### Video Transcript

In this video, we will learn how to prove that two triangles are congruent using either the side-side-side, side-angle-side, or the right-angle-hypotenuse-side criterion. We can recall that congruent means the same shape and size. Generally, if we want to show that two polygons are congruent, we need to prove that all the corresponding pairs of sides are congruent and that all the corresponding pairs of angles are congruent. So when it comes to triangles, in theory, we would need to show that there are three pairs of corresponding sides congruent and that there are three pairs of corresponding angles congruent. We can use the notation SSS to represent the three sides and AAA to represent the three angles.

However, if we want to show congruence in triangles, instead of having to show that there are three pairs of corresponding sides congruent and three pairs of corresponding angles congruent, there are some shortcuts that we can take. The first thing we’re going to check is the SSS rule to see if this by itself is enough to demonstrate congruence. So let’s take any triangle and see if we can create another noncongruent triangle that has the same length of sides as this one. Let’s start by drawing our blue line of the same length. We’re not constraining the angles in anyway, so we know that the second orange length could be in any position.

The same is true for the third pink side. It can be at any angle, but we’re just constraining the length of it. With the two lines in this position, however, they’re clearly not going to create a triangle. There is in fact just one way to create a triangle. And that triangle will be identical to the original triangle. In fact, if we were to think of constructing this as a triangle using a compass. Then once we’ve drawn our blue line at the base and put the pointed end of the compass at the end of the line, setting the distance between the point and the tip of the pencil to be the same as the orange line. Then the arc that’s created would represent all the different positions that that line could take.

In order to work out the position of the pink line, we’d put the pointed end at the other end of the blue line, set the distance between the tip of the compass and the tip of the pencil to be the same as the pink third line. And we’d have an arc that looks like this. Joining the point of intersection would create the triangle. The other triangle we could’ve created using this construction method would be the same shape and size, but it would’ve just looked like it had been flipped.

We can still say that shapes are congruent, even if one is a rotation or looks like a reflection of another one. Even though this isn’t a proof, it is a demonstration that if we were to construct the second triangle on top of the original triangle, we would have an identical triangle. Therefore, if we have a pair of triangles and we can prove that there are three pairs of corresponding sides congruent, then the triangles themselves would be congruent.

Let’s have a look at another rule. We’re going to look at the side-angle-side rule. The angle will be the angle that’s included between the two sides. If we take a different triangle and constrain two sides and the included angle, could we create a noncongruent triangle with these two sides and included angle? In the second triangle, all we’re saying is that the orange line with the two marks and the pink line with the one mark have to be the same corresponding size. And the angle in between them has to be the same size. The third line can be any size and in any angle we wanted. All what we’re saying has to be the same is the orange line, the pink line, and the green angle.

We do, of course, have to actually form a triangle. And the only way to do that is by joining the vertices and thus creating a triangle that has three pairs of corresponding sides equal. And while, once again, this isn’t a proof, it is a demonstration that the side-angle-side rule would show congruence. Let’s take a look at another rule.

This rule is one which only applies in right triangles. The letters RHS represents a right angle, hypotenuse, and side. If we constrain the hypotenuse, that’s the longest side, and either of the other two sides plus the right angle, could we create a noncongruent triangle? In the same way as before, we’re constraining the lengths of these two sides, so the pink line could be in any angle. But we know that there has to be a right angle on the other side. The line coming from this point could be of any length, as we haven’t constrained that.

As we can probably see, there’s only one way to create a triangle. And that triangle would be congruent to the original triangle. Although this isn’t a proof, the Pythagorean theorem would also help to demonstrate this, which says that the square on the hypotenuse is equal to the sum of the squares on the other two sides.

In our original triangle, we’re constraining the hypotenuse and one of the other sides. Then, as we know that we’ll still have a right triangle with 90 degrees. Then the third side must be the same size as it was in the original triangle. And we know that the SSS rule, the three sides, is sufficient to show congruence. Therefore, we can demonstrate congruence in a pair of triangles by showing that they both have right angles. The hypotenuse in each triangle is congruent. And there’s another pair of corresponding sides congruent.

So now we have seen these three rules— the SSS rule, SAS rule, and RHS rule — we could look at some questions.

Which congruency criterion can be used to prove that the two triangles in the given figure are congruent? Option (A) SAS, option (B) ASA, option (C) SSS.

In this question, we’re asked how we might determine if the two triangles are congruent. We can see some examples of some criterion that we could use. We can recall that the S represents a side and A would represent angle. So let’s look at our diagram and see if we can determine any corresponding pairs of sides or angles which are congruent.

We can see the length 𝐴𝐶 is 5.52. And this would be congruent with the line 𝐴 prime 𝐶 prime. So we have a pair of corresponding congruent sides. The line 𝐴𝐵 is marked as 1.93, and so is the line 𝐴 prime 𝐵 prime. So we have another pair of corresponding sides. The final two pairs of sides, 𝐵𝐶 and 𝐵 prime 𝐶 prime, are marked as 3.75, the same length. Showing that three pairs of corresponding sides are congruent would demonstrate that two triangles are congruent. Therefore, the congruency criterion which we could use is the SSS rule, and that was given in option (C).

