《2025年诚成教育学业评价八年级数学下册人教版》

26.(8分)如图,在▱ABCD中,BD⊥AD于点D,　A=45°,E，F分别是AB,CD上的点，且
BE=DF,连接EF交BD于点O.
　(1)求证BO=DO;
　(2)若EF⊥AB,延长EF交AD的延长线于点G,当FG=1时，求AE的长.

答案：
(1)证明：$\because$四边形$ABCD$是平行四边形，$\therefore DC// AB$.$\therefore \angle OBE=\angle ODF$.在$\triangle OBE$和$\triangle ODF$中，$\begin{cases}\angle BOE=\angle DOF\\\angle OBE=\angle ODF\\BE = DF\end{cases}$$\therefore \triangle OBE\cong \triangle ODF$.$\therefore BO = DO$.
(2)解：$\because EF\perp AB,AB// DC$，$\therefore \angle GEA=\angle GFD = 90^{\circ }$.$\because \angle A = 45^{\circ },\therefore \angle G=\angle A = 45^{\circ }$.$\therefore AE = GE$.$\because BD\perp AD$，$\therefore \angle ADB=\angle GDO = 90^{\circ }$.$\therefore \angle GOD=\angle G = 45^{\circ }$.$\therefore DG = DO$.$\therefore OF = FG = 1$.由
(1)可知$OE = OF = 1$，$\therefore AE = GE = OE + OF + FG = 3$.