For angles $A$ and $B$ in a triangle, is $\cos\frac B2-\cos \frac A2=\cos B-\cos A$ enough to conclude that $A=B$? Brief enquiry:

$$\cos\frac B2-\cos \frac A2=\cos B-\cos A$$
Optionally
$$\sqrt\frac{1+\cos B}{2}-\cos B=\sqrt\frac{1+\cos A}{2}-\cos A$$
Is above equality sufficient to prove that it implies $A=B$?


Detailed explanation and motivation for this question:
Consider a triangle with bisectors of equal length:

By definition:
$\lvert AE\rvert = \lvert BD\rvert = D\\\frac A2+\frac A2 = A,\space\space\space\frac B2 + \frac B2 = B$
By cosine law:
$$\lvert AE\rvert^2=y^2+a^2-2ya\cos B\\\lvert BD\rvert^2=x^2+a^2-2xa\cos A\\x^2=\lvert BD\rvert^2+a^2-2\lvert BD\rvert a\cos \frac B2\\y^2=\lvert AE\rvert^2+a^2-2\lvert AE\rvert a\cos \frac A2$$
By sine law:
$$\frac{x}{D}=\frac{\sin \frac B2}{\sin A};\space\space\space \frac{y}{D}=\frac{\sin \frac A2}{\sin B}$$
$$\bigl[D=\lvert AE\rvert = \lvert BD\rvert\bigr]$$
Since bisectors are equal:
$$x^2-2xa\cos A=y^2-2ya\cos B\implies \lvert BD\rvert \cos \frac B2 +x\cos A = \lvert AE\rvert \cos \frac A2 +y\cos B $$
$$D (\cos\frac B2-\cos \frac A2)=y\cos B-x\cos A $$
Dividing by D and substituting y and x we obtain:
$$\cos\frac B2-\cos \frac A2=\frac{\sin \frac A2\cos B}{\sin B}-\frac{\sin \frac B2\cos A}{\sin A}$$
Consider triangles $\Delta$ABE and $\Delta$BAD
Area of triangle $\Delta$ABE:
$$A = \frac{aD}{2}\sin\frac A2 = \frac{aD}{2}\sin B \implies \sin\frac A2 = \sin B$$
Similarly for triangle $\Delta$BAD
$$A=\frac{aD}{2}\sin\frac B2 = \frac{aD}{2}\sin A\implies \sin\frac B2 = \sin A$$
Therefore:
$$\cos\frac B2-\cos \frac A2=\cos B-\cos A$$
Optionally
$$\sqrt\frac{1+\cos B}{2}-\cos B=\sqrt\frac{1+\cos A}{2}-\cos A$$

Is above equality sufficient to prove that it implies A = B?


After more thinking, I realized that the answer is in fact "Yes".
As before, let $a=\cos\frac{A}{2},b=\cos\frac{B}{2}, 0<a,b<1$
The identity becomes:
$2a^2-a=2b^2-b$
which can be rewritten as:
$(a-b)\left(a+b-\frac{1}{2}\right)=0$
So either $a=b$, which gives $A=B$, as $\cos$ is $1:1$ on $\left[0,\frac{\pi}{2}\right]$
Or $a+b=\frac{1}{2}$
But the latter is impossible, because $A+B<\pi$, so at least one of the angles $A$ and $B$ has to be acute, which gives:
$a=\cos\frac{A}{2}>\cos\frac{\pi}{4}>\frac{1}{2}$
The whole point is that we look at the interior bisectors only. But if we consider the exterior bisectors as well, it is possible to find $A\ne B$ satisfying $\cos\frac{A}{2}+\cos\frac{B}{2}=\frac{1}{2}$