In the given quadrilateral, 𝐴𝐹 and 𝐵𝐹 have the same length and 𝐸𝐹 and 𝐶𝐹 have the same length. Which angle has the same measure as angle 𝐴𝐹𝐸? Hence, are triangles 𝐴𝐹𝐸 and 𝐵𝐹𝐶 congruent? If yes, state which congruence criterion proves this.

In this diagram, we can see that there’s a quadrilateral which has some triangles within it. We’re told that there are some line segments which have the same length. So it’s always worthwhile putting this onto a diagram if they’re not already marked. 𝐴𝐹 and 𝐵𝐹 are the same length and 𝐸𝐹 and 𝐶𝐹 are the same length. In the second question, we’ll look at congruency. But the first question asks us about angle measures. Which angle would be the same as angle 𝐴𝐹𝐸? We’re not given any angle measurements in this diagram, but we should recall that angles which are vertically opposite will be equal. So angle 𝐵𝐹𝐶 would also be the same measurement. And that’s our answer for the first part of the question.

In the second part of the question, we need to check if triangles 𝐴𝐹𝐸 and 𝐵𝐹𝐶 are congruent. So let’s note down any sides or angles that we know are congruent. We were told in the question that 𝐴𝐹 and 𝐵𝐹 are the same length. We have shown in the first part of the question that we have two congruent angles, angle 𝐴𝐹𝐸 and angle 𝐵𝐹𝐶. And we were told that sides 𝐸𝐹 and 𝐶𝐹 are the same length.

And so, we have two pairs of congruent sides equal and a pair of congruent angles. Importantly, the angle is included between the two sides, which means that we can use the SAS congruency criterion. If the angle wasn’t included between the two sides, then it wouldn’t be sufficient to show congruence. So our answer for this part of the question is, yes, these two triangles are congruent, and we use the SAS criterion.

Let’s have a look at another question where we use congruency to find a missing length.

Find the length 𝐶𝐷 in the given figure.

In this question, we need to find this missing length 𝐶𝐷 on the triangle. We might already think that we know the answer, but let’s see if we can prove this mathematically. We may be able to demonstrate congruence using one of the congruency criterion. If we can demonstrate that triangle 𝐴𝐵𝐶 is congruent to triangle 𝐷𝐵𝐶, then we could find our missing length 𝐶𝐷. So let’s write down what we know about any angles or sides in these two triangles. We can start by noting that our line 𝐴𝐵 and our line 𝐵𝐷 are both five centimeters. We have a common side. 𝐵𝐶 appears in both triangles, so it would be the same length. And we also have two congruent angles. Angle 𝐴𝐶𝐵 and angle 𝐵𝐶𝐷 are both marked as a right angle of 90 degrees.

What we have shown here, however, is that we have two sides and a nonincluded angle. SSA is not a congruency criterion. So let’s look again at what we have. This time, we’re going to note that the angle that we have is actually a right angle. The common side 𝐵𝐶 is a side. But the pair of sides that we have found, 𝐴𝐵 and 𝐵𝐷, aren’t just any sides. They’re actually the hypotenuse of the triangles. We have then found that these two triangles are congruent using the right-angle-hypotenuse-side criterion. Therefore, our missing length 𝐶𝐷 must be the same as the length 𝐴𝐶. And therefore, we’ve used congruence to prove that our length 𝐶𝐷 is four centimeters.

We’ll now look at one final question on congruency.

In the given figure, triangle 𝐴𝐵𝐶 and 𝐵𝐶𝐷 have two equal sides and share one equal angle. Are triangles 𝐴𝐵𝐶 and 𝐵𝐶𝐷 congruent?

Let’s start by highlighting our triangles. We have the slightly larger triangle 𝐴𝐵𝐶 and then we have this slightly smaller triangle marked in orange, 𝐵𝐶𝐷. As we can clearly see, these are different sizes and therefore would not be congruent. But as we indeed can see, we do have some corresponding sides of equal length. And we do have a corresponding angle of equal length. Perhaps we find some sort of problem with the congruency rules. So let’s note down what we know about each triangle and see if we can work out what’s happening.

We can see in our diagram that we have two lengths marked as two units, the length 𝐴𝐵 and the length 𝐵𝐷, which we could then write as congruent. The line 𝐵𝐶 of length four occurs in both triangles. Finally, we can see that we have a common angle. The angle 𝐴𝐶𝐵 would be equal to the angle 𝐵𝐶𝐷. It might be tempting then to say that we have a congruency because of a real SSA. But in fact, this is not a congruency rule. Because as we can see from our diagram, we could in fact create two noncongruent triangles using two corresponding pairs of sides congruent and a pair of corresponding angles congruent.

We could have used the congruency rule SAS if the angle was included between the two sides. But it’s not in this diagram. So therefore, we can happily say that our two triangles are not congruent. They didn’t look congruent. And even in the case of a badly drawn diagram, we can’t prove that they’re congruent.

We can now summarize some key points of this video. Our first congruency criterion that we saw was the SSS rule, which means that if we can see that there are three corresponding pairs of sides congruent, then two triangles would be congruent. We saw the side-angle-side criterion, remembering that the angle has to be included between the pairs of corresponding sides. We saw the RHS criterion representing right angle, hypotenuse, and side which only works for right triangles.

And as we saw in a number of examples, SSA is not a congruency criterion. Any corresponding congruent sides and a pair of angles would have to have the angle included. And finally, we can note that when we’re looking at congruency, shapes may still be congruent even though they are a rotation or a reflection of one another.