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.