Lesson video

Hello, I'm Mrs. Lashley and I'm gonna be working with you as we go through the lesson today.

I really hope you're ready to try your best as we go through the lesson.

In today's lesson, we're going to be looking at and using the criteria to prove that two triangles are congruent by SAS.

On the screen is the definition of congruent, and it means that if two shapes can fit exactly on top of each other, whether they're rotated, reflected, or translated.

Our lesson has got two parts.

The first part, we are gonna think about justifying congruence by SAS, and the second part we'll be looking at two sides and an angle.

So we're gonna make a start on justifying congruence of two triangles by using SAS.

Jun, Sam and Sofia have got three straws of various lengths, so each of the straws are different in their lengths, and they are creating triangles.

They know that they can only create one triangle with these three straw lengths, and that's because of SSS.

So Sam says, "If we've had two straws "and taped them into a fixed angle, "how many different triangles can we make?" So this is what Sam is suggesting.

So they've got two straws, they're a fixed length and they've taped them together so that this angle is created.

Sofia says, "Well, I think there'll be lots "because there are many different right angle triangles, "so there's lots of triangles that have a right angle "and they're very different to each other." Jun says, "But the straws are a fixed length though." So does that have any handle on how many we can create? "If the straws could be any length," so if these straws could change length, "but the angle stays the same so that taped angle is fixed, "then lots of different triangles would be created?" Sofia's like "Yes," she thinks we can show that through a construction.

Jun believes that even if we fixed one side length, so Sam was suggesting if both straw lengths could change, then we get lots of different triangles, but Jun's suggesting that even if we were to fix one of them, then you would still get an infinite amount of triangles.

I've just opened the link that is on the slide and it takes us to a Geogebra file, which is gonna allow us to look at what they've just both suggested.

What you can do on this Geogebra profile, and you can have a go as well, is you can decide on the angle that you want to fix, so we can change the angle between these two purple line segments, and see how many different triangles we can make from fixing the angle.

We can then fix one edge length, which Jun suggests means we still could have an infinite amount.

Okay, so I fixed my angle at 54 degrees.

It can be anything, but I fixed it as 54 degrees.

And with no other limits, so without having to have an edge at specific length, then you can see that there's many triangles that can be created.

So yes, if we only have a fixed angle, then there's an infinite amount of triangles that can be created.

Some of them will be reflections of each other, but there is an infinite amount.

So let's see what Jun suggested.

Jun suggested if we were to fix one of the edges, then we still could get an infinite amount.

So let's fix AB.

It's fixed at 11, that's fine by me.

So AB is now fixed to 11.

How many different triangles can we have where there is an angle of 54 degrees and an edge of 11? Well, I can't move B because B is now in a fixed position.

I can rotate by moving A, but that B is unchanged and actually that triangle didn't change by me just rotating and moving A, but I can move C.

So Jun's correct, we have still got an infinite amount of triangles.

So we're gonna look at this from an animation point of view.

We've seen it on the Geogebra file.

So if the angle between two edges is fixed at 50 degrees, we can draw the legs as rays starting from point A.

So a ray starting from point A going off to the right, we can use our protractor to mark our 50 degree and a ray there.

So that fixed the angle of 50.

So this is where the straws could be any length, but the angle between them 50 degrees, and we need to take a point from each ray to join up to create our triangle.

So I could have this triangle, but there are infinite number because I could have a like this or like this.

So we've sort of seen this on that Geogebra file, that just by having the angle fixed and everything else free, then we can make many triangles.

But Jun had suggested that we fix one.

So let's fix one of the lengths to be seven centimetres.

So if AC is seven centimetres, then B can still go anywhere on that ray.

So this triangle has an AC of seven centimetres, an angle of 50 degrees, but so does this triangle, as does this triangle.

So we still have an infinite number of possible triangles.

And remember we are trying to guarantee and justify congruence by SAS.

This time we're gonna fix BC.

So that's not an edge that is on the rays.

BC is the edge that's created by joining two points.

This Geogebra file has been made so that we can look at that point exactly that we've got a fixed angle.

Again, you can change what that angle is, but we've got a fixed angle and we are fixing the length BC and BC is the edge opposite the angle that you have fixed.

So we've fixed it at nine.

So by moving B along the ray, we have two triangles appearing, and that's why there is a C1 and a C2.

There are actually two different positions that the vertex C could take so that we have the angle of 61.

2 degrees and the length BC being nine.

And that's because of the circle that's creating it.

So you can see the closer I'm dragging that towards A, along the ray there's a little bit more overlap on the other side.

