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 $f(B,C)\neq 0$ ？
 A: 
Its just kind of a hint, I hope its of some help.
Using sine-rule on $\triangle ABC$, you get :
$$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}= 2R$$
$$\dfrac{\sin B}{\sin C}= \dfrac{b}{c} \implies \dfrac{\sin^2 B}{\sin^2 C}= \dfrac{b^2}{c^2}$$
Use sine-rule for $\triangle ADB$ and $\triangle AEC$
$$\dfrac{c}{\sin (\frac{B}{2}+C)}=\dfrac{AD}{\sin \frac{B}{2}}=\dfrac{BD}{\sin A}= 2R'$$
$$\dfrac{b}{\sin (\frac{C}{2}+B)}=\dfrac{AE}{\sin \frac{C}{2}}=\dfrac{EC}{\sin A}= 2R''$$
$\sin A=( \dfrac{c}{\sin (\frac{B}{2}+C) \times BD} )^{-1}$
$\sin A= (\dfrac{b}{\sin (\frac{C}{2}+B ) \times EC})^{-1}$
Equate them:
$( \dfrac{c}{\sin (\frac{B}{2}+C) \times BD} )=(\dfrac{b}{\sin (\frac{C}{2}+B ) \times EC})$
$\dfrac{c \times EC}{b \times BD} = \dfrac{\sin (\dfrac{B}{2}+C)}{\sin (\dfrac{C}{2}+B)}$
$\dfrac{c \times EC}{b \times BD}=\dfrac{b^2}{c^2} \implies \dfrac{EC}{BD}=\dfrac{b^3}{c^3}$
In $\triangle BDC$ and $\triangle BEC$
$\dfrac{a}{\sin (\dfrac{B}{2}+C)}=\dfrac{BD}{\sin C}$ .....$1$
$\dfrac{a}{\sin (\dfrac{C}{2}+B)}=\dfrac{EC}{\sin B}$......$2$
Dividing ($1$) and ($2$), you get:
$\dfrac{\sin (\dfrac{C}{2}+B)}{\sin (\dfrac{B}{2}+C)}=\dfrac{BD \cdot \sin B}{EC \cdot \sin C}$
I couldn't get the conclusion right. Maybe this kinda approach is useful.:)
A: No manipulations I tried worked, so I began to think the statement was false. So I set up the function
$$f(b,c)=\sin(b/2+c)\sin^2(c),$$
noting the sides of your equation are equal iff $f(b,c)=f(c,b).$ So if $b=c$ followed from that it should be true that for example the function
$$g(x)=f(x,x+.5)-f(x+.5,x)$$
should not have zeros at values of $x$ such that angles $x,x+.5$ could be angles in a triangle. However a root finder gave the zero $x_0 \approx 1.1685189447$. In the notation of the equation this is $b=x_0$ and $c=x_0+.5$, which correspond in degrees to roughly $66.95$ and $95.59$ degrees respectively, with sum about $162.55$ degrees, so that this $a,b$ could be angles in a triangle. The two sides of the original equation then both come out about $0.916865653972.$
A: The following picture suggests that the statement might be wrong. It shows that near $\beta=\gamma=1.417079$ there are angles $\beta\ne\gamma$ satisfying the stated equality.

A: 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.
