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## Geometry (Eureka Math/EngageNY)

### Unit 5: Lesson 4

Topic C: Secants and tangents- Proof: Radius is perpendicular to tangent line
- Determining tangent lines: angles
- Determining tangent lines: lengths
- Proof: Segments tangent to circle from outside point are congruent
- Tangents of circles problem (example 1)
- Tangents of circles problem (example 2)
- Tangents of circles problem (example 3)
- Tangents of circles problems
- Challenge problems: radius & tangent
- Challenge problems: circumscribing shapes
- Geometric constructions: circle tangent
- Geometric constructions: circle tangent (example 2)

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# Proof: Radius is perpendicular to tangent line

CCSS.Math:

Sal proves that the radius that connects the intersection point of a tangent line with the circle is perpendicular to the tangent line.

## Video transcript

- [Voiceover] So what
we have here is a circle with the center at Point O, and then we have a tangent
line to the circle. Let me actually label this line. Let's call this Line L. And we see at Point A is the point that the tangent line
intersects with the circle, and then we've drawn a radius from the center of the circle to Point A. Now what we want to do in this video is prove to ourselves that this radius and that this tangent line
intersect at a right angle. We want to prove to ourselves that they intersect at a right angle. And the first step to doing that is we're going to feel good, we're going to prove to
ourselves that Point A is the closest point on Line L to the center of our circle. So I want to prove, prove that Point A is closest, closest point on L to Point O. To Point O. And I encourage you to pause the video and see if you can prove that to yourself. Well, to think about
that, just think about any other point on Line L. Pick any other arbitrary point on Line L. It could be this point right over here. It could be this point right over here. It could be this point right over here. And you immediately see that
it sits outside of the circle. And if it's sitting outside of the circle, I'll pick this point here, just so it'll become a little
bit clearer on our diagram, if it's sitting outside of the circle, in order to get from
Point O to this point, I'm going to call this
Point B right over here, you have to go the length of the radius, you have to go the length of the radius, and then you have to go some more. So this length, the length of Segment O B, is clearly going to be longer
than the length of the radius, because you have to go to the radius to get to the circle itself and then you have to
go a little bit further for any point that sits
outside of the circle. So Point A is the only point, by definition this is a tangent line, it's the only point
that sits on the circle. Every other point on Line L
sits outside of the circle, so it's going to be further. So Point A, hopefully
this makes you feel good, because you pick any other point, it's going to sit outside of the circle, so you have to go to the
radius and then some. So hopefully this makes
you feel good that Point A is the closest point on L
to the center of the circle. Now, we're not done yet, we
now have to convince ourselves that if we have a point and a line, that the segment connecting the point to the closest point on the
line, to that original point, that, that's going to be
perpendicular to the line. So let me give ourselves some space here. We want to prove, we want to prove that if we, that the segment, segment connecting, connecting a point off the line, off the line, and closest point on the line, closest point on the line, is perpendicular, is perpendicular to the line. So what we want to do,
is we want to say, hey, if we have some line here, if we have some line here, L, and if you were to take
a point off the line, so let's say, so you wanted
a point off the line, so let's say that, that's
this point right over here, Point O, and you want the
segment connecting the point off the line to the
closest point on the line. So the closest point on the line, so let's say that this is the
closest point on the line, we want to feel good that
this segment connecting them, so let me do this in a new color, that the segment connecting them is going to be perpendicular to the line, that it goes just straight down like that, that it's going to be perpendicular. And I'm gonna prove this by contradiction. I'm going to assume that
it's not perpendicular. So assume, assume that the segment, segment connecting, connecting, this is kind of wordy, a point off line and closest point to the line, closest point to line, is not, is not perpendicular to the line. So how can I visualize that? Well, I could draw my
line right over here. So that's Line L, and let's say I have my
Point O right over here. Point O, and let's say the
closest point on Line L to Point O, let's say that it's not, so let's say it's over here, that if I were to
connect these two points, that it's not perpendicular to Line L. So this is the closest point, let's call this Point A, and let's say that the
segment connecting these two is not perpendicular to the line. So let me get, so let's assume
this is not perpendicular. So this angle, this angle, this angle right here, is
not, is not 90 degrees. So if we assume that, the reason that this is going
to be a proof by contradiction is I can show that if
this is not 90 degrees, that I can always find a point
that is going to be closer, another point on Line L
that is going to be closer to Point O, which contradicts the fact that this was supposed
to be the closest point, A was supposed to be the
closest point on Line L to O. And how do I always find a closer point? Well, I construct a right triangle. I can construct a right triangle, just like this. I can construct a right
triangle like that, and we see that this distance, let's call this distance
right over here a, and we could call the
base of this triangle b, let me do this in a different color. So a, b, that's the base
of the right triangle, and the hypotenuse is
the distance from O to A. We could call that c. We know from the Pythagorean Theorem that a squared plus b squared, plus b squared, let me do this, plus b squared, is going to be equal to c squared. Is going to be equal to c squared. And so b squared, if we have a non-degenerate
triangle right over here, this is going to be some
positive value over here, and so a is going to be less than c. So this gets us to the conclusion, because if this is some
positive value here, and a and c are positive, everything are positive distances, then this tells us that
a has to be less than c. That a non-hypotenuse
side of a right triangle, of a non-degenerate right triangle, assuming it has some area, is going to be shorter
than the hypotenuse. The hypotenuse is the longest side. So a is going to be less than c, which would tell us, if a is less than c, that we've found another point, let's call this point, I don't know, I've used a lot of letters here, let's call that Point D. D is going to be closer. D is closer. D is closer. So we've just set up a contradiction. We assumed A was the closest
point on Line L to Point O, but we assumed the segment connecting them is not at a 90-degree angle. If it's not at a 90-degree angle, then we can drop a perpendicular
and find a closer point, which is a contradiction to the fact that A was supposed to
be the closer point. So this leads to a contradiction. Contradiction. Contradiction. Because you can actually
find that this is not the closest point, you can
always find a closer point, so therefore, the segment
connecting a point off the line to the closest point to the
line must be perpendicular. Must be perpendicular. So the segment connecting
a point off the line, to a closest point on the line, that must be perpendicular to the line. Must be perpendicular to the line. And just like that, we hopefully
feel good about the idea that if you have a radius, and the point at which it intersects a tangent line to the circle, that, that forms a 90-degree angle. The radius and the tangent line.