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.

So during today's lesson, we're gonna be looking at congruent triangles and looking to prove them to be congruent by using ASA and AAS.

On the screen, there is the definition of two shapes being congruent.

It's a definition you will have met before in your learning, but you may wish to read it again.

So pause the video and then press play when you're ready to move on.

So the lesson has got two learning cycles.

The first learning cycle, we're gonna be looking at constructing triangles.

And the second learning cycle, we're gonna be justifying conference by ASA or AAS.

So let's make a start with constructing triangles.

So if we wanted to construct a triangle with an angle of 50 degrees and an angle of 35 degrees, how can we do it? So just have a think about that.

If that's all you need the triangle to have, how would you do it? Well, we can create the two angles using a protractor.

So we've got a protractor here.

Draw our baseline on the zero, mark out 50 degree and we can draw out other angle of 35.

So these two angles need to be included in my triangle.

I need to have both of these as angles of my triangle.

So if we extend the legs that I drew using in the protractor as rays, because ultimately that's what they are, and if we line up one of the rays of each angles, this is gonna form side of the triangle because these two angles will be connected by a side within my triangle.

So here we are, we've connected them up.

We can see a side.

We've got an angle of 50 degrees at one end of that edge, and we've got 35-degree at the other end of that edge.

And the other rays intersect at what would be the third vertex of our triangle.

So there we've done it.

We've got two angles that we wanted, the 50-degree and the 35-degree, and we have constructed a triangle that includes those two angles.

But a different triangle could be formed by just moving the two angles further apart as I've just done here or here.

That third vertex is still where the other rays intersect or even closer together.

We've still got a 50-degree angle and a 35-degree angle.

So check.

How many different triangles can be created using these two angles? So if I've got a 55-degree and a 30-degree angle, how many different triangles could I construct? Pause the video, and then when you're ready to check, press play.

Well, there's actually an infinite number.

I can keep moving those two angles as long as those rays are on the same level, then I can move them further apart, closer together, and I'm getting an infinite amount of triangles.

But if the side length between the two vertices was also known, then there would only be one triangle constructed.

Because if I couldn't move those two angles any further apart or put them closer together because that edge was fixed, then there would only be that one point of intersection of the other two rays.

So here's an example.

The edge between the two angles of 50 degrees and 35 degrees is six centimetres.

So if I draw my six-centimeter edge, use my protractor, mark a 50-degree angle and my ray, because we know that the vertex, the third vertex is somewhere along that ray.

At the other end of my line segment, I'm going to mark my 35-degree.

And where those two rays intersect is the third vertex.

And so this would be my triangle.

It's fixed.

There is only one triangle that could have been constructed.

Sofia wonders though, "Is that true? Does it matter which ends the angles were placed if I put the 50 degree on the left-hand end? But if I put it on the right-hand side, would I have constructed a different triangle?" Laura says "No, that would just be a reflection of this triangle." So if I had a 50-degree at the other side and the 35-on the left-hand side, the third angle is still the same, and that's because the angle's inside of a triangle some up to 180.

So it's still the same triangle with this edge of six centimetres and two angles.

So a triangle can be constructed if we know the angles at either end of a known edge length.

So let's construct triangle MNO.

MN is seven centimetres, angle OMN is 72 degrees, and angle ONM is 43 degrees.

So there is my edge MN.

That's fixing the size.

OMN is 72 degrees.

So use my protractor.

Your protractor hopefully has got increments for every degree, so you can be really accurate with that.

And I'm gonna move my protractor to the other end, and this will be the angle ONM that I am drawing of 43 degrees.

And they intersect at that point O.

So angle OMN is 72, angle ONM is 43, and the edge MN is seven centimetres.

So with that information, we have constructed one triangle.

So here's a check.

Jun has tried to construct the triangle PQR where PQ is nine centimetres, angle RPQ is 135 degrees, and angle RQP is 20 degrees.

Jun says, "But it hasn't formed a triangle." So what has Jun done wrong? Pause the video and think about that.

Maybe have a try yourself and see if you get the same result as Jun.

Press play when you're ready to check.

He hasn't actually done anything wrong through his construction.

He just needs to extend the ray lines, and they would intersect at R.

And the reason those ray lines need to be quite long is because of that obtuse angle of 135 degrees.

Let's look at this another way.

