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Dear student,
If the equations of 2 sides of the triangle (say, AB and AC) are Line1: A1x+B1y+C1=0 and Line2: A2x+B2y+C2=0 Then (at vertex A) Angle bisector1: f1(x,y)=(A1x+B1y+C1)/√(A1²+B1²) + (A2x+B2y+C2)/√(A2²+B2²)=0 ... (1) Angle bisector2: f2(x,y)=(A1x+B1y+C1)/√(A1²+B1²) - (A2x+B2y+C2)/√(A2²+B2²)=0 ... (2) To find out which one is the internal angle bisector at, say, vertex A, you'll have to substitute the coordinates of the other 2 vertices, B and C, in one of the equations (1) or (2). The internal angle bisector corresponds to the equation, which will give f(B)*f(C)<0. For the external angle bisector it holds f(B)*f(C)>0.
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