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Triangle Bisectors

A triangle is a two-dimensional shape that is bounded by three line segments. They can be classified into acute-angled, obtuse-angled, right angled, equilateral, isosceles and scalene triangle. Different other properties also are associated with the triangles. We have two types of bisectors in a triangle - angle bisectors and perpendicular bisectors. Let us go ahead and discuss them in detail.

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Perpendicular Bisectors

In general, the perpendicular bisector of a line segment is another line segment that is perpendicular to the original line and bisects it too. A triangle is made up of three line segments. Therefore, it has three perpendicular bisectors.

In order to draw a perpendicular bisector of one side of a triangle, one should take a span in the compass more than half of that side. With this span and taking an endpoint as a center, mark two arcs each on the both sides of the line. This process is repeated taking another endpoint as center using the same span in compass. Thus, we will get two points which are joined to obtain the required perpendicular bisector for a side. It is shown below: Similarly, perpendicular bisectors for other two sides are also drawn.

All three perpendicular bisectors of a triangle intersect at one point, i.e. they are concurrent. The point where they meet is known as circumcenter which is equidistant from each vertex. If we draw a circle assuming this point as center and taking the distance between this point and any vertex as radius, then we get a circle passing through all the three vertices of the triangle. Hence, this circle is called circumcircle. Take a look at the following diagram: Here, let the point where all the lines meet be O. ABC is a triangle whose perpendicular bisectors M$_{1}$O, M$_{2}$O and M$_{3}$O intersect at point O. The distances A$_{1}$O, A$_{2}$O and A$_{3}$O are equal.

Angle Bisectors

We must have known that the angle bisector of an angle is a line segment that bisects the angle. A triangle has three angle bisectors, as there are three angles of a triangle.

For the construction of an angle bisector of any angle of a triangle, we have to take a reasonable distance in compass and mark an arc assuming included vertex as the center. This arc should cut the sides that includes angle. Now, taking same distance, mark an arc each from both the points of the sides. Join vertex and this point of intersection. Thus, we obtain angle bisector of an angle. It is demonstrated in the diagram below: In exactly same way, the angle bisectors of other two angles can also be drawn as shown below : All three angle bisectors of a triangle intersect at one single point, i.e. they are concurrent. Their meeting point is referred to as the incenter. If we drop a perpendicular from the incenter to any side of the triangle and then draw a circle assuming incenter as the center and taking the distance between incenter and foot of perpendicular as radius, then we obtain a circle that just touches each side.

This circle is known as an incircle which is shown in the following diagram : If ($x_{1}, y_{1}$), ($x_{2}, y_{2}$), ($x_{3}, y_{3}$) be the coordinates of three vertices of a triangle and a, b, c be the measures of sides, then incenter will have the following coordinates:

$(\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c}, \frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c})$

Example

Have a look at the following example.
Example: If the coordinates of three vertices of a triangle are A(2, 0), B(0, 0) and C(1, 0), then find the coordinates of its incenter.

Solution: Let $x_{1}$ = 2, $y_{1}$ = 0

$x_{2}$ = 0, $y_{2}$ = 0
$x_{3}$ = 1, $y_{3}$ = 0

Distance between A and B (say a)= $\sqrt{(x_{2}-x_{1})^{2} + (y_{2} - y_{1})^{2}}$

a = $\sqrt{(0-2)^{2} + (0 - 0)^{2}}$ = 2 units

Distance between B and C (say b)= $\sqrt{(x_{3}-x_{2})^{2} + (y_{3} - y_{2})^{2}}$

b = $\sqrt{(1-0)^{2} + (0 - 0)^{2}}$ = 1 unit

Distance between C and A (say b)= $\sqrt{(x_{1}-x_{3})^{2} + (y_{1} - y_{3})^{2}}$

c = $\sqrt{(2-1)^{2} + (0 - 0)^{2}}$ = 1 unit

The formula for finding coordinates of incenter is :

$(\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c}, \frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c})$

= $(\frac{2.2+1.0+1.1}{2+1+1}, \frac{2. 0+1. 0+1.0}{2+1+1})$

= ($\frac{5}{4}$, 0)
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