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第30頁

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解：?$(1)$?如圖①，過點(diǎn)?$C$?作?$OB$?的垂線分別交仰角、俯角線于點(diǎn)?$E、$??$D，$?交水平線于點(diǎn)?$F$?

在?$ Rt △A EF $?中，?$tan ∠EAF=\frac {EF}{AF}$?

∴?$EF=AF · tan 15°≈130×0.27=35.1(\ \mathrm {cm})$?

∴?$CE=160+35.1=195.1(\ \mathrm {cm})$?

∴小杜最少需要下蹲?$208-195.1=12.9(\ \mathrm {cm})$?才能被識(shí)別

?$(2)$?如圖②，過點(diǎn)?$B$?作?$OB$?的垂線分別交仰角、俯角線于點(diǎn)?$M、$??$N，$?交水平線于點(diǎn)?$P$?

在?$ Rt △A PM $?中，?$tan ∠MAP=\frac {MP}{AP}$?

∴?$MP=AP · tan 20°≈150×0.36=54.0(\ \mathrm {cm} )$?

∵?$AP=AP，$??$∠MAP=∠NAP，$??$∠APM=∠APN=90°$?

∴?$△AMP≌△ANP(\mathrm {ASA})$?

∴?$PN=MP=54.0\ \mathrm {cm}$?

∴?$BN=160-54.0=106.0(\ \mathrm {cm})$?

∴小若踮起腳尖后頭頂?shù)母叨葹?$120+3=123(\ \mathrm {cm})$?

∴小若頭頂超出點(diǎn)?$N$?的高度為?$123-106.0=17.0(\ \mathrm {cm})> 15\ \mathrm {cm}$?

∴踮起腳尖小若能被識(shí)別

?$\frac {CD}{C'D'}=\frac {AC}{A'C'}=\frac {AD}{A'D'}$?

?$∠A=∠A'$?

解：?$(2)△ABC∽△A'B'C' ，$?理由：

如圖，過點(diǎn)?$D、$??$D'$?分別作?$DE//BC，$??$D'E'//B'C'，$?

?$DE$?交?$AC$?于點(diǎn)?$E，$??$D'E'$?交?$A'C'$?于點(diǎn)?$E'$?

∵?$DE//BC$?

∴?$△ADE∽△ABC$?

∴?$\frac {AD}{AB}=\frac {DE}{BC}=\frac {AE}{AC}$?

同理可得，?$\frac {A'D'}{A'B'}=\frac {D'E'}{B'C'}=\frac {A'E'}{A'C'}$?

∵?$\frac {AD}{AB}=\frac {A'D'}{A'B'} $?

∴?$\frac {DE}{BC}= \frac {D'E'}{B'C'} $?

∴?$\frac {DE}{D'E'} =\frac {BC}{B'C'} $?

同理可得，?$\frac {AE}{AC}= \frac {A'E'}{A'C'}$?

∴?$\frac {AC-AE}{AC}= \frac {A'C'-A'E'}{A'C}，$?即?$ \frac {EC}{AC}=\frac {E'C'}{A'C'}$?

∴?$\frac {EC}{E'C'}=\frac {AC}{A'C'}$?

∵?$\frac {CD}{C'D'}=\frac {AC}{A'C'}=\frac {BC}{B'C'}$?

∴?$\frac {CD}{C'D'}=\frac {DE}{D'E'}=\frac {EC}{E'C'}$?

∴?$ △DCE∽△D'C'E$?

∴?$∠CED=∠C'E'D'$?

∵?$DE//BC$?

∴?$∠CED+∠ACB=180°$?

同理可得，?$∠C'E'D'+∠A'C'B'=180°$?

∴?$∠ACB=∠A'C'B'$?

∵?$\frac {AC}{A'C'}=\frac {CB}{C'B'}$?

∴?$ △ABC∽△A'B'C'$?

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