Okay, so last week we focused on calculating length, right, perimeter and area. And today the focus will be on angles. We divide the lesson into three parts. The first one with triangles, the second part we will deal with circles and the last one. So in the first two parts, we review basic theorems and tools that will help us solve most problems in math kangaroo competitions. Some of them require a little bit more advanced techniques that we will talk about in part three. So for triangles, the most basic theorem is sum of angles in a triangle is 180 degrees. And if you have a polygon, you just divide it into smaller triangles, right, and sum up the angles. Then this theorem is very straightforward, but super useful if you have a triangle here, then the exterior angle at C is equal to the sum of the interior angles at A and B, right, because C, the exterior angle here is supplementary to the interior angle at C and angle C plus alpha plus beta is 180 degrees. So gamma must be equal to alpha plus beta. Very straightforward, but super useful. And then the next step when we deal with triangle is that always identify all the special triangles that you can see in the problems. And that means the isosceles triangles where the bisectors is also the median and also the altitude. Right triangles are equilateral triangles. And of course, the 90, 60, 30 degree triangles in which you have the median here equal to half of a hypotenuse. And this triangle MEC is equilateral. And of course, the right isosceles triangles, right? So when you see them, just mark all the lengths, all the angles in the figures. And the next step when we have to deal with angle is that you actually have to prove pairs of congruence or similar triangles. So for congruent triangles, most of the time we will use a side-side-side test or the side-angle-side test to prove that the two triangles are congruent. And when they are congruent right away, we have three pairs of equal angles, right? And similarly for similar triangles, then we usually use a side-side-side or the side-angle-side theorems to prove that they are similar. And again, we have three pairs of equal angles. Yes, so I think that's pretty much what we need to use to solve maybe 80, 90% of the problems that involve triangles or polygons. So let's get started. Square ABCD and equilateral triangle DCE are on a plane. See the picture here. What is the measure angle AEB? Wait. Remember to mark all the equal length and angles that you see on the pictures that will help you, allow you to see things that are not obvious. If you have the Henaoza square, if not, then try to, you know, draw and when you draw as accurately as you can, draw to scale. So I mark all the lengths as equilateral, this is square, so I have a bunch of lengths that are congruent. And then I also want to mark, if you want to, all the angles. This one, because it's equilateral, so this one is 60 degrees, and here also 60 degrees. So after we mark that, we know that ECB is isosceles, right, because EC equal to CB, that means this angle alpha here and alpha here is the same, and same on the other side, also alpha, it's due to symmetry, and this angle here is 60 plus 90, that's 150, right, 150 degrees, so that means each of these is going to be equal to 15 degrees, so that they make up into 180, and so we have a 15 here, have a 15 here, and DEC is 60 degrees, so the angles that we are looking for is 30 degrees, and a lot of time for this question, you can just, you know, just mark and everything on the pictures without having to write anything. So that's 30, okay, that's the first step of problems when we just have to identify special triangles. Great that you all got the correct answer. So now go to question number two, you know, a little bit, I don't have a, let M be the point where all the altitudes of the acute triangle ABC meet. If AB is equal to CM, what is the measure of the angle at vertex C of triangle ABC? So basically we are asked to calculate, we are asked to calculate this angle, right, and let me mark, I mean, this one is not, I really recommend you study geometry draw to scale, because that, a lot of times it helps you see the answer right away, but in this case, I probably just highlight using different colors, so CM is equal to AB, right, you want to, so these two lengths, it will be ideal if they are drawn to scale, but if not, then mark them, color them, and thanks David for marking here the altitude, so these are so good, yeah, these are perpendicular, these are 90 degrees, and these are just super helpful. So remember the first step, you use the sum of angles in a triangle 180 degrees. The second step is identify all the special triangles as much as you can. And the third step is that you have to find pairs of either congruent or similar triangles. There's a lot going on, but it's kind of methodological. That's all the tools that I have. Yeah, thank you, David. You can also do congruent as well. Because, yeah, I think, similar inside one side, the same side becomes congruent. We got the first answer. So for congruent, we use the first, usually we use a side, side, side, or side, angle, side for congruent triangle. And we mark as many. So let's, okay, good. So you already put in the answer, so we can get started. Let's see, we have this angle