this lesson's on circum centers and in centers both of them are centers of triangles but deals with a different type of Center the circum Center is whenever you have the perpendicular bis sectors of the sides of the triangle intersecting at a point that point is called the circum Center the circum Center is always equidistant from the vertices of the triangle basically it will split the sides and be congruent to the angles so all three of these segments are congruent to each other it wants you to use this diagram to the left to answer the following questions first list the perpendicular bis sectors again the perpendicular bis sectors are the segments that split the sides and are perpendicular to the sides so those would be segments DP EP and FP the circum Center is where those three segments meet which is point P now for all the congruent segments we have the segments on the sides of the triangles AF is congruent to CF the ones at the bottom we also have be is congruent to C the ones on the right A D is congruent to BD and then we have all the ones that we talked about in the middle that go from the circum Center to the vertices such as AP BP and CP all of those are the congruent ones now another thing that happens with this is the triangles that share the common sides happen to be congruent to each other so if we were looking at this we have triangle PBE e would be con to Triangle p c e notice they share the side BC these two triangles have to be congruent same thing works with the other triangles on the figure we have the two triangles here that are congruent and then the two triangles a PF and cpf that would be congruent so let's look at example number two it says if H is the circum Center of triangle BCD find the missing measure gives you a bunch of information first one says it wants you to find GD since GD is the side of the triangle and we know that this is a circum Center GH must be the perpendicular bis sector which means it splits the side in half which means since those two segments are congruent half of 24 is 12 for GD we got the next one BC C again BC is the entire side we know what EC is both those parts have to be congruent so that means that the bottom part is also going to be seven which means the entire length is 14 so BC is 14 now for eh we have to do a little bit more work on here since this is a perpendicular bis sector we know that this angle here B is a right angle we have the side seven and the hypoten being 13 we need to find the other leg if you have two sides of a right triangle you can always find the third Side by doing Pythagorean theorem this one would be set up 7^2 + x^2 = to 132 again since we're finding the leg one of the missing variables has to be on the left so we now we're going to do 7 squ which is 49 + x^2 = 2 169 we've got to get X by itself so we got to get R of that 49 so we're going to subtract 49 from both sides gives us now x^2 = to 120 problem is X is still not by itself it is being squared so we have to undo that square the re way that we undo squares is to do a square root if we square root the left we have to square root the right now the problem with this is 120 is not a perfect square so we end up having to do the factor tree which now we have to break down 120 into all the factors that we can have one of the ways to do it is 12 and 10 however 12 still needs to be factored down even more we have six and two all in 10 we have two and five six still can be broken down even more so when we do that we get three and two we have to Circle any pairs that we have happens to have the pair of twos so we would pull those twos out and then multiply everything back together so the two goes on the outside 5 * 3 * 2 ends up giving you 30 so the side that we're looking for eh is 2 square tk3 and that would be our answer here all right so what we have next is we are looking for Segment FD so segment FD has uh we have again it's going to be a right triangle we know that here HD has to be the same as BH because whenever you have the perpendicular bis sectors meaning at the circumcenter the distance from the circumcenter to the angles are always going to be the same so we have 13 on that side as well for HD we have to do another Pythagorean theorem problem because we have that right angle to find FD this time this leg is three the hypotenuse is 13 so setting this up we have 3^2 + x^2 = 132 32 is 9 132 is 69 or 169 have to subtract 9 from both sides now we have x² equals to 160 again we have to get rid of the square so we're going to do the square root of both sides okay 160 is not a perfect square so we have to factor it out one of the ways is 16 and 10 16 breaks down to four and four 10 breaks down into two and five now we could continue to break down the fours if we wanted but we have a pair there so since we have a pair there we can go ahead and Circle that and pull that on the outside so that would end up giving us 4 < TK 10 which means that we have that side being 4 < TK 10 for FD now again since those two triangles on side CD have to be the same we also would know that CF would end up being 4 < TK 10 which means that the entire length which is double that would be 8 < TK 10 now let's look at the in Center the in center deals with the angle bis sectors of the angle where those intersect is called the in Center the in Center is always equidistant from the sides another way that I'd like to use this is it will split the Angles and be congruent to the sides if we look at this the angle bisectors would be segments AP BP and CP all right the in Center would be Point p and then the congruent segments are the segments DP would be congruent to EP all right and it's also congruent to FP so all of the ones that go to the sides are congruent now what's also good about this one is this one gives us that we have these Ang or these sides congruent as well so whenever you're talking about the in Center like the other one where it had the triangle sharing the same sides being congruent this one the triangles that share the same vertexes are congruent so we have the triangle here at the top DBP would be congruent to e BP so these triangles are congruent which means that these segments DB and be are congruent works the same way with all the other triangles on it as well so with the circum Center the triangles sharing the same sides are congruent with the in Center the triangles sharing the same vertexes or vertices are congruent let's look at example number two if G is the in center of triangle ABC find the missing measures first thing it wants you to find is GD so we know that it is perpendicular to the side so we have another right triangle in order to solve this we have to do Pythagorean theorem we're going to do 352 + x^2 = 372 35 squ gives you 1,225 37 squar is 1,369 now you subtract 100 or 1,225 from both sides gives you x^2 = to 144 we have to take the square root of both sides to get X by itself and luckily 144 is a perfect square which would give us X is 12 which means means that GD is 12 now since this is the in Center it is the same distance to all of those sides as well so that also means EG is 12 and FG is 12 so now wants BG BG is a little trick here we don't actually have a number for it but again we do know that this point this angle at beeg is a right angle we have the leg of six and the other leg is 12 so those two angles or those two sides can be put into Pythagorean theorem to solve for the hypotenuse which would be BG 6^ 2 + 122 = 2 x^2 62 is 36 122 is 144 combine those we get 180 = to x^2 now we can't have X squ so we need to do the square root 180 unfortunately cannot be a perfect is not a perfect square so we're going to have to do the factor treat 180 breaks down one of the ways is 18 and 10 we can break down 18 further into nine and two we continue breaking that down and break nine down into three and three and I still got to work with that 10 on the other side so 10 is 2 and five we can notice that we have a pair of twos we also have a pair of Threes so when we do this we have to pull the two and the three out 2 * 3 is 6 leaves the five on the inside if we look here back on the right we have our answer 6 < TK 5 that is the answer for BG the next thing we want is FC again the triangles on the same verticy are congruent which means that the triangles here D CG is congruent to F CG so d C is the same as FC we just basically move that 35 up to the top there's what our answer is for FC now for BF that's the same as the triangle B EEG so BFG is the same so that six can also just basically get slid over to the other side and that is our answer for BF again whenever we're talking about in centers and circum centers the circum Center will end up splitting the side in half and those two sides are congruent the in Center splits the angle in half and then the triangles on those angles are congruent