Can a triangle be constructed if we know the length of an edge and two angles that aren't either end of that line? And we might say AAS.

So before it was ASA.

We had the angles at either end of the edge.

But what if we know the two angles that we need and we know an edge that we need but the angles are not either end? So let's have a look at this animation.

So if I've got MN is six centimetres, so that's my edge.

I've got angle OMN needs to be 50 degrees, and angle MON needs to be 35 degrees.

So at the moment, this has not created a triangle.

The ray from the 35 degrees is not passing through N, and N is fixed in its position because six centimetres.

So if I move that angle along the ray, so makes it smaller as a triangle.

So we have constructed a triangle here.

There is a point of intersection, but it's not the triangle that has got an edge MN of six centimetres.

If N was where those intersect, then it's a shorter edge.

But here we find a triangle.

And so AAS is possible to construct a triangle.

So I've just opened the link to a GeoGebra which just allows us to look at this using dynamic software.

So a similar setup on the screen.

We've got an angle of 70.

2 degrees, an angle of 34 degrees, and an edge length of 24.

And currently, the triangle has not been formed.

We need B to be a point on the ray from C.

I can move it towards point A, which is doing the opposite of what I really want because I'm really not getting close to vertex B.

And if I move it along that ray, then there is a specific point that C needs to be in order for the ray to go through vertex B.

And so one unique triangle is constructed.

So from the animation and from the dynamic software, can a triangle be constructed if we know AAS? So if you have two angles under side of the triangle, but that side is not between the two angles, yes.

And the reason we can construct it is because we can calculate the third angle.

So for example, if you were asked to construct triangle MNO, where MN was seven centimetres, angle MON was 70 degrees, and angle ONM were 45 degrees.

Well, firstly, I'd encourage a sketch.

So it's a sketch.

I'm not worrying about pencil ruler and geometry equipment.

I'm just placing the information down so I can get a vague idea of what I need to construct.

So I've got the triangle MNO.

I know the MN needs to be seven centimetres, angle MON is 70 degrees, and angle ONM is 45 degrees.

So here is AAS.

You can see two angles and a side, but the side is not between the two given angles.

But we can calculate the angle at the other end of the seven centimetre edge, and that's why using the angles in a triangle sum to 180 degrees.

So 180 minus the sum of the other two, which is 115.

So 180 minus 115 is 65 degrees.

And now we can construct it in the same way we were doing previously.

We would draw the seven centimetre edge with our ruler, would place our protractor on one end, and draw a 65-degree angle.

At the other end, we'd draw a 45-degree angle, and where they intersect would be our vertex O, and that angle would be 70 degrees.

Following on from that, just to check that we understand what information we need in order to be able to do a construction.

So which of these sketches have enough information to be able to construct them? So look at them, think about whether you could do it with your pencil, ruler, protractor.

Press pause whilst you're doing that.

And then when you're ready, press play.

A and D have enough information.

B would give you an infinite amount.

So you could draw an edge length of 9.

2 using your ruler.

You could use your protractor to mark a ray on an angle of 133 degrees, but then where would you connect them? There would be an infinite amounts of ways that you could connect them up.

C, again, you can draw a right angle and you can draw a 27 degree angle with a protractor.

But because there is no fixed edge between them, then the size of this triangle is choice, rather than a construction.

Remember that all of these are sketches, but D you could construct, and that's because you know it's an isosceles triangles from the hash markings.

So you can calculate the angle on the other end of the 87 millimetre line, and then you'd be able to construct it using your protractor.

So the task for this part of the lesson is to construct four different triangles.

So you need to construct the four different triangles that have been described there in words.

It may be that you wish to sketch them.

I would encourage you to sketch them first before you construct them accurately.

So pause the video whilst you work through those four constructions.

And then when you are ready, press play and we will go through them.

So here is part A.

And to check if you measure the angle ACB, it should be 79 degrees.

And BC, as a second check, should be 6.

2 centimetres.

So check the third angle.

It should be 79 degrees.

Use your protractor, but you can also use your ruler to check that BC is 6.

2 centimetres.

For part B, this is what your triangle should look like.

It might be that yours is a reflection.

You put F on the left-hand vertex and E on the right-hand vertex.

So it might be that it's reflected, but a check again.

