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1.如图，在矩形ABCD中，AE⊥BD于点E，对角线AC，BD相交于点O，且BE：ED = 1：3，AB = 6cm，求AC的长.
答案：
解：$\because BE:ED = 1:3$，设$BE = x$，$ED = 3x$，$\therefore BD = 4x$，$\therefore BO = 2x$，$\therefore EO = x$，$\therefore E$为$BO$的中点.$\because AE\perp BO$，$\therefore AB = AO = 6cm$.$\because O$为$AC$的中点，$\therefore OA = OC = 6cm$.$\therefore AC = 12cm$.
2.如图，▱ABCD的对角线AC，BD相交于点O，AE = CF.
(1)求证:△BOE≌△DOF;
(2)若BD = EF，连接DE，BF，判断四边形EBFD的形状，并说明理由.
(1)求证:△BOE≌△DOF;
(2)若BD = EF，连接DE，BF，判断四边形EBFD的形状，并说明理由.
答案：
(1)证明：$\because$四边形$ABCD$是平行四边形，$\therefore OA = OC$，$OB = OD$，$\because AE = CF$，$\therefore OE = OF$，在$\triangle BOE$和$\triangle DOF$中，$\begin{cases}OB = OD,\\\angle BOE=\angle DOF,\\OE = OF,\end{cases}$ $\therefore \triangle BOE\cong\triangle DOF(SAS)$.
(2)解：四边形$EBFD$是矩形.理由如下：$\because OB = OD$，$OE = OF$，$\therefore$四边形$EBFD$是平行四边形.$\because BD = EF$，$\therefore$四边形$EBFD$是矩形.
(1)证明：$\because$四边形$ABCD$是平行四边形，$\therefore OA = OC$，$OB = OD$，$\because AE = CF$，$\therefore OE = OF$，在$\triangle BOE$和$\triangle DOF$中，$\begin{cases}OB = OD,\\\angle BOE=\angle DOF,\\OE = OF,\end{cases}$ $\therefore \triangle BOE\cong\triangle DOF(SAS)$.
(2)解：四边形$EBFD$是矩形.理由如下：$\because OB = OD$，$OE = OF$，$\therefore$四边形$EBFD$是平行四边形.$\because BD = EF$，$\therefore$四边形$EBFD$是矩形.
3.如图，在矩形ABCD中，点M在DC上，AM = AB，且BN⊥AM，垂足为N.
(1)求证:△ABN≌△MAD;
(2)若AD = 2，AN = 4，求四边形BCMN的面积.
(1)求证:△ABN≌△MAD;
(2)若AD = 2，AN = 4，求四边形BCMN的面积.
答案：
(1)证明：在矩形$ABCD$中，$\angle D = 90^{\circ}$，$DC// AB$，$\therefore \angle BAN=\angle AMD$.$\because BN\perp AM$，$\therefore \angle BNA = 90^{\circ}$.$\therefore \angle BNA=\angle D = 90^{\circ}$.在$\triangle ABN$和$\triangle MAD$中，$\begin{cases}\angle BAN=\angle AMD,\\\angle BNA=\angle D,\\BA = AM,\end{cases}$ $\therefore \triangle ABN\cong\triangle MAD(AAS)$.
(2)解：$\because \triangle ABN\cong\triangle MAD$，$\therefore BN = AD$.$\because AD = 2$，$\therefore BN = 2$.又$\because AN = 4$，$\therefore$在$Rt\triangle ABN$中，$AB=\sqrt{AN^{2}+BN^{2}}=\sqrt{4^{2}+2^{2}} = 2\sqrt{5}$.$S_{矩形ABCD}=2\times2\sqrt{5}=4\sqrt{5}$，$S_{\triangle ABN}=S_{\triangle MAD}=\frac{1}{2}\times2\times4 = 4$.$\therefore S_{四边形BCMN}=S_{矩形ABCD}-S_{\triangle ABN}-S_{\triangle MAD}=4\sqrt{5}-8$.
(1)证明：在矩形$ABCD$中，$\angle D = 90^{\circ}$，$DC// AB$，$\therefore \angle BAN=\angle AMD$.$\because BN\perp AM$，$\therefore \angle BNA = 90^{\circ}$.$\therefore \angle BNA=\angle D = 90^{\circ}$.在$\triangle ABN$和$\triangle MAD$中，$\begin{cases}\angle BAN=\angle AMD,\\\angle BNA=\angle D,\\BA = AM,\end{cases}$ $\therefore \triangle ABN\cong\triangle MAD(AAS)$.
(2)解：$\because \triangle ABN\cong\triangle MAD$，$\therefore BN = AD$.$\because AD = 2$，$\therefore BN = 2$.又$\because AN = 4$，$\therefore$在$Rt\triangle ABN$中，$AB=\sqrt{AN^{2}+BN^{2}}=\sqrt{4^{2}+2^{2}} = 2\sqrt{5}$.$S_{矩形ABCD}=2\times2\sqrt{5}=4\sqrt{5}$，$S_{\triangle ABN}=S_{\triangle MAD}=\frac{1}{2}\times2\times4 = 4$.$\therefore S_{四边形BCMN}=S_{矩形ABCD}-S_{\triangle ABN}-S_{\triangle MAD}=4\sqrt{5}-8$.
4.如图，AD和BC相交于点O，∠ABO = ∠DCO = 90°，OB = OC，点E，F分别是AO，DO的中点.
(1)求证:OE = OF;
(2)当∠A = 30°时，求证:四边形BECF是矩形.
(1)求证:OE = OF;
(2)当∠A = 30°时，求证:四边形BECF是矩形.
答案：
(1)证明：在$\triangle AOB$与$\triangle DOC$中，$\begin{cases}\angle ABO=\angle DCO,\\OB = OC,\\\angle AOB=\angle DOC,\end{cases}$ $\therefore \triangle AOB\cong\triangle DOC(ASA)$.$\therefore AO = DO$.$\because$点$E$，$F$分别是$AO$，$DO$的中点，$\therefore OE=\frac{1}{2}OA$，$OF=\frac{1}{2}OD$.$\therefore OE = OF$.
(2)$\because OB = OC$，$OE = OF$，$\therefore$四边形$BECF$是平行四边形.$\because \angle A = 30^{\circ}$，$\therefore OB=\frac{1}{2}OA = OE$.$\therefore BC = EF$.$\therefore$四边形$BECF$是矩形.
(1)证明：在$\triangle AOB$与$\triangle DOC$中，$\begin{cases}\angle ABO=\angle DCO,\\OB = OC,\\\angle AOB=\angle DOC,\end{cases}$ $\therefore \triangle AOB\cong\triangle DOC(ASA)$.$\therefore AO = DO$.$\because$点$E$，$F$分别是$AO$，$DO$的中点，$\therefore OE=\frac{1}{2}OA$，$OF=\frac{1}{2}OD$.$\therefore OE = OF$.
(2)$\because OB = OC$，$OE = OF$，$\therefore$四边形$BECF$是平行四边形.$\because \angle A = 30^{\circ}$，$\therefore OB=\frac{1}{2}OA = OE$.$\therefore BC = EF$.$\therefore$四边形$BECF$是矩形.
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