Proof: Segments tangent to circle from outside point are congruent

- [Voiceover] So I have a circle here, with a center at point O and let's pick an arbitrary point that sits outside of the circle. So let me just pick this point right over here. Point A. And if I have an arbitrary point outside of the circle, I can actually draw two different tangent lines that contain A, that are tangent to this circle. Let me draw them. So one of them would look like this. Let me just start right over here. So I make a tangent to the circle. So it could look like that. And then the other one would look like this, would look like, would look like that. Now let's say that the point that the tangent lines intersect the circle, let's say this point right over here, is point B, and this point right over here is point C. This is point C. Right over here. What I want to prove What I want to prove is that the segment AB is congruent to the segment AC, or another way of thinking about it, I want to prove, let me do this in a new color that I haven't used yet, I want to prove that this segment, right over here, is congruent to this segment... is congruent to that segment. I encourage you to pause the video and try to work it out on your own before I work through it. Alright, now let's try to work through this together. And to work through this together, I'm going to actually set up two triangles. Two triangles. And they're going to be right triangles as we'll see in a second. So let me draw some lines here to set up our triangles. So I'm going to draw one line, just like that, and then we draw that. And then, let me draw that. Now what do we know about these triangles? Well, as I mentioned, we're going to be dealing with right triangles. How do I know that? Well, in a previous video, we saw that if we have a radius intersecting a tangent line, that they intersect at right angles. We've proven this. So that's a radius. That's a tangent line. They're going to intersect at a right angle. So radius, tangent line, they intersect at a right angle. We also know, since OB and OC are both radii, that they're both the length of the radius of the circle. So this side, right over here, let me do some new colors here, so this side is going to be congruent to that side. And you can see that the hypotenuse of both circles is the same side, side OA, so of course, it is equal to itself. This is equal to itself. So we see triangle ABO and triangle ACO, they're both right triangles that have two sides in common, in particular, they both have a hypotenuse that are equal to each other and they both have a base, or a leg. And we know from Hypotenuse-Leg Congruence that if you have two right triangles, where the hypotenuses are equal, and you have a leg that are equal, then the both triangles are going to be congruent. So triangle ABO is congruent to triangle ACO. And in that proof where we prove it, as well, the Pythagorean theorem tells us if you know two sides of a right triangle that determines what the third side is. So the third side, so length of AB is going to be the same thing as the length of AC. Once again, it just comes out of, these are both right triangles, if two sides, if two corresponding sides of these two right triangles are congruent, then the third side has to, that comes straight from the Pythagorean theorem. And there you go. We hopefully feel good about the fact that AB is going to be congruent to AC. Or another way to think about it, if I take a point outside of a circle, and I construct segments that are tangent to the circle, that those two segments are going to be congruent to each other.