here, right? Because I see the equal length as somehow I want to make use of this information. The two lengths are equal. And how do I do that? These are C and A and A prime B, they are right triangle, and these are hypotenuses, right? So maybe I can sort of work backward, maybe I can try to prove that they are congruent. And how do I do that? This angle equal to this angle, right? Because they are vertical. So that means, and this angle M is complementary to angle alpha here, and this one is complementary to angle beta over here. So actually, these are also alpha. Do you see that? So that means an alpha here, alpha here, that means this one is beta, and that one is also beta, because they are complementary. So now we have a triangle C, we have a C and A prime, they actually, it is actually congruent to A, A prime B, right? Because we have the side, angle, side, angle theorem. And, and from that, right away, you can get that C, A prime equals to A, A prime. It's not very pleasing to look at the picture because not to scale, but this is longer than that. So C, A prime is equal to A, A prime, that means the triangle C, A prime A is right and isosceles. And, and that means the answer is that C, 45 degrees. And I said, if you draw it to scale, you would see right away that A, A prime C is right isosceles, and then you go back what you tried to prove that. Good. Next, this one, David, could you please do this question? Okay, question three, which side of the polygon shown in the picture is the shortest? Yeah, very small problem statement, a lot of segments. So the question really becomes, how do we compare the length of the segments when we're not given any lengths, right? We need some other trick. And so once we figure that out, which is, you know, not hard at all, once we think of the way to compare angles, to compare sides, not angles, then the trick is just to just keep comparing everything listed. So this is another angle chasing problem, except it's a side chasing problem, right? Again, how do you solve it? Just keep getting stuff until you find what you want. Oh, do you want to show the side rules or? Um, I guess I could show exactly, but there's no reason to get into things like law of science, right? Okay. The main trick is to realize, yeah, the bigger the angle, the bigger the side is. Yes, because it's anchor. Okay. So I have the next slide can show and then we come back to this if you want. I mean, we may not need this, but yes. So if you draw a triangle, then you see that the bigger the length way, then the bigger the angle. So if you need to review, that is a slide you can review. All right, we have our first answer, which is some pretty quick bashing. I'm just going to start bashing slowly as we get through this, because that's what this problem is. We bash, okay? So first of all, let's start with the first triangle. Maybe you started from the other side. That's fine too. It really doesn't matter. I'm going to start with triangle EDM. First of all, if the sum of these angles is 180 degrees, that means that this angle here is also 42 degrees. Because 42 plus 42 plus 96 equals 180. That's just doable arithmetic, right? That means that this is an isosceles triangle. These two sides are equal. And EM is clearly larger. EM is bigger because the opposite angle here is very big. And by the law of sines, that means that EM is also very big. So that tells us that whatever the smallest thing is, it's not EM, right? In the next triangle, what do we see? Well, this angle over here is 59 degrees, right? Has to be 59 degrees because... No, 58 degrees. Sorry. It's 58 degrees because 58 plus 61 plus 61 is 180, right? We're just using the sum of the angles as 180 theorem. And since 58 is smaller than 61, by the way, this is equal. And this is equal too, because a bunch of isosceles triangles. That means that DC is smaller than either DM or MC. Because the opposite angle is smaller. So that means that the smallest thing is definitely not this, right? We can get rid of these two. They're too big. Now, 45, 45, what's the other angle? What triangle has 45, 45? Obviously, 45, 45, 90 triangles. That means that BC is even bigger than CM, right? This is a hypotenuse. The hypotenuse is always bigger in a right triangle. So this isn't the smallest either. Then, because we know that MB, MB is too big, just because we know MB is bigger than DC, since DM is, and these are all equal. MB is the smallest side in this final triangle, and then AMB, because it has the smallest angle opposite to it. 35 degrees is greater, so AM is bigger. And then the angle that's left here is over 100 degrees, something very big. So whatever it is, we know that AB is also huge, which tells us that the only thing we're left with is DC. And that would be our answer. That's the smallest one. And so the only thing we did is we basically did angle chasing and found the smallest angles and used that to find the small sides because we have this relationship by the famously abbreviated law of science. This is how it's typically written. Yeah, thank you. Thank you, David. Okay, good. So let's move to the next part with circles. For circles, I think we briefly talked about circles last time, about tangential circles. So this is another thing we want to review in a circle. All inscribable angles subtended