So if we go back to the animation on this, BC here is five centimetres, this is also BC of five centimetres, BC being five centimetres.

So the point where B is can change and C would be in different positions.

So this still leads to an infinite amount of possible triangles.

So having the fixed edge opposite the angle that you fix still leads to an infinite amount of possible triangles.

But if we go back to the straws being a fixed length and then an angle created, we know that that will give us just one triangle.

So if we make AC seven centimetres and AB 6.

2 centimetres, there is only one line segment that can join those up and that's because the angle has been fixed as 50 degrees.

If the angle could change, then BC would change as well, but because our angle is fixed and our two sides are fixed, then there is only one line segment that will join them up.

This Geogebra file will allow us to see this for a variety of different lengths.

So that the one I've just demonstrated with with particular lengths is not just a fluke, that this actually is true for all.

So again, I can change the angle.

So if I fix it at 54 degrees, I want to fix both of the edges.

So if I fix AB to be 11, then I have an infinite amount of triangles.

But if I also fix the other side, AC, so these are our two straws, if we go back to them at the beginning of the lesson, the two straws are a fixed length and then we take them together to cause an angle or the angle here is 54 degrees, one straw is 11, and the other straw is 6.

5, then there is only one edge or one straw that would be able to be joined up to make a triangle.

I no longer can move point B, vertex B or vertex C.

If I was to fix AB as 6.

5 and AC as 11, I have just created a reflection of the same triangle.

So this is still the same unique triangle with those two edge lengths as just a reflection.

So for that triangle, that BC would be 5.

6 centimetres.

So here's a check, when constructing, which of these conditions will always lead to one fixed triangle? So pause the video whilst you're thinking about that and then press play when you want to check.

B and C will lead you to one fixed triangle.

Part A wouldn't lead you to one fixed triangle.

That's where we've fixed one length and therefore we have an infinite amount.

BC can have a variety of different angles, whereas on B, you've got two fixed lengths, the angle between them and so that fixes the triangle.

And C is SSS, you've got three edge lengths, and therefore they are one triangle only.

So two triangles can be proved to be congruent if their corresponding edges are the same.

And we say they're congruent by SSS, side, side, side.

The angle opposite is therefore fixed too.

This triangle is fixed from having two edges and the angle between them, alternatively, by having the three edge lengths.

If we only have the three edge lengths, we know it would be a congruent triangle to another triangle that has the same three edge lengths because of SSS.

SSS means that the angles are also fixed.

Jun says, "If we know two edges and the angle between them, "there is only one possible triangle that can be made.

"All triangles with the same SAS," so the same side, angle, side, "will be congruent to it," and that's the reflections and the rotations.

One way of proving two triangles are congruent is by using SSS, and that's when you know the three edge lengths.

But if you don't know the three edge lengths but you do know two corresponding sides and the angle between them, then you can prove them congruent by SAS.

On these two triangles you can see that AB equals PR, angle BAC equals angle QPR and AC equals PQ.

And so we can say that these triangles are congruent by SAS.

Two corresponding sides are equal and a corresponding angle is equal.

These two triangles that has changed, these two triangles, we can see that AB equals XZ, AC equals XY and angle BAC equals angle XZY.

However, these two are not congruent.

The 89 degrees is not between the two known edges in both of the triangles.

So it's really important that to use SAS, that the angle between the two known corresponding edges is the same.

A check, here's a check for you.

Are these triangles congruent to each other? Pause the video, think about what you do know from the diagrams, what you could work out, and then when you're ready to check, press play.

So these are congruent.

At first glance, it's not obvious.

We've got two corresponding edges that are equal, but on the left hand triangle we had a 63 degree angle between the edges, which is what we need, and on the right hand triangle, on triangle TVU, we didn't have that angle in between those two edges, but we did have the other two angles of the triangle.

And whenever you know two angles in a triangle, you can work out the third.

So we can calculate the third, and the third is 63 degrees.

Therefore it is congruent by SAS.

So we don't need to know the actual values or the lengths or the size of the angles to be able to prove that two triangles are congruent.

Sometimes we can make use of properties of shape in order to be able to do this.

So prove that triangle ABC is congruent to triangle ACD given that ABCD is a rhombus.

So there we have a rhombus, a diagram of a rhombus and a diagonal drawn on, the diagonal's clearly split that into two triangles, and we are trying to prove that those two triangles are congruent to each other.

So AB is equal to CD as they are edges of a rhombus.

Angle ABC is equal to angle ADC as they are opposite angles in a rhombus.

