High school geometry
Geometry proof problem: squared circle
Sal finds a missing angle using triangle congruence in a diagram that contains sector of a circle inscribed within a square. Created by Sal Khan.
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- How was it proven that Line EC and Line EB and Line BC are all equal?(4 votes)
- Because ABCD is a square, AB=BC=CD=DA. Because arc AC is part of circle B, that means BE is a radius as well as BA and BC and are, therefore, all equal. When Sal drew EC, he created triangle ECG and showed it was congruent to triangle EBG by SAS. Since BE and EC are both the hypotenuse of congruent triangles, they are equal. So EC=EB=BC.(5 votes)
- So its still considered a triangle if one or more of its legs are not a straight line? As long as it still has three legs that total to 180 degrees?(2 votes)
- Nope. A triangle must have 3 straight lines with interior angles adding up to 180 degrees.
Happy to help!(9 votes)
- He said that the harder geometry problems revolve around drawing the right triangles and visualizing the right lines. How can I get there(Knowing which lines/triangles to draw)?(4 votes)
- In respect to this, a good way to train your brain is just by practicing different types of problems altogether. On KA, a good way to do this is by taking the course challenge multiple times. Of course, you will also need to know the material well enough to use it when it's needed.
You might find it helpful to work on a difficult problem backwards. Say, if you have four lines that intersect at random angles and you need to find one specific measurement that is not offered, you might work backwards from the angle trying to figure out how to get everything you need. You could also work it forwards, finding all the measurements until you know how to get what you need.
But these are the difficulties of math: finding what operations you need to perform and not making small (or large!) mistakes. Anyone else who has tips is welcome to reply, as well. This answer to this question might very well be different to everyone!(3 votes)
- At7:15, How can you just add x+x+30=180 degrees to solve for the missing angles? How would you which one would be 75 degrees?(3 votes)
- They are both 75 because it is an isosceles triangle (two of its sides and two of its interior angles are the same)(4 votes)
- How often will we use proof theorems/problems in real life?(3 votes)
- The math itself may not be too useful, but the critical thinking and logic is.(3 votes)
- At6:52wouldn't it make sense for the missing angle to be 60˚, because CE could act as a transversal to ∆DEC, so angle CEB = angle DCE. So then the equation would be 2x + 60 = 180. So the two base angles for ∆DEC would be 60˚ right? Wouldn't that be true? Because most of these diagrams aren't draw to scale.(2 votes)
- Would the triangles at2:59be using the postulate RSH?(2 votes)
- At2:07we draw the segment EC. So, my question is can you add any segments you want in the square as we did for EC to solve a problem? Or are there any rules for adding such segments?(1 vote)
- No, you can add whatever you want! There are no rules as to what you can add to a problem(2 votes)
- At4:42I did something different, why is this incorrect?
Instead of doing Sal's solution, I removed the curved line of EC. This left a triangle, that was SS congruent (therefore not congruent) with the former. I.e ∆DEC SS ∆BEC, from DC = BC and EC = EC.
From there, I figured, that since it is a square, the sides are parallel. We can see, that FG and DC are parallel, since FG is perpendicular (a flat line) and DC is a flat side of a square.This means, that BC is a transversal, which means, that ∆DEC must at least share one angle with ∆BEC, and they're therefore both kongruent, which means I got the result, that BED = 120.