by the same arc are equal, and that are also equal to half of the center circle subtended by the same arc. I think that's how they talk about designing a theater, right? If the screen is here and everyone sit here, then they would have the same angle to the screen. This is the center of the circle. And then because the half of the circle is 180 degrees, also an inscribable angle subtended by a semicircle, when AB here is diameter, is 90 degrees. So anytime you see a diagonal circle, anytime you see a diameter here, you want to mark that angle is equal to 90 degrees. So that is super helpful. And lastly, for the tangential line to a circle, if AB is tangent to the circle O, and the tangent, the C point is P, then the radius OP is perpendicular to AB, and that is something you also want to mark on the pictures as soon as you see it. So I think combined with what we had with triangle and with circles, let's do a few questions here. So the first two questions are just pretty much basic review of inscribable angles. Let's see if we can get it. What is the measure of angle five shown in the pictures? So you may want to label it. It is C, D. So ABD is 30, this angle here looking at this arc. And phi is angle looking at the arc BC. This is not the center, it's just, maybe we put it P, where AD intersect with BC. So similar to length, we have a picture with angles that try to mark as many circle, as many angles as you can find in the pictures. Again, angle chasing. There's only, you know, less than 10 angles in this entire thing, no matter how hard you try, that means you can brute force. Which is, brute force is very useful in this case. Brute force is very useful in angle problems, because at some point there's no trick. It's just, you have a bunch of angles, some of them are equal. You can find some by doing the subtractions in a triangle, just pummel it down and eventually you'll have no angles left. And you'll find them. Yes. Yeah, so this 70, not super useful. It's, you know, it's not the center of the circles right here. We look at 30 first, right? Because 30 is the inscriber angle looking at this arc AD. And then we have another angle also subtended by arc AD, which is this one. So I can right away mark that this one equal to 30 degrees. Yeah, I hope, I think you figured out. Oh, 30 degrees. And then now we look at this triangle PCD over here, right? We talk about the exterior angle theorem. And this is a classical case. We have an exterior angle, 70 degrees here at P, and an interior angle at C equal 30 degrees. So right away we can have 5 and 5 must be equal to 40 degrees, right? Of course, I mean, you can also go 70 degrees and this one is 110. And then, you know, subtract it, but just really bypass this part. So we just learn to bypass this part. It's going to speed up the calculation and have a C thing much quicker. Okay, so 40 degrees. Yeah, I quite understand where you got. Oh, some, maybe some got seven. Oh, some got 30 degrees, 35, which is half of 70, because maybe you think that 70 here is the center. Remember, right? Because remember in this one, this point here must be the center. That's why I call the central angle, because this must be the center of the circle. And in this case, it's not. Okay, good. So it's good that we review that theorem. Okay, next one. In the figure, you see points A, B, C, D on a circle, of this inner circle. Quad AB is a diameter of this circle, and the measure of angle ABC is 35 degrees. What is the measure of angle BDC? So right away, when you see that quad AB is diameter, what do you do? So just quickly draw the pictures and draw AB as a diameter. You want to do that. And then you can mark O here, you don't have to draw much, just mark O here so that you know that AB goes through the center of the circle. Okay, good, good, I think some of you realize it. So AB here, as we said earlier, yes, we want to draw that. We want to draw this line because ADB is circle, so right away, ACB has to be 90 degrees. So our old trick, right, 35 degrees here, 90 here, that means this one's going to be 55, and I just mark it. 55 degrees, and this is on a circle, so I know I have to invoke the inscriber circle triangle, angle theorem, and all of you got it, 55 subtended by this RCB, that means the same thing here, also equal to 55 degrees. So this is one of the most beautiful theorems in circles, but you should do a few examples and, you know, get a handle of that and be confident using it. Okay, good, so I hope we reviewed the basic and now we go to something a little bit more complex. So, David, could you do this question, please? Okay, here we go, right, a fear, okay, anyway, no. In the picture, PT is tangent to circle C with center O, and PS bisects angle TPR. Take the measure of angle TSP. So we're going to apply everything we've done with angles so far, both with circles and both things without circles, and we're going to start by doing what? Chasing! Angle chase. Draw your lines, chase your angles, try to find that extra thing, and hopefully, eventually the angle that we want will just fall into your lap. Yes, so I think for laboring, that means PS, bi-sec, and then PT is tangent to C, right? So we want to represent that information on the picture. So how do we do that? Okay, so probably the reason that this is hard