<s>[INST] A condition for a triangle to be isosceles in $\Delta ABC$,and 
$$\dfrac{\sin{(\dfrac{B}{2}+C)}}{\sin^2{B}}=\dfrac{\sin{(\dfrac{C}{2}+B)}}{\sin^2{C}}$$
prove that $B=C$
I think $\sin{(\dfrac{B}{2}+C)}\sin^2{C}=\sin{(\dfrac{C}{2}+B)}\sin^2{B}$
then
$$\sin{(\dfrac{B}{2})}\cos{C}\sin^2{C}+\cos{\dfrac{B}{2}}\sin^3C=\sin{\dfrac{C}{2}}\cos{B}\sin^2B+\cos{\dfrac{C}{2}}\sin^3B$$
so
$$(\sin{\dfrac{B}{2}}-\sin{\dfrac{C}{2}})f(B,C)=0$$
my question:How can prove $f(B,C)\neq 0$ ？
 [/INST] The other answers make one realize there are many counter-examples, so we can compose an answer as follows.
We shal prove that there exists a right-angled triangle $ABC$ which is not isosceles, but for which
$$\frac{\sin\left( \frac{B}{2} + C \right)}{(\sin B)^2}
-
\frac{\sin\left( \frac{C}{2} + B \right)}{(\sin C)^2}
\quad\quad\quad(*)$$
is zero. So for the entire post, fix $C=90^\circ$, and consider $60^\circ \le B \le 75^\circ$ (and $A$ is the complement which makes the triangle Euclidean, of course).
In this range, clearly $A<B<C$, so the triangles we consider are all scalene.
Direct calculation (exact or by floating-point) shows that for $B=60^\circ$ the quantity $(*)$ above is positive, but for $B=75^\circ$ it is negative. So by continuity there exists a value of $B$ in $\left[ 60^\circ , 75^\circ \right]$ which makes $(*)$ zero. So we have a counter-example as promised.
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