9. (2025·遂宁)如图,在四边形ABCD中,$AB// CD$,点E,F在对角线BD上,$BE=EF=FD$,且$AF⊥ AB$,$CE⊥ CD$.
(1)求证:$△ ABF≌△ CDE$;
(2)连接AE,CF,若$∠ ABD=30°$,请判断四边形AECF的形状,并说明理由.
答案：9.(1)证明:$\because AB// CD,\therefore ∠ABF=∠CDE.$
$\because AF⊥AB,CE⊥CD,\therefore ∠BAF=∠DCE=90^{\circ }.$
$\because BE=EF=FD,\therefore BE+EF=FD+EF$,即$BF=DE.$
在$△ ABF$和$△ CDE$中,$\{\begin{array}{l} ∠ABF=∠CDE,\\ ∠BAF=∠DCE=90^{\circ },\\ BF=DE,\end{array} $
$\therefore △ ABF≌ △ CDE(AAS).$
(2)解:四边形AECF是菱形,理由如下:
如答图.
$\because ∠ABD=30^{\circ },AB// CD,\therefore ∠CDB=∠ABD=30^{\circ }.$
$\because BE=EF,∠BAF=90^{\circ },$
∴AE是$Rt△ ABF$斜边BF上的中线,$\therefore AE=\frac{1}{2}BF.$
在$Rt△ ABF$中,$∠ABD=30^{\circ },\therefore AF=\frac{1}{2}BF,\therefore AE=AF$
$=\frac{1}{2}BF$,同理$CE=CF=\frac{1}{2}DE.$
$\because BF=DE,\therefore AE=AF=CE=CF$.又$\because ∠EAF≠90^{\circ },$
∴四边形AECF是菱形.