really is because we don't really have enough on this diagram. And this is a very common strategy. It's just draw the extra line, draw the line that looks nice, the line that matters. Often it's an altitude, we will see one of those later. Often it's a radius. And in this case, we have a circle, which means it seems very logical that one should go ahead and try and draw this and see what that's a terrible line. Actually, don't do that. That's too bad. Line. You can draw, you can draw. Perfect. Okay, so we should draw this radius, which as we know very well, is equal to this radius. And just see what we can conclude from that, right? Suddenly just drawing that line. And that's why it's important. It brings up a whole host of new opportunities, right? For one thing, we know that whatever this angle is, beta, let's say this angle is the same, because look, I saw this triangle. We also know that we have a radius and a tangent, which means that this over here, which I need, I don't have enough columns is a 90 degree angle. And at this point, we can start chasing a bit more. So I'll give a bit more time before I go to the next step before I do some more chasing. and this is the angle we're looking for obviously and if this is helpful Okay, good, we got one. Let's start going slowly. So this is just a little hint I gave at the end of the problem, which tells us that TSP is just alpha plus beta, because it's an exterior angle to the triangle. Welcome now, that's like a while to show up. Um, so the this exterior angle thing is very useful because we can use the fact that beta plus beta also sums up to this angle here, which tells us that two alpha plus two beta, these two angles, plus 90 degrees equals 180, right? They make a triangle, which means that two times alpha plus beta, I'm going to write this two alpha plus two beta equals 90 degrees. They're complementary. They're in a right triangle, which means that alpha plus beta equals half of that, right? Or 45 degrees. And suddenly, we're done, believe it or not, because once we have that, we know that TSP is just the sum of alpha and beta being an exterior angle, which we know is 45, which would give us an answer of B. So the trick here was draw the extra line. That was the new strategy, really, we didn't have that, and we needed that line. But then once we got there, we just were chasing angles, right? I chased this angle, I chased this angle, I chased alpha beta. In principle, I could have just gotten all the angles in this entire thing. I knew which ones were necessary, because I've done this problem, right? I know what angles I needed to pick, so I only showed you the necessary ones. But if you solve this problem by getting more angles than you needed, and then eventually finding the right one, that's okay too, right? As long as eventually we brute force our way to it, and we have few enough angles that we can do it in the few minutes we have in Mapping Guru, that's really all that matters. So if your solution wasn't as efficient, don't feel bad, because I had prior experience, I knew what to look for, right? Thank you, David, that's a very useful tip. So, you know, there's a lot of angles, a lot of notation, so we just decided to, you know, put the things here. You can review parts of the video and review at home if you need to, sort of nice and neat typing of what David just presented. Okay, so we've finished with part A, with triangles, polygons, part C with circles, and I would say most of the time, maybe 80-90% of the time, that's been up for you to solve Maths Kangaroo geometries, but maybe 5%, 10% of the questions that require you to do a little bit more work. So that more work is to draw the actual lines in this case. So the previous slide that we draw over here, we actually can get from the question because we know that PT is a tangential to OT, and sometimes we have to draw actual lines out of nowhere, we just have to invent that, and then we have to use, I mean, if you have advanced, know advanced theorems, sometimes you can have advantage, but I don't know, I don't think Maths Kangaroo really tests how far you know ahead of time, so I would not, like, worry too much about advanced theorems, like calculating the area of triangle, knowing its length, something like that. You should not worry too much about that. And if you study trigonometry, yes, sometimes you can, you know, have advantage, you know that, but we're going to show that you can use elementary Euclidean geometry to solve most questions. Trigonometry can be a shortcut, that's really what it is, and I've done problems where it takes, it would take, you know, four or five minutes if you did it Euclidean, drawing a bunch of weird extra lines and trying things, but then if you know the trig, it's not that you need it, it's just that if you have it, sometimes they can just go like that, and it's a useful thing to have. That's true, yeah, so in the next few sort of more challenging questions, we will present several solutions, and then for Euclidean geometry, always draw as accurately as you can. Yeah, I love, you know, construction questions when you have to just a compass and a straight edge. So I don't think this one is probably the most basic trigonometry rule. Some of the questions required, I would not worry too much, but I just