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- Geometric constructions: congruent angles
- Geometric constructions: parallel line
- Geometric constructions: perpendicular bisector
- Geometric constructions: perpendicular line through a point on the line
- Geometric constructions: perpendicular line through a point not on the line
- Geometric constructions: angle bisector
- Justify constructions
- Congruence FAQ

Geometric constructions: angle bisector

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Sal constructs a line that bisects a given angle using compass and straightedge. Created by Sal Khan.

## Want to join the conversation?

- He says that we know the two triangles are congruent, but he doesn’t prove it. How do we know?•(2 votes)
- He did prove it. Sal showed that each pair of corresponding points is equidistant, thus demonstrating that the sides are all equal. By the SSS theorem, if all three sides of two triangles are the same length, the triangles are congruent.(29 votes)

- I have been watching a couple of videos about trisecting arbitrary angles, using only an unmarked straightedge and compass. It is said to be impossible, because apparently Euclid said so.

But the girl in this video https://www.youtube.com/watch?v=XZPY86eo3XU is doing exactly that. She uses an unmarked straightedge and compass to trisect an angle. Nothing else.

Now on the wikipedia page it says the following, and I quote: “Angle trisection is a classical problem of compass and straightedge constructions of ancient Greek mathematics. It concerns construction of an angle equal to one third of a given arbitrary angle, using only two tools: an unmarked straightedge and a compass.

The problem as stated is impossible to solve for arbitrary angles, as proved by Pierre Wantzel in 1837. However, although there is no way to trisect an angle in general with just a compass and a straightedge, some special angles can be trisected. For example, it is relatively straightforward to trisect a right angle (that is, to construct an angle of measure 30 degrees).

It is possible to trisect an arbitrary angle by using tools other than straightedge and compass.

For example, neusis construction, also known to ancient Greeks, involves simultaneous sliding and rotation of a marked straightedge, which cannot be achieved with the original tools” end quote.

Note, a marked straightedge, like a ruler or something. She uses a neusis construction with an unmarked straightedge and compass.

Has she nailed it? If not, why not? She fullfills the requirement exactly. She trisects arbitrary angles using only a unmarked straightedge and compass using a neusis construction. How can this possibly not be a total win. XD•(12 votes)- No, she has not. The problem require constructing it with only the straightedge and compass. Any other devices or constructions used are not allowed.

The neusis requires marking of the straight edge. We can’t know that for these problems. It’s meant to be a way to make lines. In general, if the method used needs something else other than

: a way to make a straight line

: a way to make a circle of any size

it is not allowed.(5 votes)

- No, she has not. The problem require constructing it with only the straightedge and compass. Any other devices or constructions used are not allowed.
- Hey Sal, I loved your video but can you put some practice questions? It would really help, the video was awesome and I understood it!•(8 votes)
- Can you still access these exercises?•(4 votes)
- No because these exercises are old (before renovation) Khan Academy.(2 votes)

- Why is this nothing like the practice?•(4 votes)
- Sometimes, the three circle point does not work for an angle because the points are not able to intersect the same circle. How do you find the angle bisector in this case?•(4 votes)
- angle 75

How can we draw angle 75 with a compass•(4 votes)- So you could use a 90 degree angle and a 15 degree angle, so the 3rd angle would be 75 degrees. To get the 90, use a right triangle, and to get the 15, use an equilateral triangle, bisect the 60 degree in half, and then the 30 degree in half to get the 15 degree angle. But it’s a pretty complex shape involving both a right triangle and an equilateral triangle combined. (Same answer as above)(2 votes)

- Is the angle bisector also the side bisector? And also is the angle bisector always perpendicular to the side opposite the angle? Pls help.•(2 votes)
- Question 1:

No, unless you have equilateral or isosceles triangle.

actually name of side bisector is median it equally divides side into two parts.

Question 2:

Also No, unless you have equilateral or isosceles triangle.

*in isosceles triangle it is only true if you construct

bisector from not equal angles. so if isosceles triangle has angles: 30, 30 and 120. bisector will be median and also a perpendicular if you construct it from 120 degree angle.(5 votes)

- Question 1:
- I read that all constructible points are actually solutions to simultaneous polynomial equations whose degrees are all powers of two. The author bases his argument of off the fact that in constructions the only two allowed instruments are ruler and compass, which generate one and two degree equations respectively, but I don’t understand how you can extend this to higer powers of two(I understand that you could, I don’t understand why it is necessary, since the operations used in the equations of lines and circles would not generate higher powers as far I can see, even if done in succesively) . In addition, angles don’t really seem to work in this context, since they are defined in radians, and since pi is transcendental, this doesn’t seem to make sense in this context. Finally, I would appreciate tips on how to avoid run on sentences and large brackets, and I apologize if my question was difficult to read. Thanks!•(3 votes)
- If you treat the plane that you’re working in as the complex plane, and so treat each constructible point as a complex number, then you can construct the square root of any constructible number.

Since you can construct 2 (stick two unit segments together), you can also construct √2 (by taking its square root), and so you can also construct √√2, or the fourth root of 2, by taking its square root, and so on. This is how we get arbitrarily high powers of 2.

Constructing an angle of π/2 radians is not the same as constructing the number π/2. Constructing π/2 is impossible, as you point out, because π is transcendental. This means we cannot construct the point whose coordinates are (π/2, 0). That has no bearing on constructing a right angle (of measure π/2), which is straightforward.(3 votes)

- If you treat the plane that you’re working in as the complex plane, and so treat each constructible point as a complex number, then you can construct the square root of any constructible number.
- Where did the exercises go that used to be linked to this video?•(4 votes)

## Video transcript

We’re asked to construct an angle bisector for the given angle. So this is the angle they’re talking about. And they want us to make a line that goes right in between that angle, that divides that angle into two angles that have equal measure, that have half the measure of the first angle. So let’s first find two points that are equidistant from this point right over here on each of these rays. So to do that, let’s draw one circle here. And I can make this of any radius. Wherever this intersects with the rays, that’s where I’m going to put a point. So let’s say, here and here. Notice both of these points, since they’re both on this circle, are going to be equidistant from this point, which is the center of the circle. Now, what I want to do is construct a line that is equidistant from both of these points. And we’ve done that already when we looked at perpendicular bisectors for lines in this construction module. So let’s do that. So let’s add their compass. And so what I want to do, this circle is centered at this point. And it has a radius equal to the distance between this point and that point. And then I do that again. So this circle is centered at this point and has a radius equal to the distance between that point and that point. And then the two places where they intersect are equidistant to both of these points. And so we can now draw our angle bisector, just like that. And you might say, well, how do we really know that this angle is equal to this angle? Well, there’s a couple ways we can tell. We know this distance right over here is equal to this distance right over there. We know that this distance over here is equal to this distance over here. And both of these triangles share this line. So essentially, if you look at this point, this point, and this point, that forms a triangle. And if you look at this point, this point, and this point, that forms a triangle. We know those two triangles are congruent, so this angle must be equal to this angle. These are the corresponding angles. So they’re going to be congruent. This is an angle bisector.