搜索
22. (10分)如图，某楼的高度AB = 18.2米，小华在距该楼30米(FB = 30米)的点F处，沿FB向点B前进，当走到点H处时，恰好看到广告牌CD的顶端C和楼顶A在一条直线上，小华的眼睛到地面的距离EF = GH = 1.7米，广告牌的高度CD = 3.2米，BD = 20米，点B，D，H，F在同一水平线上，AB⊥BF，CD⊥BF，GH⊥BF，EF⊥BF. 求小华从点F处向前走了多少米恰好看到点C和点A在一条直线上(即求HF的长).
答案：
如图，过点$G$作$GM\perp AB$于点$M$，交$CD$于点$N$. 易得四边形$MBDN$，四边形$MBHG$，四边形$NDHG$为矩形，$\therefore AB// CD$，$NG = DH$，$MB = ND = GH = 1.7$米，$MN = BD = 20$米. $\therefore CN = CD - ND = 1.5$米，$AM = AB - MB = 16.5$米. $\because AB// CD$，$\therefore \triangle AMG\sim\triangle CNG$. $\therefore \frac{AM}{CN}=\frac{MG}{NG}$，即$\frac{16.5}{1.5}=\frac{20 + NG}{NG}$. $\therefore NG = 2$米. $\therefore DH = NG = 2$米. $\therefore HF = BF - BD - DH = 30 - 20 - 2 = 8$(米). $\therefore$小华从点$F$处向前走了$8$米恰好看到点$C$和点$A$在一条直线上
如图，过点$G$作$GM\perp AB$于点$M$，交$CD$于点$N$. 易得四边形$MBDN$，四边形$MBHG$，四边形$NDHG$为矩形，$\therefore AB// CD$，$NG = DH$，$MB = ND = GH = 1.7$米，$MN = BD = 20$米. $\therefore CN = CD - ND = 1.5$米，$AM = AB - MB = 16.5$米. $\because AB// CD$，$\therefore \triangle AMG\sim\triangle CNG$. $\therefore \frac{AM}{CN}=\frac{MG}{NG}$，即$\frac{16.5}{1.5}=\frac{20 + NG}{NG}$. $\therefore NG = 2$米. $\therefore DH = NG = 2$米. $\therefore HF = BF - BD - DH = 30 - 20 - 2 = 8$(米). $\therefore$小华从点$F$处向前走了$8$米恰好看到点$C$和点$A$在一条直线上
23. (12分)
(1)如图①，△ABC和△EDC都是等边三角形，则BD_______AE(填“＞”“＜”或“=”)；
(2)如图②，在△ABC和△EDC中，AB = AC = 4，BC = 5，EC = ED，∠BAC = ∠CED = α，探究AE和BD之间的数量关系；
(3)如图③，在正方形ABCD和正方形DEFG中，探究BF和AG之间的数量关系.
(1)如图①，△ABC和△EDC都是等边三角形，则BD_______AE(填“＞”“＜”或“=”)；
(2)如图②，在△ABC和△EDC中，AB = AC = 4，BC = 5，EC = ED，∠BAC = ∠CED = α，探究AE和BD之间的数量关系；
(3)如图③，在正方形ABCD和正方形DEFG中，探究BF和AG之间的数量关系.
答案：
（1）$=$ （2）$\because AB = AC$，$EC = ED$，$\angle BAC=\angle CED=\alpha$，$\therefore \angle ABC=\angle ACB=\angle ECD=\angle EDC=\frac{1}{2}(180^{\circ}-\alpha)$. $\therefore \triangle ABC\sim\triangle EDC$. $\therefore \frac{AC}{EC}=\frac{BC}{DC}$，即$\frac{AC}{BC}=\frac{EC}{DC}$. 又$\because \angle ACB+\angle BCE=\angle ECD+\angle BCE$，即$\angle ACE=\angle BCD$，$\therefore \triangle ACE\sim\triangle BCD$. $\therefore \frac{AE}{BD}=\frac{AC}{BC}=\frac{4}{5}$ （3）如图，连接$BD$，$DF$. $\because$四边形$ABCD$和四边形$DEFG$都是正方形，$\therefore AB = AD$，$\angle BAD = 90^{\circ}$，$DG = GF$，$\angle DGF = 90^{\circ}$. $\therefore \angle ADB=\angle ABD=\angle GDF=\angle GFD = 45^{\circ}$. 在$Rt\triangle BAD$中，由勾股定理，得$BD=\sqrt{AB^{2}+AD^{2}}=\sqrt{AD^{2}+AD^{2}}=\sqrt{2}AD$，$\therefore \frac{BD}{AD}=\sqrt{2}$. 同理，可得$\frac{DF}{DG}=\sqrt{2}$，$\therefore \frac{BD}{AD}=\frac{DF}{DG}$. $\because \angle ADB=\angle GDF$，$\therefore \angle ADB+\angle ADF=\angle GDF+\angle ADF$，即$\angle BDF=\angle ADG$. $\therefore \triangle BDF\sim\triangle ADG$. $\therefore \frac{BF}{AG}=\frac{BD}{AD}=\sqrt{2}$
（1）$=$ （2）$\because AB = AC$，$EC = ED$，$\angle BAC=\angle CED=\alpha$，$\therefore \angle ABC=\angle ACB=\angle ECD=\angle EDC=\frac{1}{2}(180^{\circ}-\alpha)$. $\therefore \triangle ABC\sim\triangle EDC$. $\therefore \frac{AC}{EC}=\frac{BC}{DC}$，即$\frac{AC}{BC}=\frac{EC}{DC}$. 又$\because \angle ACB+\angle BCE=\angle ECD+\angle BCE$，即$\angle ACE=\angle BCD$，$\therefore \triangle ACE\sim\triangle BCD$. $\therefore \frac{AE}{BD}=\frac{AC}{BC}=\frac{4}{5}$ （3）如图，连接$BD$，$DF$. $\because$四边形$ABCD$和四边形$DEFG$都是正方形，$\therefore AB = AD$，$\angle BAD = 90^{\circ}$，$DG = GF$，$\angle DGF = 90^{\circ}$. $\therefore \angle ADB=\angle ABD=\angle GDF=\angle GFD = 45^{\circ}$. 在$Rt\triangle BAD$中，由勾股定理，得$BD=\sqrt{AB^{2}+AD^{2}}=\sqrt{AD^{2}+AD^{2}}=\sqrt{2}AD$，$\therefore \frac{BD}{AD}=\sqrt{2}$. 同理，可得$\frac{DF}{DG}=\sqrt{2}$，$\therefore \frac{BD}{AD}=\frac{DF}{DG}$. $\because \angle ADB=\angle GDF$，$\therefore \angle ADB+\angle ADF=\angle GDF+\angle ADF$，即$\angle BDF=\angle ADG$. $\therefore \triangle BDF\sim\triangle ADG$. $\therefore \frac{BF}{AG}=\frac{BD}{AD}=\sqrt{2}$
查看更多完整答案，请扫码查看