put it here for completeness. Okay, so let's go for this question. Three squares form a rectangle, see the picture here, right, all three squares, lines AE intersect with CH at point P, what is the measure of angle CBE? This one. So we have time, so we give you some quiet time, maybe two minutes for you to And you know, say, if you have a compass and a triangle or ruler or straight edge where you draw it accurately and you can actually measure this one, not a good thing, but once you, lots of times when you know the answer, it helps you to sort of construct the solutions. And as we said earlier, if you know trigonometry, use it, but we don't need it, we can use it in an elementary way. Okay, so 2.5 minutes passed, so let's start in the first step. This angle P here, just in the middle of figure, right, doesn't look very nice, so maybe I want to break it down into other angles that might be easier to calculate. So let's use the exterior angle theorem here. We know that this angle P here equal to alpha and beta, that would be the first step. And alpha and beta are so much nicer than P, right, because alpha is an angle of this AED, this right triangle, and we know ED is 1, AED is 3, and similarly beta is an angle in this AHC triangle with length 1 and 2, so it looks like alpha and beta are more manageable than P. That would be the first step. And from here, let's see if we can find out what's alpha plus beta. Wow, great! Okay, we got the answer, so we can get started. Alpha and beta, they are separate. I need to add them up, right? So naturally, I want to bring them together. There are many different ways to do that, but I would like to bring beta next to alpha. And one way to do that, if I extend this fc to equal the length of the sine length of the square, then for example, I call this point m, then I connect a and m, then right away I see, I know that am is parallel to hc, right? So this angle is beta. Or you can see that the hac and mca, they are congruent. So now I know that alpha and beta, they're next to each other, right? So can I calculate the angle eam? Okay, it's a lot to type, so I just move to the next slide, sort of easier to explain here. So what we said earlier, I bring beta here next to alpha. So eam is the angle that I need to calculate, but I look at over here, there's a lot of squares that I haven't made use of, right? So let's connect e and m, a and m here, and let's, let me make life easier by just, you know, call the length of, the sine length of each square as one. So, and I have a bunch of right triangles, so let's just use the Pythagorean theorem. So if I look at angle acm, am square is going to be equal to one square plus two squares that equal to phi, right? I mean, that's am square, not just am. And then similarly, if I look over here, fem is a right triangle, so one square plus two square, me square is going to be equal to phi as well. This is me square, not me. And finally, I look at this right triangle, ahe, that means ae square is equal to one square plus three squares, that is 10, right? So ae square equal to 10. And that's the thing with geometry, the question gave us three right triangles, and we have to, and they are, you know, put next to each other in this particle configuration, so I have to make use of all information. If I haven't made use of all information, I know that, you know, I wouldn't be able to get to the solutions. So now it's nice to have a phi, phi, and 5 plus 5 is 10. So again, using the Pythagorean theorem, I know that ame is actually an isosceles, right isosceles triangle, so alpha plus beta equal to 45 degrees. And for this one, part of the reason you can get to the solution very quickly is because the pictures are very much drawn to scale, so once you bring beta and alpha here, I think you can quickly see right away that ame is a right isosceles triangle, and we just sort of work backward and see how can I prove that. There's also, there's a million, there's already another completely separate solution we'll show, but there are a million ways to do geometry problems, and in particular, the way that I solved this, I got to this point, and what I realized is efm is also a right triangle with this ratio 1 to 2, which means that this angle is also beta, right? It's beta because we have a congruent triangle to acm to ach, they're all the same, and since this angle is 90 minus beta, because you've got a right angle here, that means that this entire angle is 90, which tells me immediately 90 degrees, these are the same, 45. So again, there's a billion ways, and this is just one solution of multiple. Exactly, and that's just like, I feel like we do, sometimes we do with geometries, like you're playing chess, you look at this direction, like that direction is always something interesting going on, and we sort of put things together, and then as I said earlier, we can use, you can use trigonometry. Again, we have alpha and beta, and we know that tangent of alpha is 1 over 3, and tangent of beta is 1 over 2, right, h over ac, and so I'm, if you know this formula, the thing with trigonometry is that you have to know it by heart, you know it very well to be able to use it, you know, confidently, so that's why I don't think it makes sense for it to go with the theory, if you know it, you can do it, most