Check the third angle, It should be 88 degrees, and DE should be 3.

5 centimetres.

Triangle C, so this one had an obtuse angle.

So remember when Jun tried it, he didn't have his rays long enough.

So it may be that you needed to extend your rays in order to find a point of intersection.

So GI should be 9.

8 centimetres and HI should be 4.

5 centimetres.

You did need to calculate that the angle IGH was 25 degrees so that you could construct this triangle because 42 degrees was given, but that was one that was not on the side GH.

And lastly D, the information told you it was isosceles, and you knew it was isosceles because you were told that KJ was equal to KL.

So if you've got two equal edges, it's an isosceles triangle.

So if you measure KJ and KL, because they are equal, because it's isosceles, that they should be 8.

6 centimetres.

You did also need to calculate that the base angle is 71 degrees.

So making use of the fact that you knew it was an isosceles, you could calculate the two base angles which were the same, and then you would draw your 5.

6 centimetre, measure your 71 degrees, and the rays would intersect at K.

We're now at the second learning cycle of the lesson, which is justifying congruence by ASA or AAS.

Okay, so two triangles can be shown to be congruent if all corresponding edges are the same, and we use the criteria of SSS for this.

The three edges of the triangle fix the interior angles as well.

If a triangle has those three edges, then it will also have the same three interior angles, but the converse is not true.

The three angles do not fix the side lengths, and that's because enlargements will also have the same angles.

So if we take our triangle ABC and enlarge it by the scale factor of K, then the edges will now be KA, KB, and KC, but the angles will still be the same.

They will still be corresponding to the object.

So to be sure that triangles are not only similar and we say that they're similar, if their three angles are the same, at least one edge length will need to be known.

So on the right hand side, we've got our three edges known, which then implies the three angles 'cause that fixes the angles.

When two edges and the angle between them is known, SAS, the third edge is fixed, which in turn fixes the other two angles.

The criteria of congruence of SSS fixes the angles.

The criteria of congruence SAS fixes the other two angles because the third edge is fixed.

So if you know one edge length, the size is fixed.

And by knowing two of the angles, you in fact know all three.

So if you've got the edge length A and you know the angle's either end of it, it fixes one triangle.

We saw that when we constructed.

If you have one edge A and you know two of the angles, but they're not both either end, it still fixes the triangle.

And that's because the third angle can be calculated by using angles in a triangle adding up to 180 degrees.

So the left hand one is ASA.

You've got an angle, then a side, and then an angle.

And the second one is AAS, but in fact they're equivalent to each other because we can just calculate the third.

To be able to prove two triangles are congruent, you must know at least one corresponding angle.

True or false.

Pause whilst you'll think about that.

And then when you press play, we'll go through the answer whether it's true or it's false, and then we'll pull up the justifications.

So that's false.

So to prove two triangles are congruent, you must know at least one corresponding angle is false.

So is it false because knowing all three edges is enough to prove two triangles are congruent, or is it false because knowing one angle does not always allow you to calculate the other two? Pause the video, and then when you're ready to check your justification, press play.

And that one is A.

So SSS is enough information to prove two triangles are congruent.

So we now have another way of proving that two triangles are congruent, and that's if you know one side length and two interior angles.

So let's prove that triangle ABC is congruent to triangle DEF by ASA or AAS.

Remember that they are actually equivalent to each other.

So BC is equal to EF.

They're both 9.

2 centimetres.

So they are corresponding.

Angle ABC and angle DFE are both 99 degrees, and angle ACB and angle DEF are both 17 degrees.

So the triangle ABC is congruent to triangle DEF by ASA.

We're gonna go for ASA over AAS because it is angle side angle.

Both of those angles are at the end of the edge.

If the 17 degrees had been in a different position on one of the triangles, then they wouldn't be congruent.

So it's really important that they are corresponding edges and angles.

So here's another one.

Let's prove that triangle QPR is congruent to triangle TUV by ASA or AAS.

So when you look at these two initially, we can see that both of them have an edge that's 7.

1 centimetres, we can see that they both have an angle of 97 degrees, but then suddenly there's a difference.

One of them has got an angle of 38 degrees and the other one's got an angle of 45 degrees.

So at a first glance, they don't look to be congruent, and this is where we need to remember that if you have two angles in a triangle, you can calculate the third.

