stupid proof but kinda

easy and straightforward ---

denote: triangle ABC:
a --- half angle of A
b --- half angle of B
L --- length of bisection line of angle A (and B).
PI=180 degrees.
other symbol follows tradition.
Therefore:
a>0, b>0, a+b < PI/2
Now consider side AB:
L/sin(2a)=AB/sin(PI-2a-b)
L/sin(2b)=AB/sin(PI-2b-a)
Therefore,
sin(2b+a)/sin(2b)=sin(2a+b)/sin(2a)
sin(2b+a)sin(2a)=sin(2a+b)sin(2b)
cos(2b-a)-cos(2b+3a)=cos(2a-b)-cos(2a+3b)
cos(2b-a)-cos(2a-b)=cos(2b+3a)-cos(2a+3b)
sin((a+b)/2)*sin(3(b-a)/2)=sin((a-b)/2)*sin(5(a+b)/2) --- (E)

We only need to show the above (E) can be true only when b=a. We need to consider the following cases:

If 5(a+b)/2 < PI, then, sin(5(a+b)/2)>0, sin((a+b)/2)>0.
However, sin(3(b-a)/2) and sin((a-b)/2) will have opposite signs. The only way for E to hold is a=b.
If 5(a+b)/2 >= PI, then 0 <= 5(a+b)/2 - PI < 5(a+b)/2 - 2*(a+b) = (a+b)/2 < PI/4.
Therefore, |sin(5(a+b)/2)| < sin((a+b)/2)
However, let us look at the other factor in E:
|a-b|=|b-a|<|a|+|b| Thus, |3(b-a)/2|<3*PI/4, |(a-b)/2| If |3(b-a)/2|>PI/2, then |sin(3(b-a)/2)|>sin(3*PI/4) = sin(PI/4) > |sin((a-b)/2)|
If |3(b-a)/2||sin((a-b)/2)|.
Therefore, the absolute value of the lefthand side of E is always strictly greater than the absolute value of the righthand side unless a=b.
QED