of the time, you can solve it without knowing it, but it's just so powerful, all the trigonometries are the ones that relate length and angles, so definitely that is something that you want to pursue. And it is difficult to know, because those previous solutions did require ingenuity, you had to draw multiple new lines, you had to extend the figure out of the original, because it, you have to literally think outside the box, you have a box of three squares, it really feels like the solution is supposed to be in that box, but no, you make it bigger, you draw these lines outside of everything, and that's a very common theme in hard geometry problems, but it requires a lot of ingenuity and a lot of inspiration, and I guess the good thing about having trig is that if we don't have the time to think of that, or we just find ourselves not able to, we have the opportunity to just bash, and that's the only reason you'd want this, it's not as nice of a solution, it's not as elegant, but if you happen to know them by heart, then a lot, by all means, use that, and they are so powerful, we're able to do those things, because angles is 945 degrees, but you write 42.5, then no, we need to use trigonometry. Okay, so another question that also has multiple solutions, I think it's a very beautiful question, let AD be the median of triangle ABC, if ACB is 30 degrees and ADB is 45, then what is measured by AD, and of course I want to label that BD is equal to DC, always label everything in the pictures, so that was, okay, sorry, I just want to move it over here, so ADB equal to DC, and then this one is 30 degrees, right, 30 degrees, this one is 45 degrees, and by now I know that this one has to be 50, right, so I label as much as I can on the pictures, and now stop for a moment and see what can I do. So that's the typical question when we have, you know, like two lengths that are equal, not, and then we have to ask how to bring them together, how to make use of the information, because what I'm asking is not length, what I'm asking is angle, BAD, this angle. We definitely need to draw actual lines and which line is that? And definitely you can try, right? You try, if it works, if it doesn't, we try something else. Yes. Oh, good. That's interesting. So we got it very quickly. Let's see. We can, we don't know how the students solved it, but it's great that you got it so quickly. Okay, so let's go with, we have three different solutions. So let's get started and you can also just send us the answer why we are discussing this. So to me, when I look at this question, I have a 30 here and 15 here and I just thought, okay, how I can make use of that information. I want to bring this length closer to this length right together. So one thing I can think of is that if I put m over here, such that maybe m and n is any point over here, such that dm equals to dc, right? Then that serves two things. The first is that mdc is isosceles triangle. So I would have 30 degrees over here. And again, using the exterior angle theorems, I know this one equal to 15 degrees. And that means this amd is also isosceles. So right away I have amd, I have a bunch of equal lengths. And also md is median of dmc and it's also equal to half of dc. So that means this one is 90 degrees, right? So it looks like a good choice. I tried a few things and I got this. Let's move to the next pictures when I have a drawing, it's quite more neat. Okay, so that's what we do over here. We have many different lengths that are equal to each other. And also we realize that this triangle bmc is 90, 30, 60 degrees. The bmd is equilateral, right? So that one is also equal to bd. So many equal lengths. And now right away you see that amd is a right isosceles triangle. This is 15 degrees. So this one is 30 degrees. So it looks like, oh, how do I draw this length? But it really just stem from the fact that I want to make use of the fact that these two are the same. I just don't actualize that, make use of that information. And it's great because I think students figure it out. But when David look at it, he had a different, you know, he look at from a different angle, a different point of view. So he has a different solution. So David, please tell us a little bit. Probably maybe not actually what you're thinking of, because I do have to chase a lot more. I do have quite a bit of chasing to do. But what I saw when I did this problem was I was thinking, I have a bunch of angles that fit really well with right triangles, right? I have a 45 degree angle. I like right triangles with 45 degrees. I have a 30 degree angle. I love right triangles with 30 degrees. What I don't have is I don't have the right angle. So let me do that. I'm going to draw the altitudes. So I drew the altitude down to BC, which is also thinking outside the box, right? Because it's outside the triangle. And I drew the altitude H down to here. It's also important to know that this is 15 degrees. You can solve it in that same exterior angle way. What I really realized, though, is simply if this is 45 degrees, this entire angle here has to be, this entire angle is 60 degrees because 30, 60, 90. And this smaller angle is 45 because 45, 45, 90, which is how I know this is a 15 degree angle. So that's step one. Immediately having two right triangles has given me