Rules of Congruence

In the last lesson, we learned a rule for determining that two triangles are congruent without having to find a rigid motion which connects them: the Side-Angle-Side or SAS rule. In this lesson, we will look at some other rules for determining whether two figures are congruent.

Triangle congruence — ASA

Two triangles $\cl"red"{▵ABC}$ and $\cl"blue"{▵A'B'C'}$ are drawn to the left, along with controls which move the red triangle $▵ABC$. Notice that the angles $\cl"red"{∠A}$ and $\cl"blue"{∠A'}$ are equally large, as are the angles $\cl"red"{∠B}$ and $\cl"blue"{∠B'}$ and the sides $\cl"red"\ov{AB}$ and $\cl"blue"\ov{A'B'}$ which join them. We’ve also drawn rays from $\cl"red"A$ and $\cl"red"B$ through $\cl"red"C$, and from $\cl"blue"{A'}$ and $\cl"blue"{B'}$ through $\cl"blue"{C'}$.

Do the two triangles appear to be congruent (that is, do they look like they’re the same overall size and shape)?

Using only translation and rotation, find a rigid motion which takes both $\cl"red"A$ to $\cl"blue"{A'}$ and $\cl"red"B$ to $\cl"blue"{B'}$.

Does this rigid motion make $\cl"red"{∠A}$ and $\cl"blue"{∠A'}$ coincide?

Does it make $\cl"red"{∠B}$ and $\cl"blue"{∠B'}$ coincide?

Now reflect the red triangle across its side $\cl"red"\ov{AB}$. Does $\cl"red"{∠A}$ coincide with $\cl"blue"{∠A'}$?

Does $\cl"red"{∠B}$ coincide with $\cl"blue"{∠B'}$?

Do the corresponding light-colored rays coincide with each other?

In order for the triangles’ angles to have the indicated measures, the points $\cl"red"C$ and $\cl"blue"{C'}$ must both be on both of the light-colored rays. How many points are there on both of these rays?

If $\cl"red"C$ and $\cl"blue"{C'}$ are on both of these rays, do they have to coincide?

If you have two triangles with corresponding measurements equal for two of their angles and for the side between those two angles, as in $\cl"red"{▵ABC}$ and $\cl"blue"{▵A'B'C'}$, then you can always make the triangles coincide with a rigid motion in this way:

Using translations and rotations, make the equally long sides coincide, as you did for $\cl"red"\ov{AB}$ and $\cl"blue"\ov{A'B'}$.
Using a reflection if necessary, make the equal-measured angles coincide, as you did for $\cl"red"{∠A}$ and $\cl"blue"{∠A'}$, and $\cl"red"{∠B}$ and $\cl"blue"{∠B'}$.
Now the points $\cl"red"C$ and $\cl"blue"{C'}$ will both be on the same two rays. Since these two rays can intersect in at most one point, $\cl"red"C$ and $\cl"blue"{C'}$ must coincide. So the two original triangles are congruent.

This gives another rule which lets you see if two triangles are congruent. It is called the Angle-Side-Angle or ASA rule for congruence of triangles.

Three triangles are drawn to the left, along with some of their measurements. Which triangle must be congruent to the red triangle ($\cl"red"{▵ABC}$) by the ASA rule: the blue triangle ($\cl"blue"{▵A'B'C'}$) or the green triangle ($\cl"green"{▵A''B''C''}$)?

A triangle $\cl"red"{▵ABC}$ is drawn to the left, along with the rays from $\cl"red"A$ through $\cl"red"C$ and from $\cl"red"B$ through $\cl"red"C$. By sliding the sliders, you can change the lengths of the side $\cl"red"\ov{AB}$ and the measure of its surrounding angles $\cl"red"{∠A}$ and $\cl"red"{∠B}$. The rays show all the possible locations of $\cl"red"C$ for a given measurement of $\cl"red"{∠A}$ or $\cl"red"{∠B}$. The approximate length of each side and measure of each angle of the triangle are also shown.

For any length of $\cl"red"\ov{AB}$ and measures of $\cl"red"{∠A}$ and $\cl"red"{∠B}$ that you can slide the sliders to, is there always a triangle that has those measurements?

Give a length of $\cl"red"\ov{AB}$ and measures of $\cl"red"{∠A}$ and $\cl"red"{∠B}$ which do form a triangle.

Length of $\cl"red"\ov{AB}$:	units
Measure of $\cl"red"{∠A}$:	˚
Measure of $\cl"red"{∠B}$:	˚

Give a length of $\cl"red"\ov{AB}$ and measures of $\cl"red"{∠A}$ and $\cl"red"{∠B}$ which do not form a triangle.

Length of $\cl"red"\ov{AB}$:	units
Measure of $\cl"red"{∠A}$:	˚
Measure of $\cl"red"{∠B}$:	˚

Triangle congruence — SSS

The SAS and ASA rules for triangle congruence involve measuring both angles and sides. Now we’ll look at a rule where only sides are measured.

Two triangles $\cl"red"{▵ABC}$ and $\cl"blue"{▵A'B'C'}$ are drawn to the left, along with controls which move the red triangle $▵ABC$. Notice that the sides $\cl"red"\ov{AB}$ and $\cl"blue"\ov{A'B'}$ are equally long, as are the sides $\cl"red"\ov{AC}$ and $\cl"blue"\ov{A'C'}$, and the sides $\cl"red"\ov{BC}$ and $\cl"blue"\ov{B'C'}$. We’ve also drawn circles centered at the points $\cl"red"A$ and $\cl"red"B$ passing through the point $\cl"red"C$, and circles centered at the points $\cl"blue"{A'}$ and $\cl"blue"{B'}$ passing through the point $\cl"blue"{C'}$.

