# Angle bisectors meet at one point

### Triangle Proofs

For an angle bisector to be true, the angle will be 90 degrees in a quadrilateral. Therefore the angle is split into two 45 degree angles. For the diagonal thus. The three perpendicular bisectors of the sides of a triangle meet in a single point, called the circumcenter. A point where three or more lines intersect is called a. Let's prove that the three perpendicular bisectors of the sides of a triangle are concurrent which means that they intersect at one point. To do so, let's consider.

Incenter Theorem The angle bisectors of a triangle intersect at a point called the incenter of the triangle, which is equidistant from the sides of the triangle.

Point G is the incenter of? Summary While similar in many respects, it will be important to distinguish between perpendicular bisectors and angle bisectors. We use perpendicular bisectors to create a right angle at the midpoint of a segment.

Any point on the perpendicular bisector is equidistant from the endpoints of the given segment. The point at which the perpendicular bisectors of a triangle meet, or the circumcenter, is equidistant from the vertices of the triangle. On the other hand, angle bisectors simply split one angle into two congruent angles.

Points on angle bisectors are equidistant from the sides of the given angle. We also note that the points at which angle bisectors meet, or the incenter of a triangle, is equidistant from the sides of the triangle.

Let's work on some exercises that will allow us to put what we've learned about perpendicular bisectors and angle bisectors to practice. Exercise 1 BC is the perpendicular bisector of AD. Find the value of x. The most important fact to notice is that BC is the perpendicular bisector of AD because, although it is just one statement, we can derive much information about the figure from it. The fact that it is a perpendicular bisector implies that segment DB is equal to segment AB since it passes through the midpoint of segment AD.

N is the circumcenter of? Find the values of x and y. We have In order to solve for y, we have to use the information given by the Circumcenter Theorem. This theorem states that the circumcenter is equidistant from the vertices of the triangle. Exercise 3 Find the value of x. The illustration shows that points A and B are equidistant from point L.

By the converse of the Angle Bisector Theorem, we know that L must lie on the angle bisector of? BYL, so we can solve for x as shown below: Exercise 4 QS is the angle bisector of? From the information we've been given, we know that? PQS is congruent to? SQR because QS bisects the whole angle,?

We have been given the measure of the whole angle and the measure of? SQR, which is half of the entire angle since the angle has been bisected. If we were to plug in 9 for x, we would see that, indeed,? Pick any angle and consider its bisector. Circles that touch two sides of the angle have their centers on the bisector.

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. Angle Bisectors as Altitudes Altitudes of a triangle serve as angle bisectors of the associated orthic triangle.

This association can be used in reverse. Through each vertex draw a line perpendicular to the corresponding angle bisector. The same is true of the pairs of angles at vertices A and B. Let's call this a mirror property. We'll make use of this observation shortly.

Assume to the contrary, that at least one of them does not pass through the corresponding vertex. However, it can be shown that this is impossible.

### All about angle bisectors

There is only one triangle with that property. For other properties of the orthic triangle see the discussion on Fagnano's problem. The proof is pretty simple. It appears that angles in a triangle with the mirror property are not arbitrary. Count the angles in the diagram.

## Bisectors of Triangles

Two cases - of obtuse and acute-angle triangles - should be considered separately. In the former case, instead of talking of inscribed triangles, we should consider triangles with vertices on the sides, or the extensions thereof, of a given triangle. Any triangle with the mirror property must have the same angles as the orthic triangle and have its sides parallel to the latter.