问题背景:如图\(1\),等腰\(\triangle ABC\)中,\(AB=AC\),\(∠BAC=120^{\circ}\),作\(AD⊥BC\)于点\(D\),则\(D\)为\(BC\)的中点,\(∠BAD= \dfrac {1}{2}∠BAC=60^{\circ}\),于是\( \dfrac {BC}{AB}= \dfrac {2BD}{AB}= \sqrt {3}\);
迁移应用:如图\(2\),\(\triangle ABC\)和\(\triangle ADE\)都是等腰三角形,\(∠BAC=∠DAE=120^{\circ}\),\(D\),\(E\),\(C\)三点在同一条直线上,连接\(BD\).
\(①\)求证:\(\triangle ADB\)≌\(\triangle AEC\);
\(②\)请直接写出线段\(AD\),\(BD\),\(CD\)之间的等量关系式;
拓展延伸:如图\(3\),在菱形\(ABCD\)中,\(∠ABC=120^{\circ}\),在\(∠ABC\)内作射线\(BM\),作点\(C\)关于\(BM\)的对称点\(E\),连接\(AE\)并延长交\(BM\)于点\(F\),连接\(CE\),\(CF\).
\(①\)证明\(\triangle CEF\)是等边三角形;
\(②\)若\(AE=5\),\(CE=2\),求\(BF\)的长.