# Bisectors of a triangle meet at point.

Prove that internal angle bisectors of $\triangle ABC$ meet at a point.

The problem is that I have to prove this using the locus of a straight line and its properties, I can't use vectors.

The proof I can think of is very simple but extremely tedious.

Let the coordinate of triangle be $(0,0), (a,0), (x_0,y_0)$

Let side connecting $(0,0)\ \& \ (a, 0)$ be C, $(0,0) \ \& \ (x_0, y_0)$ be A and $(x_0,y_0) \ \& \ (a, 0)$ be B.

Then the equation to sides are,

$$C : y = 0 \ ; \ B : y(a - x_0) + xy_0 - ay_0 = 0\ ; \ A: xy_0 - yx_0 = 0$$

The general form of angle bisectors between two angles is $${Ax + By + C\over \sqrt{A^2 + B^2}} = \pm {A_0x + B_0y + C_0\over \sqrt{A_0^2 + B_0^2}}$$

Using this equation, and some very tedious math I got,

$$\operatorname{bisector(AC)} : y\left(\sqrt{x_0^2 + y_0^2} + x_0\right) -xy_0 = 0 \ ; \ \\ \operatorname{bisector(BC)} : y\left(\sqrt{(a -x-0)}- (a-x_0)\right)- xy_0+ay_0 = 0 \ ; \ \\ \operatorname{bisector(AC)} : x\left(\left(\sqrt{(a -x-0)}- (a-x_0)\right)y_0 - \sqrt{y_0^2 + x_0^2}y_0\right) - y\left(x_0\left(\sqrt{(a -x-0)}- (a-x_0)\right) + \sqrt{y_0^2 + x_0^2}(a-x_0)\right) + \sqrt{y_0^2 + x_0^2}ay_0 = 0$$

Now I just need to prove that these three lines are concurrent, for which I to prove that the determinant of the coefficient of these three lines equal $0$ .

I tried that but it does not come to zero, I probably lost somewhere in find the equations.

There has to be some other way of doing this, either by using some clever method of calculating the bisectors or by choosing the coordinates of the triangle such that we don't get such horrific equations.

Any help is appreciated.

## Edit

Though the answers by @Joffran and @Mark is what I needed, I want to see if somebody can prove this by the method I described.

Consider the bisectors of two of the angles. Note that for any point on each bisector, the perpendicular distance to the two sides of the angle it bisects is equal, so at the intersection, the perpendicular distance to all three sides of the triangle is equal. Therefore this point is also on the third bisector line.

This intersection point is the center of the incircle - the common distance to each side is the radius of that circle.

• So I need to find intersection of bisectors and substitute it in third bisector and I should get 0 and the answer. Is this correct ? – A---B Jan 20 '17 at 18:14
• sounds plausible; I have to admit I didn't look through all your calculations. – Joffan Jan 20 '17 at 18:16

The angle bisector is the locus of points equidistant from the sides which contain the angle. What can you say about the point where two of the bisectors meet?

• It is equidistant from all three sides. – A---B Jan 20 '17 at 18:05
• @A---B Thus what do you now know? – Jan Jan 20 '17 at 18:06
• @A---B and which points are on the third bisector? – Mark Bennet Jan 20 '17 at 18:08
• @MarkBennet Got it I guess, I need to find intersection of two bisectors and put it in third bisector and I will get 0 as the end result. Am I correct ? – A---B Jan 20 '17 at 18:10
• @A---B I think you should be able to use just the properties - you don't need the calculations - simply the observation that if a point is not on the bisector it is closer to one line than the other. This is enough to give you zero or non-zero. – Mark Bennet Jan 20 '17 at 18:15

Suppose that the triangle is ABC -- the vertices are named. Consider the locus of points that are equidistant from AB and AC, and also that of points that are equidistant from AB and BC. These two lines are the angle bisectors of the angles at A and B. At the point where these meet -- call it I -- The distance to BC = the distance to AB = he distance to AC. That implies that I lies on the angle bisector of angle C. I.e., the angle bisectors of any triangle (non-degenerate) have a common point.

Cf: Siddons and Hughes: Trigonometry (sic), Book 2

• +1; But I would like to recommend to you not to answer a question that is already been answered unless you something new share. – A---B Jan 21 '17 at 10:37
• Also try to use mathjax, It makes math look pretty. meta.math.stackexchange.com/questions/5020/… – A---B Jan 21 '17 at 10:39