If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Geometric constructions: perpendicular bisector

Sal constructs a perpendicular bisector to a given line segment using compass and straightedge. Created by Sal Khan.

## Want to join the conversation?

• can someone prove why those 2 circle intersection points create a perpendicular bisector?
• You might have to get a piece of paper and draw this out to see it, but if you draw the line and the 2 circles, and then connect each of the two points of intersection with the center of each circle, you get a rhombus(all the sides are radii and therefore equal). Because the original line & the line created by the intersection are both diagonals of the constructed rhombus, they are perpendicular bisectors to each other (a property of rhombuses is that the diagonals are perpendicular bisectors to each other).
• Can these be made exercises so students can practice the constructions? Right now I only see that they're videos.
• As with many things on Khan Academy the content is experiencing constant "tweaking" in terms of what resources are available, which videos and exercises are found in a given catagory or sub-catagory, and how you would find and access them. During the last academic year there were Compass Construction exercises that I had my students (I am a Geometry teacher) practice as part of what they needed to do when we covered constructions. As tools appear similar/identical to the tools that are used on the PARCC test it was time well spent. This year I looked and discovered the exercises I used are no longer available :-( Hopefully will be coming back someday soon.
• Can someone explain this because I don't get it
• We want to find a line that is perpendicular to the given line. Just drawing any line through the given line won't do, we want to find one that is perfectly perpendicular to it. In order to find it, we have to use a number of tools: a compass (which in real life helps us draw perfect circles) and a straight edge (anything that would help you draw a straight line, like using the side of a ruler).

To find the perpendicular line we use two compasses to draw two circles, both on the given line. The places where the two circles meet can hep us draw a perpendicular line. Simply use those two point to draw a line, and that line will be perpendicular to the given line.
• Why is it called a Straightedge? I've never heard of it til now.
• Not trying to be smart with you, but it's because it has a straight edge. While typically you would use a ruler as your straight edge, you could use anything that is long and straight. So the side of a book, or the edge of you desk, or the side of a Monopoly board are all perfectly good straight edges. The main thing to remember is that you are only using it to make lines using points from your construction so you can't cheat by measuring how long a certain line should be.
• My question is not about this video per se but generally answering questions regarding compass constructions. I frequently get my answer, it is marked incorrect, I then scroll through all the hints and the correct answer is right on top of mine. Why why why? Also, does anyone know a way of getting the line segments to stop changing length? I create a segment, drag it somewhere else and it "snaps" onto the closest object, thus changing length. Super frustrating.
• This isn't just you. No matter how I do it, either an equivalent way that produces the same result or the exact way shown in the hints it takes a dozen tries to get it 'right'. SO FRUSTRATING. Apparently I'm never going to get 100% on geometry.
• When drawing the angle bisect does it have to pass the point?
• From what I know, yes. The definition of bisect is to intersect a line at the middle, which is also a useful tool for proofs I may add.
• can someone prove why those 2 circle intersection points create a perpendicular bisector
• This is what user lsathi19 wrote:
"You might have to get a piece of paper and draw this out to see it, but if you draw the line and the 2 circles, and then connect each of the two points of intersection with the center of each circle, you get a rhombus(all the sides are radii and therefore equal). Because the original line & the line created by the intersection are both diagonals of the constructed rhombus, they are perpendicular bisectors to each other (a property of rhombuses is that the diagonals are perpendicular bisectors to each other)."
(1 vote)
• Is there a video about why does this work?
Or can someone take time to explain?
• This is what user lsathi19 wrote:
"You might have to get a piece of paper and draw this out to see it, but if you draw the line and the 2 circles, and then connect each of the two points of intersection with the center of each circle, you get a rhombus(all the sides are radii and therefore equal). Because the original line & the line created by the intersection are both diagonals of the constructed rhombus, they are perpendicular bisectors to each other (a property of rhombuses is that the diagonals are perpendicular bisectors to each other)."
(1 vote)
• can we place the circles anywhere ?
(1 vote)
• I am not sure exactly what you mean because the center of the circles will always have to be at the endpoints, so we cannot place them anywhere. If you are talking about the size of the radius It has to be bigger than 1/2 of the line segment, so if you estimate it to be about 3/4 of the line, you will be in good shape. The radius of the circle can be different sizes, but you still have to use the same radius from both endpoints.
Note that for a perpendicular bisector, every point on it will be equidistant from the two endpoints (If I drew two triangles, they would be congruent by SSS).