Why is this incorrect?(1 vote)
- Several issues in the beginning you say SS congruent (therefore not congruent) which appears to conflict with each other, so I have no clue what you are talking about there. Secondly, there is no such thing as SS congruence, it has to be SSS congruence and it appears to be obvious that BE is not equal to DE.(2 votes)
- Could you make Tri EBG and Tri ECG congruent by using Hypotenuse leg ?(1 vote)
So we're told that quadrilateral ABCD is a square, which tells us that the four sides have equal length and that all the interior angles are 90 degrees. We also know that FG is a perpendicular bisector of BC. So we've already shown that it's perpendicular, that this is a 90-degree angle. But it also bisects BC, so this length is equal to that length right over there. Then they tell us that arc AC-- that's a curve on top-- arc AC is part of circle B. So this is a circle centered at B. So if this is the center of the circle, this is part of that circle. It's really kind of the bottom left quarter of that circle. And then given that information, they want us to find what the measure of angle BED is. So what is BED? So we need to figure out the measure of this angle right over here. And I encourage you to pause it and try it out. And you might imagine-- well, you could pause it and try it without any hints. And now I will give you a hint if you tried it the first time and you weren't able to do it. And you should pause again after this hint-- try to draw some triangles that maybe split up this angle into a couple of different angles. And it might be a little bit easier. You might be able to use some of what we know about triangles. Now, with that said, I will try to solve it. And you should pause at any point where you think that you know exactly how to do this and try to do it yourself. So the trick here to realize that this is a circle. And so any line that goes between B and any point on this arc is going to be equal to the radius of the circle. So AB is equal to the radius of the circle. BE is equal to the radius of the circle. And we can keep drawing other things that are equal to the radius of the circle. BC is equal to the radius of the circle. So let's think about it a little bit. If we were to draw it-- and a lot of trickier geometry problems really all revolve around drawing the right lines or visualizing the right triangles. And I'll do one right here that might open up a lot for you in terms of thinking about how to do this problem. So let me draw segment EC. I'll draw that as straight as possible. I can draw a better job of that, so segment EC. Now something becomes interesting. Because what is the relationship between triangle EBG and triangle ECG? Well, they both definitely share this side right over here. They both share side EG. And then BG is equal to GC, and they both have 90-degree angles. You have a 90-degree angle here. You have a 90-degree angle there. So you see by side-angle-side that these two triangles are going to be congruent. So we know that triangle EBG is congruent to triangle ECG-- I should emphasize the C, not the E-- by side-angle-side congruency. And that also tells us that all of the corresponding angles and sides are going to be the same. So that tells us right there that EC is equal to EB. So we know that EB is equal to EC. And what also is equal to that length? Well, once again, this is the radius of the circle. BE is one radius of the circle going from the center to the arc. But so is BC. It is also a radius of the circle going from the center to the arc. So this is also equal to BC. So I could draw the three things right here. I'm referring to the entire thing, not just one of the segments, all of BC. So what kind of a triangle is this right over here, triangle BEC? Triangle BEC is equilateral. And we know that because all three sides are equal, so that tells us that all of its angles are equal. So that tells us that the measure of angle BEC-- we're not done yet, but it gets us close-- is 60 degrees. So the measure of angle BEC right over there is 60 degrees. So that gives us part of the problem. BEC is part of the angle BED. If we can just figure out the measure of angle CED now, if we can figure out this angle right over here, we just add that to 60 degrees, and we're done. We've figured out the entire BED. Now, let's think about how we can do this right over here. So there's a couple of interesting things that we already do know. We know that this right over here is equal to the radius of the circle. And we also know that this length down here-- this is a square-- we know that this length down here is the same as this length up here, that these are the exact same length. And this is equal to the radius of the circle. We already put these three slashes here. BC is the same as that length is the same as that length. And so all four sides are going to be that same length because this is a square. So let me write this down. I'll just write it this way. Because it's a square, we know that CD is equal to BC, which is equal to-- and we already established this-- is equal to EC, which is equal to EB, which is equal to EC. But the important thing here is to realize that this and this are the same length. And the reason why that is interesting is it lets us know that this is an isosceles triangle. So isosceles triangle, if you have your two legs of it, the two base angles are going to be congruent. So whatever angle this green angle is, this angle is going to be as well. So if somehow we can figure out this angle right over here, we can subtract that from 180, and then divide by 2 to figure out these two. Because we know that they're the same. So how can we figure out this angle? Well, we know all the angles of this thing. We can figure out the angles of all of this larger one up here. We know this is an equilateral triangle. So this over here has to be 60 degrees, as well. That's 60 degrees. And that is also 60 degrees. In fact, I could write over here, which is equal to the measure of angle BCE. So if this is 60 degrees, we know we're dealing with a square, so this whole angle over here is a right angle. What is the measure of angle ECD? What is this angle right over here? Let me do this in a new color. This right over here is going to have to be 30 degrees, so that is going to be 30 degrees. And now, we're ready to solve for these two base angles. If we call these x-- and we know they have to be the same-- we have x plus x plus 30 degrees. x plus x plus 30 degrees is going to be equal to 180 degrees. That's the sum of all of the interior angles of a triangle. So you get 2x plus 30 is equal to 180 degrees. Now, you can subtract 30 from both sides. And we are left with 2x is equal to 150. Divide both sides by 2, you get x is equal to 75. So we've figured out that x is equal to 75. And now we're at the home stretch. We need to figure out angle BED. So x is equal to the measure of angle CED. So BED is CED plus BEC, so the 60 degrees plus 75 degrees. So it's going to be-- ready for the drum roll-- this is going to be equal to 75 degrees plus 60 degrees, which is equal to 135 degrees. And we are done.