That will allow us to see if they are congruent or not.

Angle TVU can be calculated as 38 degrees.

So now we've got 38 degrees in both of the triangles.

So now let's just check that they are actually corresponding.

So the 7.

1 centimetre has got a 38-degree angle at one end and a 45-degree at the other end.

On triangle PQR, there is a 38-degree at one end and the other end would be 45 if we calculated the missing angle.

The other way we could have checked this is that the 7.

1-centimeter is the opposite edge to the 97-degree angle, and that is the same on both triangles.

So we can see that they are corresponding.

So PR equals UV.

They're both 7.

1 centimetres.

Angle PQR and angle UTV are both 97 degrees, and angle QPR and an angle TVU are both 38 degrees.

So we can say that these are congruent by AAS.

Again, we're gonna go for AAS because the 97 and the 38 are not both at the end of the edge.

So it's more angle-angle-side, rather than angle-side-angle.

But in fact, either is okay.

So here's a check.

Triangle IJK and triangle TUV are congruent by AAS.

So which edge corresponds with IK? Pause the video.

And then when you're ready, press play.

Hopefully you went for UV.

So how do we know it's UV? Well, IK is the edge that is between the 28-degree angle and the 118-degree angle.

On triangle TUV, there is a third angle that's not labelled.

We can calculate it to be 118, but the fact that we were told they are congruent, we know it's 118.

And so then which edge on triangle TUV has 118 degrees at one end and 28 at the other end? And that would be UV.

If you wrote VU, that's the same and that's correct.

When we look at congruent triangles or we try to prove that two triangles are congruent, it might be that we don't actually have the size of the angles or the lengths of the edges or we still can use properties of shape in order to prove that two triangles are congruent.

So here is a regular hexagon, A, B, C, D, and F, with three diagonals drawn on.

Which triangles are congruent to each other? So we can see triangles have been formed by drawing these diagonals into this regular hexagon.

So which triangles are congruent to each other? Well angle A, B, C, and angle DEF are equal, and they're both 120 degrees in fact because it's a regular hexagon, and they are the interior angles of a regular hexagon.

AB and DE are equal as they are edges of a regular hexagon.

Angle BAC, angle ACF, angle CFD, and angle FDE, are all equal to each other as they are interior alternate angles in parallel lines.

There are a lot of sets of parallel lines in this one diagram, so just use those feathers.

If there's one feather, the arrow, then that's one set.

If it's got three, then that's a set itself.

So all of those angles that have now been marked with double arcs are equal to each other, and we can use interior alternate angles to know that they are equal.

So we can say that that triangle ABC and triangle DEF are congruent by ASA because we know they have corresponding angles and an edge between them that are equal.

However, we can go on to say that angle AFC is the same as angle FCD as they are interior alternate angles in parallel lines.

So they've been marked onto the diagram with the three arcs, so just locate them.

Angle ACF and CFD are equal.

They were already labelled with the double arcs.

And AF equals CD as they are edges of a regular hexagon.

So we can say that triangle AFC and FCD are congruent by AAS.

Two angles and a side are known to be the same in both triangles.

So a check on a proof using properties.

So complete the proof that triangle ABD and triangle BCD are congruent by AAS or ASA, given that ABCD is a rhombus.

So pause the video, read through it, and you need to figure out what those blanks should say.

Press play when you're ready to check.

So firstly, angle DAB and angle BCD are equal because they are opposite angles in a thrombus.

Angle ABD is equal to angle BDC as they are interior alternate angles in parallel lines.

So this is making sure you know that a thrombus has two sets of parallel sides.

The opposite edges are parallel.

And then AB equals DC as they are edges of a thrombus.

So a thrombus, all the edges are equal, hence triangle ABD and triangle BCD are congruent by ASA.

We have shown that both of those triangles have two corresponding angles and an edge that is the same.

So here is your task, the second learning cycle.

Here is question one.

Question one, there's two parts.

You need to prove each pair of triangles are congruent by ASA or AAS.

Remember that ASA and AAS are equivalent to each other because you can always just work out the third angle.

So pause the video whilst you're doing that.

And then when you're ready for questions two, three, and four, press play.

Question two and three are on the slide now.

So question two, why are these two triangles not congruent by ASA? And question three, given of parallelogram ABCD, prove that triangle ABC is congruent to ACD by ASA or AAS.

