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## Inscribed shapes problem solving

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# Proof: perpendicular radius bisects chord

CCSS Math: HSG.C.A.2

## Video transcript

In a previous
video, we've already shown that if we have some
circle centered at O right over here, and that
if OD is a radius, and it's a radius that
bisects chord AC-- so bisects means that it kind of
splits it in two, that AB is equal to BC-- we've
proven in a previous video that OD will be
perpendicular to AC. So we've proven that we can
assume that is perpendicular. And that video, if you
want to look it up, and you might want to
prove it yourself just out of interest, but
the video-- if you do a search on Khan
Academy for radius is perpendicular
to a chord, you'll hopefully find
the proof of that. What I want to do in this
video is go the other way. If we know that OD is
a radius and that it is perpendicular to chord AC,
what I want to do in this video is prove that it's bisecting it. So we're not assuming
that it's bisecting it in this video right over here. We're just assuming
that it's perpendicular. So we're essentially
going to go the other way. Here, we started with the
fact that it bisected it, and we established that
they were perpendicular. That was in the previous video. Now we're going to start
with the assumption that they're perpendicular and
then prove that they bisect. And just like we did
in that previous proof, we'll set up some
triangles here since we know a lot about triangles now. And we'll set up the
triangles by drawing two more radii, radius OC and radius OA. And that's useful
for us because we know that they're both
radii for the same circle. So they have to be
the same length. The radius doesn't
change on a circle. So those two things
are the same length. And you might already
see where this is going. Let me label this point here. Let me call this M
because we're hoping that ends up being
the midpoint of AC. Triangle AMO is
a right triangle. This is its hypotenuse. AO is its hypotenuse. Triangle OMC is
a right triangle, and this is its hypotenuse
right over there. And so already showed
that the hypotenuses have the same length, and both
of these right triangles share segment or side OM. So OM is clearly
equal to itself. And in a previous video,
not the same video where we explained this thing. In a previous video where
we talk about-- and I think you can look it up. I think the video is
called "More on why SSA is not a postulate." In that video, we say that
SSA is not a postulate. So SSA, not a
congruency postulate. But we did establish
in that video that RSH is a
congruency postulate. And RSH tells us that if we
have a right triangle-- that's where the R comes from-- if
we have a right triangle, and we have one of the
sides are congruent, and the hypotenuse
is congruent, then we have two congruent triangles. And if you look at
this right over here, we have two right triangles. AMO is a right triangle. CMO is a right triangle. They have one leg that's
congruent, right over here, MO, and then both of
their hypotenuses are congruent to each other. So, by RSH, we know
that triangle AMO is congruent to triangle
CMO by the RSH postulate. And so if we know that
they're congruent, then their corresponding
sides have to be congruent. So based on that,
we then know that AM is a corresponding side. Let me do that in
a different color. AM is a corresponding
side to MC. So we know that AM
must be equal to MC because they're
corresponding sides. These are corresponding sides. So congruency implies
that these are equal. And if those are equal, then we
know that OD is bisecting AC. So we've established
what we need to do. Another way that we could
have proven it without RSH, is just straight up with
the Pythagorean theorem. We already know, just by
setting up these two radii right over here, we know that OA--
so we draw a little line here. This is another way that
we could have proven it. We already know that
OA is equal to OC. And we also know that
OM is equal to itself. OM is clearly equal to OM. It's equal to itself. And we also know from
the Pythagorean theorem that AM-- do this
in a new color. We know that AM squared plus OM
squared is equal to OA squared. The length of the
two legs squared summed up is equal to the length
of the hypotenuse squared. So we know that for the left
triangle right over here, for AMO. And we can set up the
same relationship for CMO. We know in CMO
that-- and I'm going to try to do it corresponding--
that CM squared plus OM squared is equal to OC squared. Now we know a few things. We know that OA is equal to OC. So, for example, right over
here, where we have OA squared, we could replace this with OC. And then you can already
see where this is going. You can already see that CM and
AM are going to be the same. But if you want to do it a
little bit more formally, you can subtract OM squared from
both sides of this equation, and you'd get AM squared
is equal to OC squared-- I've replaced this with OC
squared-- minus OM squared. So that's on the left-hand
side right over here. And then on the
right-hand side, if we subtract OM squared from both
sides, we have CM squared. CM squared is equal to OC
squared minus OM squared. And then we could take
the principal root of both sides of this
because we really care about the positive
root because we don't want to have
negative distances. So if you take the principal
root of both sides, this becomes AM is equal to
the principal root of that. And we also get that CM is equal
to the principal root of that. Well, these two
quantities are the same. So AM must be equal to CM. They both equal to this
quantity right here. So AM is equal to CM, and
it might be a bisector. And this is really common sense. If you know that two sides of
two different right triangles are going to be
congruent to each other, you can always use the
Pythagorean theorem to get the third side. And that third side is
uniquely constrained by the length of
the other two sides, because it's a right triangle. And so these are all
ways of really getting at the same thing. But now we can feel
pretty good about the fact that if OD is
perpendicular here, then it's definitely going
to bisect this chord.