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### Course: Class 10 (Foundation) > Unit 12

Lesson 1: Chords in circles# Proof: perpendicular radius bisects chord

Simple proof using right triangle-side-hypotenuse (RSH) congruence criterion to show that a radius perpendicular to a chord bisects it. Created by Sal Khan.

## Want to join the conversation?

- Where can this be used in the real world?(56 votes)
- Computer programming, software, carpentry, and other things(6 votes)

- Is RSH at2:39a real theorem? Or just another name for the HL postulate?(15 votes)
- RSH is actually the HL congruence Theorem(5 votes)

- is the principal root the same as a square root?(4 votes)
- A positive number has two square roots; a negative root and a positive root. The principal square root (that is, the one denoted by the radical sign) is the positive one.(13 votes)

- So a "chord" is just a line, or a line segment or a ray within a circle?...(1 vote)
- It's a line segment who's ends are on the circumference of the circle.(13 votes)

- I've always wondered this...

At4:50, Sal uses the Pythagorean Theorem, A squared + B squared = C squared

My question is, why can't you just find the square root of both sides of the equation and get rid

of the squared ?(2 votes)- Because the square root of A^2 + B^2 does not equal A + B. (A + B)^2 = A^2 + 2AB + B^2, which does not equal C^2. Look at https://www.khanacademy.org/math/algebra/quadratics/factoring_quadratics/v/factoring-quadratic-expressions.(7 votes)

- I understand how am=mc but how does this prove that the radius bisects the chord?(3 votes)
- To "bisect" something means to divide it into two exactly equal parts. Therefore if you "bisect" a line you are cutting it in half giving you two smaller lines of equal measure. Hope that helps.(3 votes)

- Where does the s and the h stand for in this postulate?(2 votes)
- He is calling it RSH, meaning
`R`

ight triangle`S`

ide is congruent`H`

ypotenuse is congruent

I have seen it called HL more often, meaning:`H`

ypotenuse is congruent`L`

eg is congruent(4 votes)

- At6:00, what is the principle root of both sides? How can you take the square root of everything? If you can do that, cant Pythagorean theorem be A + B = C? It doesn't seem algebraically correct.(1 vote)
- The principal root of both sides is the non-negative square root of both sides.

Taking the square root of A^2+B^2 means adding A^2 and B^2 together then taking the square root of the sum. You only get A+B=C if you take the square root of A^2 and B^2 separately then adding it together. If you take A+B=C and square both sides you would get A^2+2AB+B^2=C^2 instead of A^2+B^2=C^2.(4 votes)

- the radius of a circle is 20cm . which of the follwing cannot be the length of a chord of the circle?(1 vote)
- Hello,

Remember that a chord is a line that touches a circle at two points. The longest chord a circle can have passes through the center of the circle. Since this chord passes through the center of the circle and touches it on both sides, it is also a diameter. You know that a diameter is equal to two radii. Therefore, any length that is greater than 0 and less than or equal to the diameter would be a valid chord length.

Hope that helps :-)(3 votes)

- Is there a video where sal does a formal proof of some of the postulates?(2 votes)

## Video transcript

- [Instructor] So we have this circle called circle O, based on the point at its center, and we have the segment OD, and we're told that segment OD, is a radius of circle O, fair enough. We're also told that segment OD is
perpendicular to this chord to chord AC, or two segment AC. And what we wanna prove is
that segment OD bisects AC. So another way to think about it, it intersects AC at AC's midpoint. So pause this video and see if
you can have a go with that. All right, now let's go
through this together. And the way that I'm going to do this, is by establishing two
congruent triangles. And let me draw these triangles. So I'm gonna draw one
radius going from O to C and another from A to O. Now we know that the
length AO is equal to OC because AO and OC, both radii. In a circle, the length of
the radius does not change. So I can put that right over there. And then we also know that OM is going to be congruent to itself, it's a side in both of these triangles. So let me just write it this way. OM is going to be congruent to OM and this is reflexivity. Reflexivity kind of obvious. It's going to be equal to itself, it's going to be congruent to itself. So you have it just like that. And now we have two right triangles. How do I know they're right triangles? Well, they told us that segment OD is
perpendicular to segment AC and our assumptions in our given. Now if you just had two triangles, that had two pairs of congruent sides that is not enough to establish
congruency of the triangles. But if you're dealing
with two right triangles, then it is enough. And there's two ways to think about it. We had thought about the RSH postulate where if you have a right triangle or two right triangles, you have a pair of sides are congruent, a pair and the hypothesis are congruent that means that the two
triangles are congruent. But another way to think about it, which is a little bit of common sense is using the Pythagorean theorem. If you know two sides of a right triangle, the Pythagorean theorem would tell us that you could determine
what the other side is. And so what we could say is and let's just use RSH for now, but you could also say we can
use the Pythagorean theorem to establish that AM is
going to be congruent to MC, but let me just write it this way. I will write that triangle, AMO is congruent to triangle CMO by RSH. And if the triangles are congruent then the corresponding
sides must be congruent. So, therefore, we know that AM, AM, segment AM is going to be, I'm having trouble writing congruent is going to be congruent to segment CM, that these are going to
have the same measure. And if they have the same measure, we have just shown that
M is the midpoint of AC or that OD bisects AC. So, let me just write it that way. Therefore, OD bisects AC. Segment OD bisects
segment AC and we're done.