Let $\triangle ABC$ be an isosceles triangle with two equal sides that $AB = AC$ than $\angle AB = \angle AC$ so with law of cosines we can calculate: $$\cos\theta_{1} = \frac{b^2+c^2-a^2}{2bc}$$ $$\cos\theta_{2} = \frac{a^2+c^2-b^2}{2ac}$$ which than we can prove: $$\frac{-a^2+b^2+c^2}{2bc} = \frac{a^2-b^2+c^2}{2ac}$$ $$2ac(-a^2+b^2+c^2) = 2bc(a^2-b^2+c^2)$$ $$ac(-a^2+b^2+c^2) = bc(a^2-b^2+c^2)$$ $$c(ab^2-a^3)+ac^3 = c(-b^3+a^2b)+bc^3$$ $$c(ab^2-a^3)+ac^3 - c(-b^3+a^2b)+bc^3 = 0$$ $$c(ab^2-a^3) - c(-b^3+a^2b)+ac^3-bc^3 = 0$$ $$c(b^3+ab^2-a^2b-a^3)+c^3(-b+a) = 0$$ $$-c(-b+a)(a+b-c)(a+b+c) = 0$$ $$c(a+b-c)(a+b+c) = 0$$ $$c= 0 \lor a+b-c = 0 \lor a+b+c=0$$ $$\Longrightarrow$$ $$\frac{-a^2+b^2+0^2}{2b\cdot0} \neq \frac{a^2-b^2+c^2}{2ac}$$ $$\Longrightarrow$$ $$c\neq0$$ $$\Longrightarrow$$ $$c = \pm(a+b)$$
what's wrong in my prove?
any help is appreciated!
Just to get this out of the Unanswered queue (there's an answer in the comments):
$$-c(-b+a)(a+b-c)(a+b+c) = 0$$ $$\color{red}{c(a+b-c)(a+b+c) = 0}$$
You divided by $-b+a$ after assuming the triangle has $a=b$, so your divisor would be zero.