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Common orthocenter and centroid

Video transcript
We're asked to prove that if the orthocenter and centroid of a given triangle are the same point, then the triangle is equilateral. So I have a triangle over here, and we're going to assume that it's orthocenter and centroid are the same point. Just as a review, the orthocenter is the point where the three altitudes of a triangle intersect, and the centroid is a point where the three medians. So we can do is we can assume that these three lines right over here, that these are both altitudes and medians, and that this point right over here is both the orthocenter and the centroid. So if we assume that these lines are altitudes, and that tells us that they are perpendicular to the opposite sides-- so that is a 90 degree angle, those are both 90 degree angles. These are both 90 degree angles. These are both 90 degree angles. And the fact that it's a centroid means that each of these lines bisect at the opposite sides. So it tells us that this length is equal to this length, that this length is a different color, this length is equal to that length, and it tells us that this length-- this length is equal to that length. And you might also say, well, gee. Each of these lines are also, then, the perpendicular bisectors of each of the sides. So not only is this the orthocenter in the centroid, it is also the circumcenter of this triangle right over here. But with that out of the way, we've kind of marked up everything that we can assume, given that this is an orthocenter and a center-- although there are other things, other properties of especially centroids that we know. But now let's prove that this has to be an equilateral triangle. So the first thing that you might see-- and let me label some letters here, so we can refer to things a little bit better. So let's call that A, B, C, D, E, and F-- and we could label the centroid right over here, G. So the first thing, let's look at triangle AFG and triangle-- with E-- and triangle EFG. So let's compare triangle A, F-- triangle A, F, G-- Let's compare that to triangle EFG. Triangle E,F,G. So they definitely have one side-- side EF is congruent to side AF-- these are both the same. And we have angle EFG is the same as angle AFG, they're both 90 degrees. And then they both clearly share this side FG-- they both share this side right over here-- they both share that over there. So they have a congruent side, a corresponding angle in between another congruent side, and they're all-- So congruent side, congruent corresponding angle, and another congruent corresponding side. So by side, angle, side, these two characters are going to be congruent. By side, angle, side congruency. Now can you make that exact same argument to say that all of these pairs that have the kind of share-- that both have these 90 degree angles next to each other are going to be congruent. By the same exact argument, we can say triangle EDG triangle E,D,G, is going to be congruent to triangle CDG. Triangle C,D,G. Same exact thing-- side, angle-- and then we have a side right over here. And then we can use the exact same argument for this one over here. Triangle C-- that looks like an A-- triangle CBG is congruent to triangle ABG. A, B, G. That by itself is interesting. But we know that if two triangles are congruent, all of their corresponding sides and angles are going to be congruent. So for example, if we know the measure of this angle is blue, the corresponding angle on this triangle is also going to have the same measure. I'll just make it with that same blue angle. And we know if this angle right over here is magenta, the corresponding angle on triangle AFG is also going to have that same measure. And I'll just mark it with that magenta again. Now we also know from our properties of vertical angles, that whatever angle measure AFG is, DGC is going to have the same measure, because they are vertical angles. But we know whatever angle measure this is, this triangle, triangle CDG, is congruent to triangle EDG, so corresponding angles have to be congruent. So this angle is this magenta measure, then this angle also has to be the magenta measure. And then once again, you see vertical angles. If this is magenta, then this is also going to have that same measure. And if this has a measure, then that is also going to have the same measure. So by using a little bit of argument of congruent triangles, corresponding angles are going to be congruent, and vertical angles, we can see that all of these inner angles right over here are going to have to be the same measure. And I'm using that with this little magenta arc right over there. Now all of these triangles, that we split this triangle into, they all have a 90 degree angle, they all have a magenta arc. So whatever's left over is going to be 180 minus 90 minus magenta. Or it's really 90 minus this magenta angle right over here. And that's what this blue angle must be. This blue angle is essentially 90 minus the magenta angle. The blue angle is 90 minus the magenta angle. And so that angle must be the third angle for all of these. So once again, this blue angle, there's going to be 90 minus the magenta angle, or 180 minus the magenta minus the 90. So that's going to be a blue angle. Essentially what we're saying is, if you know two angles of a triangle, it forces what the other angle is going to be. We already know that all six of these triangles have two angles in common-- the 90 degree angle and the magenta angle. So the third angle must be the same as well. And we're specifying that with this blue angle, this blue angle right over here. And now you see if we were to look at angle EAC, so angle E,A,C. We see that that's just two of these blue angles. That is congruent to angle ACE, A,C,E, which is just two of those blue angles, which is congruent to angle CEA. Angle C,E,A. Which is once again, just two of those blue angles. So we have a triangle here where the three angles in that triangle all are congruent, so it is an equilateral triangle. It's a 60 degree. We've proven before if all three of your angles are the same, then the lengths of all three sides are the same as well.