And so that's a property of a rhombus.

Any rhombus, the opposite angles are equal.

So it doesn't matter what size they are, we just know that they have this relationship of being equal.

And AD is equal to BC as they are edges of a rhombus.

So we can say that the two triangles are congruent by SAS.

We know that two corresponding edges are equal and the angle between them is also equal.

We don't know the size of this one at all, but we do know the relationships so that will always be the case.

This is the same diagram, it's a rhombus.

We're trying to prove that the two triangles are congruent again, but we're doing it in a different way.

And you need to pause the video and decide what is missing as a justification on that line of the proof.

Press play when you're ready to check it.

So we've used the AB equals CD, they are edges of a rhombus and then angles DAC and angles ACB are equal.

Why are they equal? Well, they're equal because they are interior alternate angles in parallel lines.

Another property of a rhombus is that it has two sets of parallel lines.

And as soon as we've got parallel lines, then we can start to think about alternate angles, corresponding angles, co-interior angles.

So there's so much more to the rhombus that we can use from its sort of first glance.

So you're onto your first task of the lesson.

And so question one and two is on the screen.

Question one, you need to construct the triangles.

It might be that you want to sketch them first, you'll need a pencil, a ruler, a protractor, and a pair of compasses to do those.

And then question two, you need to measure the edges and angles that you have constructed in question one and decide if they are congruent triangles.

Press pause whilst you are working through those two questions.

And then when you're ready for the next question, press play.

So question three and four are both on the screen.

Question three, you need to prove that triangle ABC, which is the one on the left, is congruent to triangle PQR, the one on the right by SAS.

And question four is given a parallelogram ABCD, prove that triangle ABC is congruent to triangle ACD by SAS.

Again, I would encourage you to draw yourself a parallelogram, label up the vertices, make sure that you're going clockwise or anti-clockwise.

Don't just sort of jump around the vertices.

Press pause whilst you're having a go.

And then when you're ready for question five, press play.

Here we've got question five.

So point E is the midpoint of AC and BD.

Prove that triangle AEB is congruent to triangle ECD by SAS.

Pause the video whilst you're working through question five.

And then when you press play, we are gonna go through questions one, two, three, four and five.

Question one, you need to construct the triangle.

So this is triangle ABC, AB needed to be six centimetres, BC needed to be nine centimetres and angle ABC needed to be 42 degrees.

So you should have constructed this by drawing AB with a ruler, marking 42 degrees and drawing the ray and then using your ruler to mark where nine centimetres is to locate your vertex C.

And then finally you can join A and C.

You may have drawn a reflection of this or a rotation.

Part B, triangle PQR where PQ is six, QR is nine and angle QPR is 42.

So here you needed to find a point of R along that ray, and I've used a pair of compasses, you can see the arc there.

So I've set up my pair of compasses to nine centimetres, and therefore found where it intersects.

Otherwise it can can be quite a challenge with a ruler to keep rotating it to find exactly nine centimetres.

Question two, you needed to measure the edges and angles in the two triangles and decide if they were congruent.

So triangle ABC, all of the information is on the left hand side.

You will not have measured your angles to the same degree of accuracy.

So probably your angle BAC is about 97 degrees and angle ACB is probably 41 degrees or about 42 degrees.

They are not congruent.

And so you can see that they don't have the same three edges.

They've both got a six centimetre and they've both got a nine centimetre, but their third edge is different.

So automatically we can say that they're not congruent by SSS.

As well, because the edges then fix the angles by their edges being different, their interior angles are different.

So they do both have a 42 degree angle, but that's the only angle they have shared, the only common angle.

Question three, you needed to prove that the two triangles were congruent by SAS, and this is by using interior angles, adding up to 180 degrees in any triangle to calculate the missing angle on PQR.

And that way you can see that actually the 128 degrees is corresponding in both.

So AB equals PQ, angle ABC equals angle PQR, and BC equals QR.

So the triangle is congruent by side, angle, side.

Question four, I would've really encouraged you to sketch yourself, draw yourself a parallelogram, label it up to be able to start your proof.

I'm gonna go through a couple of ways that you may have done this.

So this is the first way that you could have approached this question.

AD is equal to BC as they are opposite edges on a parallelogram and are equal in length.

So that is a property of a parallelogram.

I've used a hash mark.

Angle ABC is equal to angle ADC as opposite angles in a parallelogram are equal in size.

Again, that is a property of a parallelogram.

AB equals DC as opposite edges on a parallelogram are equal in length, property of a parallelogram.

So you can see that those two triangles both have a one hash, a two hash edge, and the angle between them is the equal angle.

