are congruent in two triangles, those triangles are said to be congruent. But with the SAS congruence theorem, we only consider two sides and one angle to establish the congruence between the triangles.

Here as the name suggests, SAS stands for Side-Angle-Side. When using the SAS congruence theorem, we consider two corresponding adjacent sides and the angle included between those two sides.

Mathematical we represent as, if $AB=XY,\angle A=\angle X,AC=XZ,$ then $△ABC\cong △XYZ.$

SAS congruent triangles, Mouli Javia - StudySmarter Originals

If the SAS congruence theorem satisfies for any two triangles, then we can directly say that all the sides and angles of one triangle will be equal to the other triangle respectively.

SAS similarity theorem

We can conclude two triangles are similar using the SAS similarity theorem. Usually, we need information about all the sides and angles of both the triangles to prove them similar. But with the help of the SAS similarity theorem, we only consider two corresponding sides and one corresponding angle of these triangles.

As SAS triangles have two sides, we can take the proportion of these sides to show the similarity between the two triangles.

Mathematically we say that, if $\frac{AB}{DE}=\frac{BC}{EF}$ and $\angle B=\angle E,$ then $△ABC~△DEF.$

SAS similarity theorem, Mouli Javia - StudySmarter Originals

SAS theorem proof

Let us look at the SAS theorem proof for both congruence and similarity.

SAS congruence theorem proof

Let's perform an activity to prove the SAS congruence theorem. From the statement of the SAS congruence theorem, it is given that $AB=XY,AC=XZ,$ and $\angle A=\angle X.$

SAS congruence theorem, Mouli Javia - StudySmarter Originals

To prove: $△ABC\cong △XYZ$

We take a triangle $△XYZ$ and place it over the other triangle $△ABC.$ Now it can be seen that the B coincides with Y as $AB=XY.$ As $\angle A=\angle X$ and when both triangles are placed over each other, AC and XZ will fall alongside each other. As $AC=XZ,$ point C coincides with Z. Also BC and YZ will coincide with each other.

So clearly, both triangles $△ABC$ and $△XYZ$ coincide with each other. Hence $△ABC\cong △XYZ$.

ASA similarity theorem proof

It is given from the statement of similarity theorem that $\frac{AB}{DE}=\frac{BC}{EF}$ and $\angle B=\angle E.$

To prove: $△ABC~△DEF$

SAS similarity triangle with constructed line PQ, Mouli Javia - StudySmarter Originals

Consider a point P on DE at such a distance such that $AB=EP.$ Then join a line segment from P to EF at point Q such that $PQ\parallel DF.$ As $PQ\parallel DF,$ we get that $△PEQ~△DEF.$

Then by Basic Proportionality Theorem,

$\frac{EP}{DE}=\frac{EQ}{EF}\left(1\right)$

We are already given that,

$\frac{AB}{DE}=\frac{BC}{EF}\left(2\right)$

As $AB=EP$ and from equation (1) and equation (2),

$$\begin{gathered} \frac{EP}{DE}=\frac{AB}{DE}=\frac{EQ}{EF}=\frac{BC}{EF} \\ \Rightarrow EQ=BC \end{gathered}$$

And we also have that $\angle B=\angle E.$

So using the SAS congruence theorem from the above information we get that $△ABC\cong △PEQ.$

$\Rightarrow △ABC~△PEQ$

We already have that $△PEQ~△DEF.$ So from the both obtained similarity we get that $△ABC~△DEF.$

SAS theorem formula

The SAS theorem is not only used to show congruence and similarity between two triangles, but we get the SAS theorem formula from it. This SAS formula can be very helpful in trigonometry to calculate the area of a triangle. This formula uses trigonometry rules to find the area of the triangle.

The SAS theorem formula for the triangle is expressed as :

Area of triangle$=\frac{1}{2}\times a\times b\times \sin x,$ where a and b are the two sides of SAS triangle and x is the measure of the included angle.

SAS triangle, Mouli Javia - StudySmarter Originals

Let us derive the SAS theorem formula. In the SAS $△ABC$ construct a perpendicular from point A onto the line BC at D. Now as $△ABD$ forms a right triangle, we can use trigonometric ratios with $\angle B$ as the angle.

$$\begin{gathered} \Rightarrow \sin x=\frac{AD}{AB} \\ \Rightarrow p=a\times \sin x\left(1\right) \end{gathered}$$

Also, we know that the general formula to calculate any triangle is

Area $=\frac{1}{2}\times base\times height$

Now in the $△ABC,$ b is the base and p is the height. Then substituting this value in the formula of are we get,

Area $=\frac{1}{2}\times b\times p\left(2\right)$

From equation (1) we know the value of p, so we substitute that in the above-obtained equation (2) of area.

Area $=\frac{1}{2}\times b\times a\times \sin x$

Hence the formula for the area of the triangle using the SAS theorem is $\mathit{A}\mathit{r}\mathit{e}\mathit{a}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\times }\mathit{a}\mathbf{\times }\mathit{b}\mathbf{\times }\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{x}.$

SAS theorem examples

Here are some of the SAS theorem examples to understand the concept better.