SOLUTION: Let K = (0,0), L = (12,0), and M = (0,9). Find equations for the three lines that bisect the angles of triangle KLM. Show that the lines are concurrent at a point C, the incenter o

Question 1158586: Let K = (0,0), L = (12,0), and M = (0,9). Find equations for the three lines that bisect the angles of triangle KLM. Show that the lines are concurrent at a point C, the incenter of KLM. Why is C called this?
Answer by greenestamps(13216) (Show Source): You can put this solution on YOUR website!

Whenever a geometry problem talks about bisectors of angles of a triangle, it is almost certain that solving the problem will involve the use of the angle bisector theorem.

That theorem says that the bisector of an angle in a triangle divides the opposite side into two parts whose lengths are in the same ratio as the lengths of the two sides of the triangle that form the angle.

In this problem, the numbers work out nicely, making it relatively easy to work all parts of the problem.

Let P, Q, and R be the points on LM, KM, and KL such that KP, LQ, and MR are the angle bisectors.

Clearly the equation of angle bisector KP is y=x, since it bisects the 90 degree angle K. So there is no need to use the angle bisector theorem to determine the coordinates of P.

Angle L is formed by sides KL and ML. KL:ML = 12:15 = 4:5, so KQ:QM = 4:5; that makes Q (0,4).

Angle M is formed by sides KM and LM. KM:LM = 9:15 = 3:5, so KR:RL = 3:5; that makes R (4.5,0).

Having the coordinates of Q and R makes it easy to find the equations of LQ and MR; then having the equations of the three angle bisectors makes it easy to show that they are all concurrent at a point C.

Every point on each angle bisector is equally distant from the two sides of the triangle that form the angle. So the point where the three angle bisectors intersect is the one point that is equidistant from all three sides of the triangle.

That means a circle with center C can be inscribed in the triangle -- hence the name incenter.

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