So hence triangle ABC is congruent to triangle ACD by SAS.

So that is one way that you may have proved those two triangles are congruent by SAS.

This is the second way.

So once again, starting the same, the AD equals BC because they're opposite edges of a parallelogram, then using interior alternate angles.

So angle DAC equals angle ACB, because they are equal interior alternate angles, and that's because the two edges are parallel to each other, a property of being a parallelogram.

AC is a shared side, it's a side that is on both of the triangles, so it must be the same length.

So triangle ABC is congruent to triangle ACD by SAS.

So that's an alternative way of you proving that those two triangles were equal.

This one could have you been done again, you could have a repeat of this, but instead using AB and DC as equal edges and also angle DCA and angle CAB being equal by alternate angles, and then AC being shared.

So there is sort of the third way that you could have done that.

Question five, point E is the midpoint of AC and BD, and that was the important word here, that it was the midpoint.

If it's a midpoint, that means that it is exactly halfway along.

So the distance from A to E is the same distance as E to C.

So DE equals EB as E is the midpoint of BD.

So making use of the fact that you were told explicitly that point E was a midpoint.

Angle AEB is equal to angle DEC because they're vertically opposite angles, and AE equals AC as E is the midpoint of AC.

So again, making use of that midpoint, we know that those two edges are equal.

So the two triangles are therefore congruence by SAS.

We've got a side, an angle, and a side that are corresponding and equal.

So we're now at the second part of the lesson, and this time we're gonna look at two sides and an angle.

You might be thinking we've just been looking at two sides and an angle.

So let's have a look at what do I mean by two sides and an angle.

Aisha and Jacob are discussing the amount of information we need to prove two triangles are congruent to each other.

So Aisha says, "If you know the corresponding angles "and edges are equal, then you can be sure "that they are congruent." So on these two triangles, if you know that C equals R and B equals Q and A equals P, and the angle A is equal to the angle P, and the angle B is equal to the angle Q, and the angle C is equals to the angle R, if you know that the information on the left hand triangle is exactly the same as the information on right hand triangle, then you can be completely sure that they are congruent.

Jacob says, "But you don't need all of that information." So you don't need all the edges and all the angles to be sure that two triangles are congruent.

So, "If you only know the three corresponding edge lengths, "then you can prove they're congruent by SSS." And Jacob says, "Yes, the angles are fixed "by the edge lengths, SSS." So already we've removed some of the information you need.

You don't need all angles and all edges, you just need all the edges.

That is one way of proving congruence.

Aisha says, "If you know two corresponding sides "and the angle between them, "then you can prove the two triangles are congruent." And that's SAS, Jacob has told us, so side, angle, side.

So that's what we were just working with, side, angle, side, two corresponding edges and the angle between them.

Aisha asks, "Do these two corresponding sides "and angle prove that they are congruent to each other?" So if we have this information, if we have two edges, but they're not the edges that are either side of the angle, Jacob's not sure, and he says, "There isn't one of each pair in this situation." Do you understand what he means by that? One of each pair.

Aisha doesn't, she said, "What do you mean?" He says, "Each angle is marked by a capital letter "and its opposite edge is the lowercase "of that same letter." So you can see here there's the capital B, which is angle B, and its opposite edge is lowercase b.

And on the right you've got the capital Q and its opposite edge is the lowercase q.

So when you know all three edges, you have the case of each letter, A, B, and C on one of these triangles and P, Q and R on the other.

And Jacob says, "So there is one of each letter," A, B, and C, P, Q, and R.

"Similarly, when you know two sides and the angle between." Aisha's starting to recognise what Jacob has said now, so, "Oh yes, side c and b and the angle A "on one of the triangles," so that means you've got an A, b and a c, "and sides r and q and the other angle P on the other." So P, Q and R.

Whereas this set, the one that's still on the screen, there's no C on the triangle on the left and there is no R on the one on the right.

So we don't have a letter from each.

So these two triangles both have a 50 degree angle, an edge that is five centimetres and an edge that is six centimetres, but they're not congruent.

You can see quite clearly that these are not congruent triangles.

The angles inside the other two angles are gonna be different.

The third edge is a different length.

So this is what we've just seen in terms of a circumstance of not having one of each pair.

We've got the angle 50 degrees and the opposite edge.

So that is the same letter, whatever letter, whatever triangle we label this to be, ABC, PQR, XYZ, that angle and its opposite edge are both known.

So they are the same letter.

So here we've got the two triangles we just saw, but now overlaid on each other.

