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.