this angle. I didn't use an exterior angle. I used the right triangles. The second thing I did, and this is a bit ugly, there's probably another solution which uses this diagram, which is neater, that will be shown. But I did a bunch of chasing with special triangles, right? I know this is one. I know this whole thing is two because you have a one, two, root three. I know, really. Let's say AE is one, right? It's just one because why not? Then I have a one root three, two triangle because that's what 30, 60, 90 triangles are. We've looked at that. These are the side length ratios. I also know that this is root two. I know that since these are the same... What did I use? Okay, now I'm going to be... Yeah, no, yeah. This is one, so this whole thing is also one, right? Which means that this is root three minus one, right? So I'm just chasing sides here. It's really not that complicated, which tells me that this is root three minus one over two. Root three minus one over two because you have half of this hypotenuse. This is 30, 60, 90 triangle, right? This is half of this because the angle of 30. Similarly, this has also got to be root three minus one, which means that this is one minus root three minus one, which is two minus root three, this angle here. And what you realize is that if I do some more angle chasing, just real quick, I'm almost done here. I know this is a bit messy, but every single step I'm doing makes sense, right? I just multiply by root three. I multiply by two. It looks like a lot, but every single line, I'm just plugging in the same rules. If it's 45, 45, 90, I have one, one root two. Otherwise, I have one root three and two. Two minus three minus root three over two. And the reason that I do all this is because you realize something super important. The ratio of DH to AH is the same as the ratio of EB to AE. We did all that to realize that AEB is actually similar to AHD, which means that this angle is also 15 degrees, which means that since this angle is 45 degrees, the angle we're looking for is 30 degrees. So instead of angle chasing, I did line chasing. Every single step is simple, but there are a lot of them. But the reason I really wanted to show the solution is because I proved something really cool. I proved in the coolest way I've ever seen that the tangent of 15 degrees is two minus root three, and that's a good thing to remember. It's another special angle that you can figure out with all of this. So just as a quick summary, you chase all the lines. AHD and AEB are similar, and that gives you two 15-degree angles. That is really nice. A lot of the time, a simple solution may work for this thing, but it's really we don't learn much. I really like this question, and especially the way you think that, oh, I have a 30, I have a 40, but I like 90. So that's how you draw this to actualize, right? I mean, we learn so much just by doing different things. That's what I love about geometries. So thank you for that. It's really nice. I really like it. And lastly, we have the solution using trigonometry over here. Also, last time, we had the formula using the sum of the tangent of the sum, and here's the tangent of the difference, pretty much the same formula, just different side here and here. So again, we leave it for you to review at home if you would like to, but also a very beautiful solution. So, I mean, because you got the question very quickly, maybe you can send us to the chat what solution you used, one, two, or three, or maybe a different one. That would be very good to know. I am really curious. It was either one, which is fast if you see the lines, or three, which is... Yeah, I think two has a lot of ingenuities and it takes time, so I would tell that it's less... Two has not really that much ingenuity, but it does take a lot of time because once you draw the two, yeah, you have to draw the two altitudes. It's just that in the competitions that I do, that's way more common. I draw altitudes all the time. That tends to work. So that's kind of ingrained now. In your problem solving, yes. So just looking back, there's a lot, but really looking at triangles, go through all the special ones, going through similar... A lot of the time, actually, we use only congruent triangles with circles, the inscriber, angle theorems, tensions, draw everything. The most important thing is that when you have a lot of information written in words there in the problem, you want to represent it in the picture. And then these advanced techniques just require a little bit more practice, but if we go through, if we know the basics very, very well, like what may be the solution, you know the 90-30-60 triangle very well, know a little bit of trigonometry, then most of the time, I believe that you can find out the answer to probably the most challenging geometry questions in math kangaroo. And as we said earlier, math kangaroo doesn't seek to test you complicated formulas or you would have an advantage because you know some theorems ahead of time, just knows the basics very well. Practice and I think you're going to be doing well. And yeah, one student replied that he used solution number one. So that's great. That means you are very proficient at drawing actualized. Okay, thank you, everyone.

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