9. 已知点$A$，$B$，$C$，$D$在同一平面内，有下列条件：①$AB // CD$；②$AB = CD$；③$BC // AD$；④$BC = AD$；⑤$AC ⊥ BD$；⑥$AC$平分$\angle DAB$与$\angle DCB$。其中，能满足四边形$ABCD$是菱形的$3$个条件是
答案不唯一,如①②⑤
(填一组序号即可)。答案：9.答案不唯一,如①②⑤
10. (2025·启东期末)如图，把$3$个相同的矩形填充到菱形$ABCD$中，如果测得每个矩形的周长为$4\sqrt{2}\mathrm{ cm}$，那么菱形$ABCD$的周长为
16
$\mathrm{cm}$。答案：10.16
11. 如图，在$□ ABCD$中，$AB = 2$，$\angle ABC$的平分线与$\angle BCD$的平分线交于点$E$。若点$E$恰好在边$AD$上，则$BE^{2} + CE^{2}$的值为
16
。答案：11.16
解析：
证明：
∵四边形$ABCD$是平行四边形，
∴$AD// BC$，$AB=CD=2$，$AD=BC$，$\angle ABC+\angle BCD=180°$。
∵$BE$平分$\angle ABC$，$CE$平分$\angle BCD$，
∴$\angle ABE=\angle EBC=\frac{1}{2}\angle ABC$，$\angle DCE=\angle ECB=\frac{1}{2}\angle BCD$。
∵$AD// BC$，
∴$\angle AEB=\angle EBC=\angle ABE$，$\angle DEC=\angle ECB=\angle DCE$，
∴$AE=AB=2$，$DE=CD=2$，
∴$AD=AE+DE=4$，即$BC=4$。
∵$\angle EBC+\angle ECB=\frac{1}{2}(\angle ABC+\angle BCD)=90°$，
∴$\angle BEC=90°$。
在$\mathrm{Rt}\triangle BEC$中，由勾股定理得：
$BE^2+CE^2=BC^2=4^2=16$。
16
∵四边形$ABCD$是平行四边形，
∴$AD// BC$，$AB=CD=2$，$AD=BC$，$\angle ABC+\angle BCD=180°$。
∵$BE$平分$\angle ABC$，$CE$平分$\angle BCD$，
∴$\angle ABE=\angle EBC=\frac{1}{2}\angle ABC$，$\angle DCE=\angle ECB=\frac{1}{2}\angle BCD$。
∵$AD// BC$，
∴$\angle AEB=\angle EBC=\angle ABE$，$\angle DEC=\angle ECB=\angle DCE$，
∴$AE=AB=2$，$DE=CD=2$，
∴$AD=AE+DE=4$，即$BC=4$。
∵$\angle EBC+\angle ECB=\frac{1}{2}(\angle ABC+\angle BCD)=90°$，
∴$\angle BEC=90°$。
在$\mathrm{Rt}\triangle BEC$中，由勾股定理得：
$BE^2+CE^2=BC^2=4^2=16$。
16
12. 如图，在四边形$ABCD$中，$BD$平分$\angle ABC$，$E$，$F$分别是$AD$，$BC$的中点。若$CD = 2AB = 4$，$\angle ABC = 2\angle C = 60^{\circ}$，则$EF$的长为
$\sqrt{5}$
。答案：
12.$\sqrt{5}$ 解析：$\because \angle ABC=2\angle C = 60^{\circ},\therefore \angle C = 30^{\circ}.\because BD$平分$\angle ABC,\therefore \angle ABD = \angle DBC = 30^{\circ}.\therefore \angle BDC = 120^{\circ}$. 如图，取$BD$的中点$G$，连接$EG$，$FG$.又$\because E$，$F$分别是$AD$，$BC$的中点，$\therefore EG // AB$，$EG =\frac{1}{2}AB$，$FG // CD$，$FG =\frac{1}{2}CD$.
$\therefore \angle EGD = \angle ABD = 30^{\circ}$，$\angle FGD = 180^{\circ} - \angle BDC = 60^{\circ}$.
$\therefore \angle EGF = 90^{\circ}.\because CD = 2AB = 4,\therefore AB = 2.\therefore EG =\frac{1}{2}AB = 1$，$FG =\frac{1}{2}CD = 2.\therefore EF = \sqrt{EG^{2} + FG^{2}} = \sqrt{1^{2} + 2^{2}} = \sqrt{5}$.