So we can see that they both have that 50 degree angle, that edge length of six centimetres, and it's the position of their five centimetre edge that is different.

And that's due to the circle that that would be drawn from intersecting with the ray at two distinct points.

And those two distinct points are there marked.

Because we need that 50 degree angle, then that ray of 50 degrees is going off, the five centimetres intersects with it two different points.

So knowing two sides and an angle is not enough information to prove congruence.

The angle must be between the two known sides.

So SAS is a criteria where we've got side, angle, side and the angle is in between.

If you have two sides and an angle, so SSA, that isn't sufficient to prove congruence.

There is a link to a Geogebra file which will allow you to explore that a little bit more if you wanted to pause the video and click on that link.

So Aisha is now agreeing with Jacob, "We need a minimum "of three pieces of information and one from each pair." So there's a box that shows the SSS.

If you know all three edge lengths are the same, the corresponding edges are equal, then you've got an A, a B and a C.

They don't have to be A, B, and C, but you've got one of each of the letters.

And if we can prove congruence by SAS, we've got one of each letter, you've got the two edges and the angle between them.

It's really important though that they are the corresponding edges and angles.

So here are another two triangles, triangle MNO and triangle XYZ.

Are they congruent to each other? Oh, it's not between the two known sides.

So it doesn't satisfy SAS.

Remember, SAS needs that angle between the two corresponding edges.

Can you calculate the angle between them? So just think about that, can you calculate the angle between them? To be able to prove by SAS, we need the angle between the two sides.

So can we calculate it? Well, we actually can in this example because this example is an isosceles triangle.

We know it's an isosceles because both of those edges are eight centimetres.

It can't be an equal equilateral because the angle was given us 79 degrees.

If it was given us 60 degrees, then it would be an equilateral triangle.

But this is an isosceles, both edges are eight centimetres.

Now we can calculate that the angle in between the two eight centimetres would be 22 degrees.

So this is an example of SSA.

We were given an angle, a side and another side, the angle was not between them.

So why does it work? Well, the radius would intersect the ray twice except from one of those intersections doesn't create a triangle because basically it intersects back on N, it just becomes the line segment.

And so this is a special case of SSA and it does create a fixed triangle.

But the reason it's a special case is because actually it's isosceles and therefore we have all the information to confirm that it is congruent by SAS.

So a check, which three pieces of information is sufficient to prove congruence of two triangles? Pause the video whilst you're working with that, and then when you're ready to check it, press play.

So B and C are sufficient to prove congruence.

If you know all three sides of two triangles and they match, then they are congruent.

If you know two corresponding sides and the angle between them, then they are congruent.

Knowing all three angles just confirm similarity but not congruence because their size could be different.

Two sides and another angle, SSA on its own, is not enough to guarantee congruence.

There are special cases like we just saw, but on the whole, if you know two sides and another angle, the angle that is not between them, that isn't sufficient to prove congruence.

Okay, so we're at the last part of the lesson and this is your one question for this task.

You need to decide which of the following triangles are definitely congruent to the triangle XYZ.

XYZ is the one in the box and you've got five other triangles.

Go through each triangle and make a decision whether they are congruent or not to triangle XYZ.

Pause the video whilst you're doing that, and then when you press play, we'll go through the answers.

Triangle ABC is congruent to XYZ and we can use SAS to prove it.

We can ignore the 68 degree that they've given us because we've got two corresponding edges and the angle between them to be the same.

Triangle DEF, we do not have sufficient information to be able to prove them to be congruent.

The two sides, the 6.

8 centimetres, we do not know if XZ is 6.

8 centimetres.

And so we do not have sufficient information.

Triangle GHI, there is sufficient information and that's because we have two angles outta three, so we can calculate the third.

And when you do calculate the third, it is 81 degrees and therefore we have SAS.

Triangle JKL, those three angles might be the same three angles as XYZ, but we don't have any edge lengths.

And without an edge length we cannot fix the size, and therefore we cannot prove that they are congruent.

And lastly, triangle MNO is not congruent because we have two corresponding edges and the angle between them is different.

So if that 68 said 81, then they would be congruent.

It doesn't say 81 though, so they are not congruent.

So in summary of today's lesson where we were looking at congruent triangles by SAS.

Two triangles can be proved to be congruent if two side lengths and the angle between them are the same.

So that's the corresponding sides and the angle.

If two side lengths and an angle not between them are known, this does not prove congruence and that's because there are many occasions where this will give you two triangles, and therefore they are different, they're not unique.

Well done today and I look forward to working with you again in the future.