Congruence

Congruent figures

Using the sliders to the left, what combination of translations, rotations, and reflections do you need to make the red sailboat coincide with (be the same as) the black sailboat? (When giving the measure of a rotation, remember that we call counterclockwise rotations positive and clockwise rotations negative.)

Two figures are congruent if there is some rigid motion (some combination of translations, rotations, and reflections) which makes them coincide with each other. Are the red sailboat and the black sailboat congruent?

The controls to the left control the motion of the red sailboat. The blue, green, and purple sailboats remain in the same place.

Which of the other three sailboats is the red sailboat congruent to: the blue sailboat, the green sailboat, or the purple sailboat?

Give the rigid motion which makes the red sailboat coincide with the other sailboat it is congruent to.

Triangle congruence — SAS

Say you have two triangles $\cl"red"{▵ABC}$ and $\cl"blue"{▵A'B'C'}$, and you know some of their measurements (side lengths or angle measures). (A symbol like $\cl"blue"{A'}$ — pronounced “$A$ prime” — is another way to name a point, like $A$ or $B$.) Sometimes it’s possible to decide that the triangles are congruent just by looking at those measurements, without actually finding a rigid motion which makes them identical.

Two triangles $\cl"red"{▵ABC}$ and $\cl"blue"{▵A'B'C'}$ (pronounced “triangle $A$ prime $B$ prime $C$ prime”) are drawn to the left, along with rays from $\cl"red"A$ through $\cl"red"C$ and from $\cl"blue"{A'}$ through $\cl"blue"{C'}$. There are also controls which move the red triangle $▵ABC$. Notice that the length of $\cl"red"\ov{AB}$ is equal to the length of $\cl"blue"\ov{A'B'}$, and the length of $\cl"red"\ov{AC}$ is equal to the length of $\cl"blue"\ov{A'C'}$. Also, the angles $\cl"red"{∠A}$ and $\cl"blue"{∠A'}$ between those two sides are equally large.

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

Translate the red triangle so that the point $\cl"red"A$ coincides with the point $\cl"blue"{A'}$. How far in each direction do you have to translate it?

Now rotate the red triangle so that the point $\cl"red"B$ coincides with the point $\cl"blue"{B'}$. How much do you have to rotate it?

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You have made $\cl"red"\ov{AB}$ coincide with $\cl"blue"\ov{A'B'}$. This is possible because they have the same length. Does $\cl"red"{∠A}$ coincide with $\cl"blue"{∠A'}$? (That is, do those angles exactly cover each other? Don’t reflect across $\ov{AB}$ yet.)

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

Are the points $\cl"red"C$ and $\cl"blue"{C'}$ the same distance from $\cl"red"A$ and $\cl"blue"{A'}$ along the light-colored ray?

Do the points $\cl"red"C$ and $\cl"blue"{C'}$ coincide?

If you have two triangles with corresponding measurements equal for two sides and the angle between them, 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, make the equal-measured angles’ vertices (corners) coincide, as you did for $\cl"red"A$ and $\cl"blue"{A'}$.
• Using a rotation, make one pair of 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 $\cl"blue"{∠A'}$.
• The other pair of equally long sides will now also coincide, like the sides $\cl"red"\ov{AC}$ and $\cl"blue"\ov{A'C'}$ here. Therefore all the pairs of corresponding vertices of the two triangles must coincide, like the pairs $\cl"red"A$ and $\cl"blue"{A'}$, $\cl"red"B$ and $\cl"blue"{B'}$, and $\cl"red"C$ and $\cl"blue"{C'}$ here. So the two original triangles are congruent.

This is called the Side-Angle-Side or SAS rule for congruence of triangles. It gives you one way to know that two triangles are congruent without actually finding a rigid motion connecting them.

Now you can see a different pair of triangles $\cl"red"{▵ABC}$ and $\cl"blue"{▵A'B'C'}$, along with rays from $\cl"red"A$ through $\cl"red"C$ and from $\cl"blue"{A'}$ through $\cl"blue"{C'}$, and controls which move the red triangle $▵ABC$.

Using the same procedure as in sas-qn-1, find a rigid motion that makes the red and blue triangles coincide.

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 SAS rule: the blue triangle ($\cl"blue"{▵A'B'C'}$) or the green triangle ($\cl"green"{▵A''B''C''}$)? ($A''$ is pronounced “A double-prime.”)

A triangle $\cl"red"{▵ABC}$ is drawn to the left, along with the ray from $\cl"red"A$ through $\cl"red"C$. By sliding the sliders, you can change the lengths of the sides $\cl"red"\ov{AB}$ and $\cl"red"\ov{AC}$, and the measure of the angle $\cl"red"{∠A}$ between them. The ray shows all the possible locations of $\cl"red"C$ for a given measure of $\cl"red"{∠A}$. The approximate length of each side and measure of each angle of the triangle are also shown.

For any lengths of $\cl"red"\ov{AB}$ and $\cl"red"\ov{AC}$ and measure of $\cl"red"{∠A}$ that you can slide the sliders to, is there always a triangle that has those measurements? (Don’t worry about measurements that are outside the range of the sliders.)

As the length of $\cl"red"\ov{AB}$ increases (while the other sliders stay fixed), does the measure of $\cl"red"{∠B}$ increase or decrease?

As the measure of $\cl"red"{∠A}$ increases (while the other sliders stay fixed), does the length $\cl"red"\ov{BC}$ increase or decrease?

As the length of $\cl"red"\ov{AC}$ increases (while the other sliders stay fixed), does the measure of $\cl"red"{∠B}$ increase or decrease?