12.$\sqrt{5}$ 解析：$\because \angle ABC=2\angle C = 60^{\circ},\therefore \angle C = 30^{\circ}.\because BD$平分$\angle ABC,\therefore \angle ABD = \angle DBC = 30^{\circ}.\therefore \angle BDC = 120^{\circ}$. 如图，取$BD$的中点$G$，连接$EG$，$FG$.又$\because E$，$F$分别是$AD$，$BC$的中点，$\therefore EG // AB$，$EG =\frac{1}{2}AB$，$FG // CD$，$FG =\frac{1}{2}CD$.
$\therefore \angle EGD = \angle ABD = 30^{\circ}$，$\angle FGD = 180^{\circ} - \angle BDC = 60^{\circ}$.
$\therefore \angle EGF = 90^{\circ}.\because CD = 2AB = 4,\therefore AB = 2.\therefore EG =\frac{1}{2}AB = 1$，$FG =\frac{1}{2}CD = 2.\therefore EF = \sqrt{EG^{2} + FG^{2}} = \sqrt{1^{2} + 2^{2}} = \sqrt{5}$.
13. 如图，$E$是边长为$8$的正方形$ABCD$的对角线$BD$上的一个动点(不与点$B$，$D$重合)，连接$AE$，以$AE$为边向左侧作正方形$AEFG$，$P$为$AD$的中点，连接$PG$，$DG$，$DG$的延长线与$BA$的延长线交于点$H$，在点$E$的运动过程中，线段$PG$长的最小值是
$2\sqrt{2}$
。答案：13.$2\sqrt{2}$ 解析：$\because$四边形$ABCD$、四边形$AEFG$均为正方形，
$\therefore \angle DAB = \angle GAE = 90^{\circ}$，$AD = AB$，$AG = AE$，$\angle ABD = 45^{\circ}$.
$\therefore \angle GAE - \angle DAE = \angle DAB - \angle DAE$，即$\angle GAD = \angle EAB$.
在$\triangle GAD$和$\triangle EAB$中，$\begin{cases} AG = AE, \\ \angle GAD = \angle EAB, \\ AD = AB, \end{cases}\therefore \triangle GAD \cong \triangle EAB.\therefore \angle PDG = \angle ABD = 45^{\circ}.\therefore$点$G$在线段$DH$上.
$\therefore$当$PG ⊥ DH$时，$PG$的长最短.$\because$正方形$ABCD$的边长为$8$，$P$为$AD$的中点，$\therefore DP = 4.\because PG ⊥ DH$，$\angle PDG = 45^{\circ}$，
$\therefore \triangle PDG$为等腰直角三角形，$\therefore$易得$PG = 2\sqrt{2}$.
$\therefore \angle DAB = \angle GAE = 90^{\circ}$，$AD = AB$，$AG = AE$，$\angle ABD = 45^{\circ}$.
$\therefore \angle GAE - \angle DAE = \angle DAB - \angle DAE$，即$\angle GAD = \angle EAB$.
在$\triangle GAD$和$\triangle EAB$中，$\begin{cases} AG = AE, \\ \angle GAD = \angle EAB, \\ AD = AB, \end{cases}\therefore \triangle GAD \cong \triangle EAB.\therefore \angle PDG = \angle ABD = 45^{\circ}.\therefore$点$G$在线段$DH$上.
$\therefore$当$PG ⊥ DH$时，$PG$的长最短.$\because$正方形$ABCD$的边长为$8$，$P$为$AD$的中点，$\therefore DP = 4.\because PG ⊥ DH$，$\angle PDG = 45^{\circ}$，
$\therefore \triangle PDG$为等腰直角三角形，$\therefore$易得$PG = 2\sqrt{2}$.
14. 如图，四边形$ABDE$是平行四边形，$C$为边$BD$的延长线上一点，连接$AC$，$CE$，$AD$，其中$AB = AC$。
(1) 求证：$\triangle BAD \cong \triangle ACE$；
(2) 若$\angle ABD = 30^{\circ}$，$\angle ADC = 45^{\circ}$，$BD = 10$，求$□ ABDE$的面积。
(1) 求证：$\triangle BAD \cong \triangle ACE$；
(2) 若$\angle ABD = 30^{\circ}$，$\angle ADC = 45^{\circ}$，$BD = 10$，求$□ ABDE$的面积。
答案：
14.(1)$\because$四边形$ABDE$是平行四边形，$\therefore BD // AE$，$BD = AE$.
$\therefore \angle ACB = \angle CAE.\because AB = AC$，$\therefore \angle ABD = \angle ACB$.
