1. 某数学社团遇到这样一个题目：
(1)如图①,在$△ ABC$中,点$O$在线段$BC$上,$∠ BAO = 30°$, $∠ OAC = 75°$, $AO = 3\sqrt{3}$, $BO:CO = 1:3$,求$AB$的长.
经过社团成员讨论发现,过点$B$作$BD // AC$,交$AO$的延长线于点$D$,通过构造$△ ABD$就可以解决问题(如图②).
请回答:
$∠ ADB =$$°$,$AB =$.
(2)请参考以上解决问题的思路,解决问题:
如图③,在四边形$ABCD$中,对角线$AC$与$BD$相交于点$O$, $AC ⊥ AD$, $AO = 3\sqrt{3}$, $∠ ABC = ∠ ACB = 75°$, $BO:OD = 1:3$,求$DC$的长.

(1)75 $4\sqrt{3}$ 解析: $\because BD// AC,\therefore ∠ ADB=∠ OAC=75°.$
又$\because ∠ BOD=∠ COA,\therefore △ BOD∽△ COA,\therefore \frac{OD}{OA}=\frac{OB}{OC}=\frac{1}{3}.$又
$\because AO=3\sqrt{3},\therefore OD=\frac{1}{3}AO=\sqrt{3},\therefore AD=AO+OD=4\sqrt{3}.$
$\because ∠ BAD=30°,∠ ADB=75°,\therefore ∠ ABD=180°-∠ BAD-∠ ADB=75°=∠ ADB,\therefore AB=AD=4\sqrt{3}.$
(2)过点$B$作$BE// AD$交$AC$于点$E$,如图所示.

$\because AC⊥ AD,BE// AD,\therefore ∠ DAC=∠ BEA=90°.$又$\because ∠ AOD=∠ EOB,$
$\therefore △ AOD∽△ EOB,\therefore \frac{BO}{DO}=\frac{EO}{AO}=\frac{BE}{DA}.$
$\because BO:OD=1:3,\therefore \frac{EO}{AO}=\frac{BE}{DA}=\frac{1}{3}.$
$\because AO=3\sqrt{3},\therefore EO=\sqrt{3},\therefore AE=4\sqrt{3}.\because ∠ ABC=∠ ACB=75°,$
$\therefore ∠ BAC=30°,AB=AC,\therefore AB=2BE.$在$\mathrm{Rt}△ AEB$中,
$BE^2+AE^2=AB^2$,即$BE^2+(4\sqrt{3})^2=(2BE)^2$,解得$BE=4$,
$\therefore AB=AC=8,AD=12.$在$\mathrm{Rt}△ CAD$中,$AC^2+AD^2=DC^2$,即$8^2+12^2=DC^2$,解得$DC=4\sqrt{13}.$

2. (2026·青岛月考)如图,在$△ ABC$中,$∠ ABC=90°,BC=6,AB=8,D$为边$AC$的中点,点$P$从点$A$出发,以每秒1个单位长度的速度沿$AB$运动到点$B$停止,同时点$Q$从点$B$出发,以每秒2个单位长度的速度沿折线$BC—CD$运动到点$D$停止,当点$Q$停止运动时,点$P$也停止运动.过点$Q$作$QM // AB$交$△ ABC$的边于点$M$,以$PM$和$QM$为边作$□ PMQN$.设点$Q$的运动时间为$t$(秒).
(1)用含$t$的代数式表示$CQ$的长；
(2)点$Q$在$BC$边上运动时(不含$B,C$),若$□ PNQM$的面积为$S$,求出$S$关于$t$的函数关系式,并求出$t$为何值时,$S:S_{△ ABC}=1:2$；
(3)当点$Q$不与$△ ABC$的顶点重合时,连接$BD$,直接写出$BD$将$□ PMQN$分成面积相等的两部分时$t$的值.

