All about angle bisectors
In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the Every triangle has three distinct excircles, each tangent to one of the triangle's can be found as the intersection of the three internal angle bisectors. .. of trilinear coordinates; in this group, the incenter forms the identity element. parallel sides, “one pointing up” and In general, angle bisectors of a triangle where the three angle bisectors concur. meet the circumcircle again at P. . Find with proof the minimum total number of performances by these groups. description of the lesson, and a wrap-up activity. Actual student .. For example, the three angle bisectors of a triangle always meet in exactly .. example, any triangle will have three medians and these three medians will intersect in . Activity: 1. Brainstorming & Analogizing: Working in small groups, the class brainstorms.
So we've drawn a triangle here, and we've done this before. We can always drop an altitude from this side of the triangle right over here. So we can set up a line right over here. Let me draw it like this. So let's just drop an altitude right over here. Although we're really not dropping it. We're kind of lifting an altitude in this case. But if you rotated this around so that the triangle looked like this, so this was B, this is A, and that C was up here, you would really be dropping this altitude.
And so you can construct this line so it is at a right angle with AB, and let me call this the point at which it intersects M. So to prove that C lies on the perpendicular bisector, we really have to show that CM is a segment on the perpendicular bisector, and the way we've constructed it, it is already perpendicular.
We really just have to show that it bisects AB. So what we have right over here, we have two right angles. If this is a right angle here, this one clearly has to be the way we constructed it. It's at a right angle. And then we know that the CM is going to be equal to itself. And so this is a right angle. We have a leg, and we have a hypotenuse.
Common orthocenter and centroid
We know by the RSH postulate, we have a right angle. We have one corresponding leg that's congruent to the other corresponding leg on the other triangle. We have a hypotenuse that's congruent to the other hypotenuse, so that means that our two triangles are congruent. Well, if they're congruent, then their corresponding sides are going to be congruent. So that tells us that AM must be equal to BM because they're their corresponding sides.
So this side right over here is going to be congruent to that side. So this really is bisecting AB. So this line MC really is on the perpendicular bisector. And the whole reason why we're doing this is now we can do some interesting things with perpendicular bisectors and points that are equidistant from points and do them with triangles.
So just to review, we found, hey if any point sits on a perpendicular bisector of a segment, it's equidistant from the endpoints of a segment, and we went the other way. If any point is equidistant from the endpoints of a segment, it sits on the perpendicular bisector of that segment.
So let's apply those ideas to a triangle now. So let me draw myself an arbitrary triangle. I'll try to draw it fairly large. So let's say that's a triangle of some kind. Let me give ourselves some labels to this triangle. That's point A, point B, and point C. You could call this triangle ABC. Now, let me just construct the perpendicular bisector of segment AB. So it's going to bisect it. So this distance is going to be equal to this distance, and it's going to be perpendicular.
So it looks something like that. And it will be perpendicular. Actually, let me draw this a little different because of the way I've drawn this triangle, it's making us get close to a special case, which we will actually talk about in the next video. Let me draw this triangle a little bit differently.
Let a denote half the angle BAC. Combining the two gives the required identity: A dynamic illustration of a different proof can be found elsewhere. And there is another one. Via Transitivity of Equality An angle bisector of an angle is known to be the locus of points equidistant from the two rays half-lines forming the angle. Existence of the incenter is then a consequence of the transitivity property of equality. Angle Bisectors as Axes of a 2-line If we adopt Frank Morley's outlooktransitivity of equality will still be present but only implicitly.
An angle bisector can be looked at as the locus of centers of circles that touch two rays emanating from the same point. In a triangle, there are three such pairs of rays.
Pick any angle and consider its bisector. Circles that touch two sides of the angle have their centers on the bisector.
Incircle and excircles of a triangle - Wikipedia
Conversely, any point on the bisector serves as the center of a circle that touches both sides of the angle. Consider two bisectors of angles formed by the pair a and b and by the pair b and c.
The circle with the center at the point of intersection of the two bisectors touches all three sides. In particular, it touches the sides a and c and, therefore, has its center on the bisector of the angle formed by these two sides.
But with that out of the way, we've kind of marked up everything that we can assume, given that this is an orthocenter and a center-- although there are other things, other properties of especially centroids that we know.
But now let's prove that this has to be an equilateral triangle. So the first thing that you might see-- and let me label some letters here, so we can refer to things a little bit better. So they definitely have one side-- side EF is congruent to side AF-- these are both the same.
And then they both clearly share this side FG-- they both share this side right over here-- they both share that over there. So they have a congruent side, a corresponding angle in between another congruent side, and they're all-- So congruent side, congruent corresponding angle, and another congruent corresponding side.
So by side, angle, side, these two characters are going to be congruent. By side, angle, side congruency. Now can you make that exact same argument to say that all of these pairs that have the kind of share-- that both have these 90 degree angles next to each other are going to be congruent.
Same exact thing-- side, angle-- and then we have a side right over here. And then we can use the exact same argument for this one over here. That by itself is interesting. But we know that if two triangles are congruent, all of their corresponding sides and angles are going to be congruent.
So for example, if we know the measure of this angle is blue, the corresponding angle on this triangle is also going to have the same measure. I'll just make it with that same blue angle. And we know if this angle right over here is magenta, the corresponding angle on triangle AFG is also going to have that same measure. And I'll just mark it with that magenta again.
Now we also know from our properties of vertical angles, that whatever angle measure AFG is, DGC is going to have the same measure, because they are vertical angles.