 In the last video of section 10.5 on tangents, we're given a diagram and we're told that triangle ACE is isosceles and it circumscribes the circle. This word circumscribes might be new. All that means is that triangle ACE is on the outside of the circle and it creates these three points of tangency at B, D and F. It's the same thing as saying that the circle is inscribed in the triangle. Circumscribes is just the point of view from the outside shape. Regardless, it just means we're going to have these three points of tangency here that will help us set up the problem. Let's go ahead and set up the diagram with what we know because the triangle is isosceles, that means side AC is going to be equal to side CE and because we have the points of tangency, that just means that segment DE will be congruent to AB and then separately CD will be congruent to BC. We also know that we had that theorem that states if we have two segments that come from the same exterior point, such as FE and DE, I'll start at point E here and they are a tangent to a circle, which they are at these points, then those segments will be congruent to each other. That allows us to say that DE will be congruent to FE and then at the same time on the other side AB will be congruent to AF. It's important to note then that we have these four segments that are each going to have the same measure and then separately, these two little segments up here, CD and CB, will have the same measure as well. Let's finish up filling in what we know of the given information. We are told that CD equals 2 and so then of course BC will equal 2 as well. And then this last piece, FC equals 6 is the segment right here. And note that that's not the diameter of a circle. It actually goes from the base of the triangle all the way up to the top vertices, C. And because we know that point F is a point of tangency, we can go ahead and put that right angle symbol in and we can outline the triangle CFE. Anytime we have points of tangency, we should be looking for right angles that will create right triangles and that's what we did here. We're told that F, that segment FC equals 6. So that gives me the short side of my right triangle and now I'm going to fill in the missing information for the long side of the right triangle and the hypotenuse as well. I'm going to assign a value of X to segment FE for that missing piece and because FE is congruent to DE, that's why we put those congruency marks on there to help us remember that, we can assign a value of X to DE as well. Now I have values for all the pieces of my right triangle and because the hypotenuse is going to be 2 plus X, I'm just going to write that out to show the total value of that hypotenuse piece. Hypotenuse then is going to be the sum of those two pieces and now I have the short side, the long side, and the hypotenuse of a right triangle and I of course can use the Pythagorean theorem to set up my equation. Remember when we're setting up, I'm just going to switch that around to X plus 2. When we're setting that up and we're squaring a binomial, keep in mind that that means we want to expand it out and foil if that helps to expand it out like that. Please do so and then we're going to simplify. 6 squared is 36 plus X squared. When we foil, we get X squared plus 2 X plus 2 X equals 4. Combining like terms on each side, we're going to get X squared plus 4 X plus 4 on the right and notice we have the X squareds on both sides of the equation which are going to cancel out. We don't have a quadratic equation because those cancel out and now we're left with 4 X plus 4 equals 36 and we can go ahead and solve for X and we'll get 4 X equals 32 or X equals 8. We're not asked to solve for X here. We're asked to find AC in the perimeter of triangle A, C, E. So this is where we're going to use X equals 8 and fill in the missing values to answer the questions. If we want to find segment AC, we don't have any values over on this side for AC, but remember this is an isosceles triangle. So AC is the same thing as CE and we do know that CE is 2 plus X. So we're going to plug in the fact that X equals 8 and we get 10 over here and I'm just going to write this down over on the side that AC equals 10. That's one of the answers for the problem. And then the last piece is asking us to find the perimeter of the whole triangle and this is where if we know that this whole side is 10 and this whole side is 10, all we have to do is find the base of the triangle. The bottom piece, FE equals X and we know that X equals 8, so this piece equals 8 and then of course AF is congruent, so this piece equals 8 as well. And so the perimeter of the triangle, we're just going to add up those values of the sides. 10 plus 10 plus 8 plus 8. So P for the perimeter, 10 plus 10 plus 8 plus 8. That gives me 36 and I'll just put that up on top as well. The perimeter equals 36. So we're going to add up those values of the sides and we're just going to add up those values of the sides.