a geometric proof involves comparing the lengths, and an algebraic proof i remember using Stewart's theorem:

We first invoke the angle bisector theorem: if AB=c and AC=b, and D is the point of intersection of the angle biestor and BC, then BD=ac/(b+c), DC=ab/(b+c)
we then use stewarts theorem (can be drived from cosine law at angle ADB and ADC)
AD=k
ab^2c/(b+c)+abc^2/(b+c)=a(k^2+a^2bc/(b+c)^2)
b^2c/(b+c)+bc^2/(b+c)=k^2+a^2bc/(b+c)^2
k^2=bc-a^2bc/(b+c)^2
k^2=bc*(1-a^2/(b+c)^2)


i divided by a bunch of positive number below, so dont bash me for it
so if two angle biestors are equal,

bc*(1-a^2/(b+c)^2)=ac*(1-b^2)/(a+c)^2)
b*(1-a^2/(b+c)^2)=a*(1-b^2)/(a+c)^2)
b(a+c)^2((b+c)^2-a^2)=a(b+c)^2((a+c)^2-b^2))
b(a+c)^2(b+c+a)(b+c-a)=a(b+c)^2(a-b+c)(a+b+c)
b(a+c)^2(b+c-a)=a(b+c)^2(a-b+c)
which factors into

(a-b)(c^3+ac^2+bc^2+3abc+ab^2+a^2b)
ie a=b

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