如图,在\(\triangle ABC\)中,\(∠ACB=90^{\circ}\),\(AC=BC\),\(D\)是\(AB\)边上一点\((\)点\(D\)与\(A\),\(B\)不重合\()\),连结\(CD\),将线段\(CD\)绕点\(C\)按逆时针方向旋转\(90^{\circ}\)得到线段\(CE\),连结\(DE\)交\(BC\)于点\(F\),连接\(BE\).
\((1)\)求证:\(\triangle ACD\)≌\(\triangle BCE\);
\((2)\)当\(AD=BF\)时,求\(∠BEF\)的度数.