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what I want to do in this video is prove that the circumcenter of a right triangle is actually the midpoint of the hypotenuse and to do that I'm going to take first take a look at the the perpendicular bisector of one of the legs of this of this right triangle so let me construct the perpendicular bisector of leg BC right over here so it's going to look something like this it's going to look something like this this it intersects at a right angle that's perpendicular and it bisects it so be the distance from B to this point which we'll call em maybe m4 midpoint is the same as the distance from M to C so those two distances are going to be equal and let's call the point where this perpendicular bisector intersects the hypotenuse let's call this o and we're going to prove that o is the circumcenter of this right triangle now the first thing that you might realize and this is what we've seen in many problems the triangle OB m looks similar to triangle ABC and it's actually not too hard to prove they both already have a 90-degree angle so if we show that they they both have another angle another set of corresponding angles that are congruent to each other then we know that they're similar by a a similarity and they both clearly share this angle right over here o BC is part of the smaller triangle and ABC which is really the same angle as part of the larger triangle and so on they also obviously share a 90-degree angle so by a a triangle similarity we have triangle OB m OB M is similar is similar to triangle a b c is similar to triangle a b c and what's useful about this is we know that similar triangles the ratios between corresponding sides are constant so for example we know that the ratio between side BM which is on the smaller triangle we know that the ratio between BM let me do this in a different color just to just for the sake of it we know that the ratio between BM and bc b and BC the ratio of this side on the smaller triangle to the corresponding side on the larger triangle is going to be the same as the ratio of the hypotenuse on the smaller triangle bo to the hypotenuse of the larger triangle because they are similar well we know what the ratio of BM to BC is BM is half of BC so this ratio over here the ratio of BM to BC is going to be equal to 1/2 this is M is the bi is the midpoint of these things so this is exactly the same distance as this so this is one half of the entire BC so if one half is equal to BM / bc is equal to bo / ba we then know if we just kind of ignore this middle part right over here that 1/2 is equal to Bo over ba / ba and if you cross multiply it if you cross multiply you see that well there's multiple ways to think about it but you could just cross multiply and you say ba is equal to 2 Bo or if you divide both sides by 2 and they're really equivalent statements one half ba is equal to Bo so Bo is 1/2 of B a so this is 1/2 ba and so this other length AO righty over here this is going to be B this is going to be this is going to be ba minus 1/2 ba so this is also going to be 1/2 ba and so this segment right over here AO AO is going to be congruent to OB so we just shown first of all is that this perpendicular bisector right over here the perpendicular bisector of segment BC it intersects the hypotenuse of our right triangle at the midpoint so we've already established so we one thing that we've re-established is au is the midpoint is the midpoint of the hypotenuse of the hypotenuse of the hypotenuse a B well that by itself is interesting but we also know that if a point sits on a perpendicular bisector of a segment it's equal distant it's equidistant from the end points of the segment we'd show that in a previous video so we also know that Oh oby it's equidistant from the endpoints of this segment right over here that OB is equal to OC but we know from this first statement right over here that OB is also equal to oay OB is also equal to oay and so if OB is equal to OC OB is equal to oay that means OC must be equal to oay OC must be equal to oay or another way to think about it is that this point O is equidistant from all of the points on our truck all of the vertices I should say this point O is equidistant from all of the vertices of our triangle of our triangle so this distance this distance which is really going to become our circumradius is the same as this distance right over here which is the same as this distance right over there so that we know that all is equidistant equidistant to all all vertices which is another way of saying that o is the circumcenter o is the circum Center so we've just proven that if you have the circumcenter of a right triangle it is the midpoint of the hypotenuse of the right triangle or the other way around that the hypotenuse of the right triangle is the circumcenter because you only have one circumcenter of any of any triangle