(1)在$\mathrm{Rt}△ ABC$中,$BC=6,AB=8$,故$AC=\sqrt{AB^2+BC^2}=\sqrt{8^2+6^2}=10.\because D$为边$AC$的中点,$\therefore CD=\frac{1}{2}AC=5.$当$0≤ t≤ 3$时,点$Q$在$BC$边上运动,则$BQ=2t$,故$CQ=BC-BQ=6-2t$;当$3＜t≤5.5$时,点$Q$在$CD$上运动,则$BC+CQ=2t$,故$CQ=2t-6.$综上所述,$CQ$的长为$6-2t(0≤ t≤3)$或$2t-6(3＜t≤5.5).$
(2)点$Q$在$BC$边上运动时(不含$B,C$),此时$0＜t＜3$,故$BQ=2t,CQ=BC-BQ=6-2t.\because$四边形$PMQN$是平行四边形,
$\therefore MQ// AB,\therefore △ CMQ∽△ CAB,\therefore \frac{MQ}{AB}=\frac{CQ}{CB}$,即$\frac{MQ}{8}=\frac{6-2t}{6}$,
$\therefore MQ=-\frac{8}{3}t+8,\therefore S=MQ· QB=(-\frac{8}{3}t+8)· 2t=-\frac{16}{3}t^2+16t.\because S_{△ ABC}=\frac{1}{2}× AB· BC=\frac{1}{2}×8×6=24,\therefore \frac{-\frac{16}{3}t^2+16t}{24}=\frac{1}{2}$,
解得$t=\frac{3}{2}$,故当$t=\frac{3}{2}$时,$S:S_{△ ABC}=1:2.$
(3)$t$的值为$\frac{24}{11}$或$\frac{48}{11}$. 解析:①当$0＜t＜3$时,连接$BD$,$PQ$交于点$O$,如图①.$\because BD$将$□ PMQN$分成面积相等的两部分,$PQ$是$□ PMQN$的对角线,$\therefore BD$经过平行四边形$PMQN$的中心,$\therefore PO=OQ$.又$\because ∠ PBQ=90°$,
$\therefore BO=PO,\therefore ∠ OPB=∠ OBP.\because D$为边$AC$的中点,$∠ ABC=90°,\therefore BD=DA,\therefore ∠ A=∠ OBA,\therefore ∠ OPB=∠ A,\therefore PQ// AC$,
$\therefore \frac{BP}{BA}=\frac{BQ}{BC}$.根据题意可得$BP=AB-AP=8-t,BQ=2t$,故$\frac{8-t}{8}=\frac{2t}{6}$,解得$t=\frac{24}{11}$.②当$3＜t＜5.5$时,连接$BD$,$MN$交于点$O$,如图②.


$\because$四边形$PMQN$是平行四边形,
$\therefore MQ// PN,MQ=PN,\therefore △ CMQ∽△ CBA,\therefore \frac{QM}{AB}=\frac{CQ}{CA}=\frac{CM}{CB}.\because CQ=2t-6,AB=8,CA=10,BC=6,\therefore \frac{QM}{8}=\frac{2t-6}{10}=\frac{CM}{6},\therefore QM=\frac{8t-24}{5},CM=\frac{6t-18}{5},\therefore NP=\frac{8t-24}{5}$,
$\therefore NB=NP+PB=NP+AB-AP=\frac{8t-24}{5}+8-t,BM=BC-CM=6-\frac{6t-18}{5}.\because BD$将$□ PMQN$分成面积相等的两部分,$MN$是$□ PMQN$的对角线,$\therefore BD$经过平行四边形$PMQN$的中心,$\therefore ON=OM$.又$\because ∠ NBM=90°,\therefore BO=NO,\therefore ∠ ONB=∠ OBA.\because D$为边$AC$的中点,$∠ ABC=90°,\therefore BD=DA$,
$\therefore ∠ A=∠ OBA,\therefore ∠ ONB=∠ A,\therefore MN// AC,\therefore \frac{NB}{AB}=\frac{BM}{BC}$,
$\therefore \frac{\frac{8t-24}{5}+8-t}{8}=\frac{6-\frac{6t-18}{5}}{6}$,解得$t=\frac{48}{11}$.综上所述,满足条件的$t$的值为$\frac{24}{11}$或$\frac{48}{11}.$

3. 如图，BD为$△ ABC$的高，点E在AB边上，$∠ BEC=60°$，$BE=2CD$，CE与BD相交于点F，则$\frac{BF}{FC}$的值为。

（第3题）（第4题）

$\sqrt{3}$ 解析:过点$B$作$BH⊥ CE$,垂足为$H.\because ∠ BEC=60°$,
$\therefore ∠ EBH=30°,\therefore EH=\frac{1}{2}BE$,由勾股定理得$BH=\frac{\sqrt{3}}{2}BE=\sqrt{3}CD$.易证$△ BHF∽△ CDF,\therefore \frac{BF}{FC}=\frac{BH}{CD}=\frac{\sqrt{3}CD}{CD}=\sqrt{3}.$

4. 如图,在$△ ABC$中,$∠ BAC=60°$,$AB=6$,$AC=4$,AD平分$∠ BAC$交BC于点D,则$BD=$.

$\frac{6\sqrt{7}}{5}$ 解析:过点$B,C$分别作$AD$的垂线,垂足分别为$E,F$,
易证$BE=\frac{1}{2}AB=3,CF=\frac{1}{2}AC=2,AE=3\sqrt{3},AF=2\sqrt{3}$,
$\therefore EF=\sqrt{3}$,易证$△ BED∽△ CFD,\therefore \frac{DE}{DF}=\frac{BE}{CF}=\frac{3}{2},\therefore DE=\frac{3}{5}EF=\frac{3\sqrt{3}}{5},\therefore BD=\sqrt{BE^2+DE^2}=\frac{6\sqrt{7}}{5}.$