Sketch yourself a parallelogram, add on the properties you know, and that will be a good way to start question three.

Press pause whilst you're working on question two and three.

When you're ready for question four, press play.

And question four, point E is the midpoint of AC and BD.

Prove that triangle AEB is congruent to triangle ECD by ASA or AAS.

Press pause whilst you are working through that question.

Again, add onto the diagram any information you know from angle theorems, properties of shape, et cetera.

Press pause.

And then when you're ready for the answers to these questions, press play.

Question one, part A.

Prove that these are congruent.

You can prove they're congruent by AAS from the given information.

So 71 degrees was opposite the 4.

5, and that was the case on both triangles.

So they were corresponding.

91 degrees was one of the other angles.

And so you've got two equal angles and a side length, so AAS.

Part B was a less obvious proof, and that's because if you checked the angles, they were not the same.

However, if you calculated the third angle in the triangle, then you saw that actually they were the same three angles, And the edge 29 millimetres was opposite the 34 degrees in both of the triangles.

So they were in the relative positions as well.

So you needed to work out the missing angle in order to check whether there was any chance they could be congruent.

But remember, just because they've got the three same angles does not mean that they are congruent.

That just means that they are similar, and that's why it's important that that edge is in the right position.

So the 29 millimetres, the GI, is opposite the 34-degree angle in both of the triangles.

Question two, you needed to say why these were not congruent.

So although the three interior angles of both triangles are the same, the two edges of 4.

3 centimetres are not corresponding angles, and that is about the position.

So if you look on triangle ABC, the edge 4.

3 centimetres is opposite the 80 degree.

Whereas on the triangle DEF, the 4.

3 centimetres is opposite the angle 47 degrees.

So it's not in the same position to the angles, and therefore the not congruent by ASA.

Question three, I'm gonna go through it in two different ways.

So if the first way is not the way you did it, just wait.

It might be that the second way is the way that you did it.

Given a parallelogram ABCD, prove that triangle ABC is congruent to triangle ACD by ASA or AAS.

So the first way that I did this is to say that angle DAC and angle ACB are equal because they are interior alternate angles.

We know they're equal interior alternate angles because opposite edges are parallel on a parallelogram, AC is a shared side.

Whatever that length of that edge is, it's an edge on both triangles.

And then angle DCA is equal to angle BAC as they are equal interior alternate angles, again, because opposite edges are parallel and so hence triangle ABC is congruent to ACD by ASA.

So I've shown that both triangles have got equal angles and a side length between them.

However, you may have gone about it this way, so AD is equal to BC as they are opposite sides of a parallelogram.

Angle DAC is equal to angle ACB as they are equal interior alternate angles.

We used that on the last one.

And angle DCA equals angle BAC as they are equal interior alternate angles.

And so this one is congruent as well.

This one is more about AAS because the side that we've shown to be the same is not in between the two angles.

Ultimately, it means the same thing because I could work out the third angle and say that they are equal.

Lastly, question four.

Point E is the midpoint of AC and BD.

Prove that triangle AEB is congruent to triangle ECD by ASA or AAS.

So, firstly, we can see that DE is equal to EB as E is the midpoint of BD.

So the midpoint means that it's exactly halfway along.

Angle AEB is equal to angle DEC as they are vertically opposite angles.

And angle ABE is equal to angle EDC as they are equal interior alternate angles.

We've got a set of parallel lines here so we can use interior alternate angles being equal, and hence we can say that AEB is congruent triangle ECD by ASA.

I've only gone through it the one way here, but actually there are again multiple ways that you could have done this because you could have used angles BAE and angles ECD are equal interior turn angles, and then you've got two angles again.

You could have used AE is equal to EC as edges because of the midpoint.

So there was a few different variations, but ultimately you could show that two angles were equal, two angles were same, and there was an edge length that you knew to be the same in both triangles.

To summarise today's lesson which was to do with proving congruency using ASA and AAS.

So two triangles can be proved to be congruent if two corresponding angles and a corresponding side are known to be the same.

If the corresponding side is between the two known angles, then we tend to say they're congruent by ASA, angle-side-angle.

If the corresponding side is not between the two known angles, then we tend to say AAS, which is angle-angle-side.

Well done today.

I look forward to working with you again in the future.