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 $f(B,C)\neq 0$ ？


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.:)