Do the two triangles appear to be congruent?

Using only translation and rotation, find a rigid motion which takes both $\cl"red"A$ to $\cl"blue"{A'}$ and $\cl"red"B$ to $\cl"blue"{B'}$.

After this rigid motion, are the points $\cl"red"C$ and $\cl"blue"{C'}$ on the same side of $\cl"red"\ov{AB}$?

Do you need to use a reflection to make $\cl"red"{▵ABC}$ and $\cl"blue"{▵A'B'C'}$ coincide?

Do the corresponding light-colored circles coincide with each other?

In order for the triangles’ sides to all have the indicated lengths, the points $\cl"red"C$ and $\cl"blue"{C'}$ must both be on both of the light-colored circles. How many points are there on both of these circles?

How many of those points are on each side of $\cl"red"\ov{AB}$?

If $\cl"red"C$ and $\cl"blue"{C'}$ are on both of these circles, and on the same side of $\cl"red"\ov{AB}$, do they have to coincide?

If you have two triangles with all three corresponding side measurements equal, as in $\cl"red"{▵ABC}$ and $\cl"blue"{▵A'B'C'}$, then you can always make the triangles coincide with a rigid motion in this way:

Using translations and rotations, make one pair of sides coincide, as you did for $\cl"red"\ov{AB}$ and $\cl"blue"\ov{A'B'}$.
Using a reflection if necessary, put the remaining vertex of each triangle on the same side of the coinciding sides, as you did for $\cl"red"{C}$ and $\cl"blue"{C'}$.
Now the points $\cl"red"C$ and $\cl"blue"{C'}$ will both be on the same two circles, and on the same side of $\cl"red"\ov{AB}$ (or $\cl"blue"\ov{A'B'}$). These two circles can intersect in at most one point on each side of $\cl"red"\ov{AB}$, so $\cl"red"C$ and $\cl"blue"{C'}$ must coincide. So the two original triangles are congruent.

This gives you the Side-Side-Side or SSS rule for congruence of triangles.

Three triangles are drawn to the left, along with some of their measurements. Which triangle must be congruent to the red triangle ($\cl"red"{▵ABC}$) by the SSS rule: the blue triangle ($\cl"blue"{▵A'B'C'}$) or the green triangle ($\cl"green"{▵A''B''C''}$)?

A triangle $\cl"red"{▵ABC}$ is drawn to the left, along with the circles centered at $\cl"red"A$ and $\cl"red"B$ passing through $\cl"red"C$. By sliding the sliders, you can change the lengths of the sides of the triangle. The circles show all the possible locations of $\cl"red"C$ for a given measurement of $\cl"red"\ov{AC}$ or $\cl"red"\ov{BC}$. The approximate length of each side and measure of each angle of the triangle are also shown.

For any side lengths of $\cl"red"{▵ABC}$ that you can slide the sliders to, is there always a triangle that has those measurements?

Give lengths for the sides of $\cl"red"{▵ABC}$ which do form a triangle.

Length of $\cl"red"\ov{AB}$:	units
Length of $\cl"red"\ov{AC}$:	units
Length of $\cl"red"\ov{BC}$:	units

Give lengths for the sides of $\cl"red"{▵ABC}$ which do not form a triangle.

Length of $\cl"red"\ov{AB}$:	units
Length of $\cl"red"\ov{AC}$:	units
Length of $\cl"red"\ov{BC}$:	units

Non-congruent figures

The SSS rule lets you see that two triangles are congruent by only looking at their sides. Let’s see if there could be a way to see that two triangles are congruent by only looking at their angles.

The two triangles drawn to the left have the same angle measurements as each other, for all their angles. (That is, $\cl"red"{∠A}$ and $\cl"blue"{∠A'}$ have the same measurements, as do $\cl"red"{∠B}$ and $\cl"blue"{∠B'}$, and $\cl"red"{∠C}$ and $\cl"blue"{∠C'}$.)

Do the two triangles appear to be congruent?

Find a translation of the red triangle which makes the points $\cl"red"A$ and $\cl"blue"{A'}$ coincide.

How much do you need to rotate the red triangle so that the light-colored rays (along the sides $\cl"red"\ov{AB}$ and $\cl"blue"\ov{A'B'}$) coincide?

Once you have done that, are the points $\cl"red"B$ and $\cl"blue"{B'}$ the same distance along the ray?

Is there a rigid motion of the red triangle which makes it coincide with the blue triangle?

Are the red and blue triangles congruent?

Because we can find triangles which have the same angle measurements but are not congruent, there can be no Angle-Angle-Angle (AAA) congruence rule for triangles.

A quadrilateral (four-sided figure) $\cl"red"{ABCD}$ is drawn to the left. The slider below the quadrilateral changes the measure of $\cl"red"{∠A}$.

As you slide the slider, do the lengths of the sides of the quadrilateral change or stay the same?

Are all the quadrilaterals that you get by sliding the slider congruent to each other? (Rigid motions don't change angle measurements, so congruent quadrilaterals will have the same angle measurements as each other.)

Could there be an SSSS (Side-Side-Side-Side) rule for congruence of quadrilaterals?

For figures that are not triangles, it is difficult to come up with a congruence rule like the ones from this lesson. Usually, if you want to show that two non-triangular figures are congruent, you need to find a rigid motion which makes them coincide.