$\therefore \angle ABD = \angle CAE$.在$\triangle BAD$和$\triangle ACE$中，$\begin{cases} AB = CA, \\ \angle ABD = \angle CAE, \\ BD = AE, \end{cases}$
$\therefore \triangle BAD \cong \triangle ACE$ (2) 如图，过点$A$作$AG ⊥ BC$，垂足为$G$.
设$AG = x(x ＞ 0)$.在$Rt\triangle AGD$中，$\because \angle ADC = 45^{\circ},\therefore \angle DAG = 45^{\circ} = \angle ADC.\therefore DG = AG = x$.在$Rt\triangle AGB$中，$\because \angle ABD = 30^{\circ},\therefore$易得$BG = \sqrt{3}x$.又$\because BG - DG = BD$，$BD = 10$，
$\therefore \sqrt{3}x - x = 10$，解得$x = 5\sqrt{3} + 5.\therefore AG = 5\sqrt{3} + 5.\therefore S_{□ ABDE} = BD · AG = 10 × (5\sqrt{3} + 5) = 50\sqrt{3} + 50$
14.(1)$\because$四边形$ABDE$是平行四边形，$\therefore BD // AE$，$BD = AE$.
$\therefore \angle ACB = \angle CAE.\because AB = AC$，$\therefore \angle ABD = \angle ACB$.
$\therefore \angle ABD = \angle CAE$.在$\triangle BAD$和$\triangle ACE$中，$\begin{cases} AB = CA, \\ \angle ABD = \angle CAE, \\ BD = AE, \end{cases}$
$\therefore \triangle BAD \cong \triangle ACE$ (2) 如图，过点$A$作$AG ⊥ BC$，垂足为$G$.
设$AG = x(x ＞ 0)$.在$Rt\triangle AGD$中，$\because \angle ADC = 45^{\circ},\therefore \angle DAG = 45^{\circ} = \angle ADC.\therefore DG = AG = x$.在$Rt\triangle AGB$中，$\because \angle ABD = 30^{\circ},\therefore$易得$BG = \sqrt{3}x$.又$\because BG - DG = BD$，$BD = 10$，
$\therefore \sqrt{3}x - x = 10$，解得$x = 5\sqrt{3} + 5.\therefore AG = 5\sqrt{3} + 5.\therefore S_{□ ABDE} = BD · AG = 10 × (5\sqrt{3} + 5) = 50\sqrt{3} + 50$
15. (2024·雅安)如图，$O$是$□ ABCD$对角线的交点，过点$O$的直线分别交$AD$，$BC$于点$E$，$F$。
(1) 求证：$\triangle ODE \cong \triangle OBF$；
(2) 连接$BE$，$DF$，当$EF ⊥ BD$时，$DE = 15$，求此时四边形$BEDF$的周长。
(1) 求证：$\triangle ODE \cong \triangle OBF$；
(2) 连接$BE$，$DF$，当$EF ⊥ BD$时，$DE = 15$，求此时四边形$BEDF$的周长。
答案：15.(1)$\because$四边形$ABCD$是平行四边形，$\therefore AD // CB$.
$\therefore \angle OED = \angle OFB.\because O$是$□ ABCD$对角线的交点，$\therefore OD = OB$.在$\triangle ODE$和$\triangle OBF$中，$\begin{cases} \angle OED = \angle OFB, \\ \angle DOE = \angle BOF, \\ OD = OB, \end{cases} \therefore \triangle ODE \cong \triangle OBF$ (2) 由(1)，得$\triangle ODE \cong \triangle OBF$，$\therefore DE = BF.\because DE // BF$，$\therefore$四边形$BEDF$是平行四边形.$\because EF ⊥ BD$，$\therefore$四边形$BEDF$是菱形.$\therefore DF = BF = BE = DE = 15.\therefore DF + BF + BE + DE = 4 × 15 = 60$，即四边形$BEDF$的周长为$60$
$\therefore \angle OED = \angle OFB.\because O$是$□ ABCD$对角线的交点，$\therefore OD = OB$.在$\triangle ODE$和$\triangle OBF$中，$\begin{cases} \angle OED = \angle OFB, \\ \angle DOE = \angle BOF, \\ OD = OB, \end{cases} \therefore \triangle ODE \cong \triangle OBF$ (2) 由(1)，得$\triangle ODE \cong \triangle OBF$，$\therefore DE = BF.\because DE // BF$，$\therefore$四边形$BEDF$是平行四边形.$\because EF ⊥ BD$，$\therefore$四边形$BEDF$是菱形.$\therefore DF = BF = BE = DE = 15.\therefore DF + BF + BE + DE = 4 × 15 = 60$，即四边形$BEDF$的周长为$60$