yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

Proof: perpendicular radius bisects chord


2m read
·Nov 10, 2024

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 to segment AC.

What we want to prove is that segment OD bisects AC. So another way to think about it is that it intersects AC at AC's midpoint.

Pause this video and see if you can have a go at 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. Let me draw these triangles. So I'm going to 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 are 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. This is reflexivity. Reflex, 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. 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 and, or two right triangles, you have a pair of sides are congruent and the hypotenuses 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—segment AM—is going to be, I'm having trouble writing, 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!

More Articles

View All
Analyzing problems involving definite integrals | AP Calculus AB | Khan Academy
The population of a town grows at a rate of ( r(t) = 300 e^{0.3t} ) people per year, where ( t ) is time in years. At time ( t = 2 ), the town’s population is 1200 people. What is the town’s population at ( t = 7 )? Which expression can we use to solve t…
Jack Bogle: How to Tell if the Stock Market is Overvalued (Rare Interview)
That if you go back to 1949 and read Benjamin Graham’s “The Intelligent Investor,” he said never less than 25 or more than 75 percent in either of the two asset classes, bonds and stocks. So you can be 25% stocks and 75% bonds and work 75% stocks and 25% …
How 3D Printing Can Preserve History - Tech+Art | Genius: Picasso
The genius is a word that gets used so much more feminine. I’ve always found that word very problematic. I’m here to change that. Here we are. I was doing a lot of 3D animation and 3D modeling, but just like seeing something that you modeled in a virtual …
Worked example: Maclaurin polynomial | Series | AP Calculus BC | Khan Academy
We’re told that ( f(x) ) is equal to one over the square root of ( x + 1 ), and what we want to figure out is what is the second degree Maclaurin polynomial of ( f ). And like always, pause this video and see if you could have a go at it. So, let’s remin…
Pick a Business Model With Leverage
One more question about leverage. Do you think a choice of business model or a choice of product can also bring a kind of leverage to it? For example, pursuing a business that has network effects, pursuing a business that has brand effects, or other choic…
New Hampshire Summer Learning Series Session 4: Data Informed Instruction
And all right everybody, welcome back or welcome, and hello! My name is Danielle Sullivan, and Barbara Campbell is my co-host today. We are going to be presenting to you on how to enhance teaching with data-informed planning with Khan Academy. Oh, there …