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10. 如图，在$△ ACD$中，$AD=6$，$BC=5$，$AC^2=AB(AB+BC)$，且$△ DAB ∽ △ DCA$，若$AD=3AP$，点$Q$是线段$AB$上的动点，则$PQ$的最小值是
$\frac{\sqrt{7}}{2}$
。答案
10. $\frac{\sqrt{7}}{2}$ 【点拨】$\because △DAB∽△DCA,\therefore \frac{AD}{DC}=\frac{BD}{AD}=\frac{AB}{AC}$.
$\because AD=6,BC=5,\therefore \frac{6}{5+BD}=\frac{BD}{6}. \therefore BD=4$(负值已舍去). $\therefore DC=9. \therefore \frac{AD}{DC}=\frac{AB}{AC}=\frac{2}{3},\therefore AC=\frac{3}{2}AB$.
$\because AC^2=AB(AB+BC),\therefore (\frac{3}{2}AB)^2=AB(AB+BC)$.
$\therefore AB=4. \therefore AB=BD$. 如图,过B作$BH⊥ AD$于点H,
$\therefore AH=\frac{1}{2}AD=3. \therefore BH=\sqrt{AB^2-AH^2}=\sqrt{4^2-3^2}=\sqrt{7}. \because AD=3AP,AD=6,\therefore AP=2$. 当$PQ⊥ AB$时,$PQ$的值最小.此时$∠ AQP=∠ AHB=90°$.
又$\because ∠ PAQ=∠ BAH$,
$\therefore △APQ∽△ABH. \therefore \frac{AP}{AB}=\frac{PQ}{BH}$.
$\therefore \frac{2}{4}=\frac{PQ}{\sqrt{7}}. \therefore PQ=\frac{\sqrt{7}}{2}$.
三、解答题(共40分)
11.(12分)图①②均是$6×6$的正方形网格,每个小正方形的边长均为1,每个小正方形的顶点称为格点,$△ ABC$的顶点在格点上.只用无刻度的直尺,在给定的网格中,分别按下列要求画图,保留适当的作图痕迹.
(1)在图①中画线段$EF$,点$E$在$AC$边上,点$F$在$AB$边上,且$EF=\frac{1}{2}BC$;
(2)在图②中的线段$AB$上找一点$O$,使$AO: BO=2:5$.
11.(12分)图①②均是$6×6$的正方形网格,每个小正方形的边长均为1,每个小正方形的顶点称为格点,$△ ABC$的顶点在格点上.只用无刻度的直尺,在给定的网格中,分别按下列要求画图,保留适当的作图痕迹.
(1)在图①中画线段$EF$,点$E$在$AC$边上,点$F$在$AB$边上,且$EF=\frac{1}{2}BC$;
(2)在图②中的线段$AB$上找一点$O$,使$AO: BO=2:5$.
答案
11. 【解】(1)如图①,线段EF即为所求.
(2)如图②,点O即为所求.
12.(14分)[枣庄模拟] 如图,在四边形ABCD中,AC,BD相交于点E,点F在BD上,且∠BAF=∠DBC,$\frac{AB}{AF}=\frac{BC}{FD}$.
(1)求证:△ABC∽△AFD;
(2)若$AD=2,BC=5,△ ADE$的周长为20,求△BCE的周长.
(1)求证:△ABC∽△AFD;
(2)若$AD=2,BC=5,△ ADE$的周长为20,求△BCE的周长.
答案
12. (1)【证明】$\because ∠ BAF=∠ DBC$,
$\therefore ∠ BAF+∠ ABF=∠ DBC+∠ ABF$,
即$∠ AFD=∠ ABC$.
又$\because \frac{AB}{AF}=\frac{BC}{FD},\therefore △ABC∽△AFD$.
(2)【解】由(1)得$△ABC∽△AFD,\therefore ∠ ADE=∠ ACB$.
又$\because ∠ AED=∠ BEC,\therefore △AED∽△BEC$.
$\because AD=2,BC=5,\therefore \frac{C_{△ AED}}{C_{△ BEC}}=\frac{AD}{BC}=\frac{2}{5}$.
又$\because △ADE$的周长为20,
$\therefore △BCE$的周长为50.
$\therefore ∠ BAF+∠ ABF=∠ DBC+∠ ABF$,
即$∠ AFD=∠ ABC$.
又$\because \frac{AB}{AF}=\frac{BC}{FD},\therefore △ABC∽△AFD$.
(2)【解】由(1)得$△ABC∽△AFD,\therefore ∠ ADE=∠ ACB$.
又$\because ∠ AED=∠ BEC,\therefore △AED∽△BEC$.
$\because AD=2,BC=5,\therefore \frac{C_{△ AED}}{C_{△ BEC}}=\frac{AD}{BC}=\frac{2}{5}$.
又$\because △ADE$的周长为20,
$\therefore △BCE$的周长为50.
13.(14分) [天津南开区自主招生]某数学社团遇到这样一个题目:如图①,在△ABC中,点O在线段BC上,∠BAO=30°,∠OAC=75°,AO=3√3,BO:CO=1:3,求AB的长.
(1)经过社团成员讨论发现,在图①中,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,请回答:∠ADB=
(2)请参考以上解决思路,解决问题:
如图②,在四边形ABCD中,对角线AC与BD相交于点O,AC⊥AD,AO=3√3,∠ABC=∠ACB=75°,BO:OD=1:3,求AC的长.
(1)经过社团成员讨论发现,在图①中,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,请回答:∠ADB=
75°
,AB=$4\sqrt{3}$
;(2)请参考以上解决思路,解决问题:
如图②,在四边形ABCD中,对角线AC与BD相交于点O,AC⊥AD,AO=3√3,∠ABC=∠ACB=75°,BO:OD=1:3,求AC的长.
答案
13. 【解】(1)$75°$;$4\sqrt{3}$ 【点拨】$\because BD// AC,\therefore ∠ ADB=∠ OAC=75°$. 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$\because ∠ AOD=∠ EOB$,
$\therefore △EOB∽△AOD. \therefore \frac{BO}{DO}=\frac{EO}{AO}$.
$\because BO:OD=1:3,\therefore \frac{EO}{AO}=\frac{1}{3}$.
$\because AO=3\sqrt{3},\therefore EO=\sqrt{3}. \therefore AE=4\sqrt{3}$.
$\because ∠ ABC=∠ ACB=75°$,
$\therefore ∠ BAC=30°,AB=AC$.
$\therefore AB=2BE$.
在$Rt△AEB$中,$AE^2+BE^2=AB^2$,即$(4\sqrt{3})^2+BE^2=(2BE)^2,\therefore BE=4$.
$\therefore AC=AB=8$.
