三、解答题(本大题共8小题,共49分.解答应写出必要的文字说明、证明过程或演算步骤)
16.(6分)某天早上,小伟在家里发现阳光通过窗口照射到室内,会在地面上留下亮区.小伟想通过已经掌握的知识求出家里窗户的高度.于是小伟利用家里的工具测得,此时阳光通过窗口照射到地面上留下2.7 m宽的亮区,测得亮区到窗口下的墙脚距离EC=8.7 m,整个窗口高度AB=1.8 m,请你帮小伟通过已学知识求出窗口底边离地面的高BC的长度.

16. 解:
∵光线是平行的,即$BD// AE,\therefore△ BCD∽△ ACE,\therefore\frac{AC}{BC}=\frac{CE}{CD},\therefore\frac{BC+1.8}{BC}=\frac{8.7}{8.7-2.7},\therefore BC=4\ \mathrm{m}.$

17.(6分)如图,在四边形ABCD中,AD//BC,直线EF分别交DA,BC的延长线于点E,F,分别交AB,BD,DC于点G,H,I,已知EG=GH=HI=IF,求$\frac{AE}{AD}$的值.

解:
∵$EG=GH=HI=IF,\therefore\frac{EG}{FG}=\frac{1}{3},EH=FH.$
∵$AD// BC,\therefore∠ EDH=∠ FBH.$在$△ DEH$和$△ BFH$中,$\begin{cases}∠ EDH=∠ FBH\\∠ DHE=∠ BHF\\EH=FH\end{cases},\therefore△ DEH≌△ BFH(\mathrm{AAS}),\therefore DE=BF.$
∵$AD// BC,\therefore△ AEG∽△ BFG,\therefore\frac{AE}{BF}=\frac{EG}{FG}=\frac{1}{3},\therefore AE=\frac{1}{3}BF=\frac{1}{3}DE,\therefore\frac{AE}{AD}=\frac{1}{2}.$

18.(6分)如图,某无人机于空中A处探测到目标B,D的俯角分别是$30°,60°$,此时无人机的飞行高度AC为60 m,随后无人机从A处继续水平飞行$30\sqrt{3}$ m到达$A'$处,求从无人机$A'$上看目标D的俯角的正切值.

解:根据题意可知,$AC=60\ \mathrm{m},∠ B=30°,∠ ADC=60°,\therefore∠ BAD=30°,AB=2AC=120\ \mathrm{m},\therefore∠ B=∠ BAD=30°,\therefore AD=BD.$
∵$AD=\frac{AC}{\sin60°}=60÷\frac{\sqrt{3}}{2}=40\sqrt{3}(\mathrm{m}),\therefore BD=40\sqrt{3}\ \mathrm{m},CD=\frac{1}{2}AD=20\sqrt{3}\ \mathrm{m}.$如图,作$DH⊥ A'A$交$A'A$的延长线于点H,易得四边形ACDH为矩形,$\therefore DH=AC=60\ \mathrm{m},AH=CD=20\sqrt{3}\ \mathrm{m},\therefore A'H=A'A+AH=50\sqrt{3}\ \mathrm{m},\therefore\tan∠ A'=\frac{DH}{A'H}=\frac{60}{50\sqrt{3}}=\frac{2\sqrt{3}}{5}.$

19.(6分)如图,在四边形ABCD中,DC⊥BC,E,F为BC边上两点,且AB=AE,∠AEF=∠DFE.若BE=CF,求$\frac{AE}{DF}$的值.

19. 解:如图,过点A作$AH⊥ BE$于点H.
∵$AB=AE,\therefore HE=\frac{1}{2}BE.$
∵$BE=CF,\therefore\frac{HE}{CF}=\frac{1}{2}.$
∵$DC⊥ BC,\therefore∠ C=∠ AHE=90°.$
∵$∠ AEF=∠ DFE,\therefore180°-∠ AEF=180°-∠ DFE,\therefore∠ AEH=∠ DFC,\therefore△ AHE∽△ DCF,\therefore\frac{AE}{DF}=\frac{HE}{CF}=\frac{1}{2}.$

20.(6分)如图，$△ ABC$内接于$\odot O$，$AB=AC$，点D在$\overset{\frown}{BC}$上，点E在弦AD上，$∠ BED=∠ BAC$，连接BD，CD.
(1)求证：$\frac{AE}{BE}=\frac{CD}{BD}$.
(2)连接CE，若$∠ BED=2∠ DEC$，求$\frac{AE}{BE}$的值.

20. (1)证明:
∵$∠ BAC+∠ BDC=180°,∠ BED+∠ BEA=180°,∠ BED=∠ BAC,\therefore∠ BEA=∠ BDC.$
∵$∠ BAE=∠ BCD,\therefore△ ABE∽△ CBD,\therefore\frac{AE}{CD}=\frac{BE}{BD},即\frac{AE}{BE}=\frac{CD}{BD}.$
(2)解:如图,作$EF⊥ BD$于点F.
∵$∠ BED=∠ BAC,∠ BDE=∠ BCA,\therefore△ ABC∽△ EBD,\therefore\frac{AB}{EB}=\frac{AC}{ED}.$
∵$AB=AC,\therefore EB=ED,\therefore∠ BEF=∠ DEF,BD=2DF.$
∵$∠ BED=2∠ DEC,\therefore∠ DEC=∠ DEF.$
∵$AB=AC,\therefore∠ CDE=∠ EDF.$又
∵$DE=DE,\therefore△ DCE≌△ DFE(\mathrm{ASA}),\therefore DC=DF,\therefore\frac{CD}{BD}=\frac{1}{2},\therefore\frac{AE}{BE}=\frac{1}{2}.$