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第52頁

信息發(fā)布者：

證明：∵?$△ABC∽△A'B'C'$?

又?$AD、$??$BE$?是?$△ABC$?的高，?$A'D'、$??$B'E'$?是?$△A'B'C'$?的高

∴?$\frac {AD}{A'D'}=\frac {AB}{A'B'}，$??$\frac {BE}{B'E'}=\frac {AB}{A'B'}$?

∴?$\frac {AD}{A'D'}=\frac {BE}{B'E'}$?

解：連接?$AP$?并延長交?$BC$?于點(diǎn)?$D$?

∵點(diǎn)?$P$?為?$△ABC$?的重心

∴?$\frac {AP}{AD}=\frac 23$?

∵?$EF//BC$?

∴?$△AEP∽△ABD，$??$△AEF∽△ABC$?

∴?$\frac {AE}{AB}=\frac {AP}{AD}=\frac 23$?

∴?$\frac {EF}{BC}=\frac {AE}{AB}=\frac 23$?

解：∵?$DG//AB、$??$QH//BC $?

∴?$△PKQ∽△DPE$?

∴?$\frac {S_{△KQP}}{S_{△PDE}}=(\frac {KP}{PE})^2=\frac 4{16}$?

∴?$\frac {KP}{PE}=\frac 12 $?

∴?$\frac {KP}{KE}=\frac 13$?

又∵?$△KQP∽△KBE$?

∴?$\frac {S_{△KQP}}{S_{△KBE}}=(\frac {KP}{KE})^2=(\frac 13)^2=\frac 19$?

∴?$\frac {4}{S_{△KBE}}=\frac 19 $?

∴?$S_{△KBE}=36$?

∴?$S_{四邊形BDPQ}=S_{△KBE}-S_{△KQP}-S_{△PDE}=36-4-16=16$?

同理可求得?$S_{四邊形CEPH}=24，$??$S_{四邊形AKPG}=12$?

∴?$S_{△ABC}=16+12+24+16+9+4=81$?

證明：如圖四邊形?$ABCD∽$?四邊形?$A'B'C'D'$?

設(shè)相似比為?$k，$?則?$\frac {AD}{A'D'}=\frac {DC}{D'C'}=k，$??$∠D=∠D'$?

∴?$△ADC∽△A'D'C'$?

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

命題得證

解：它們對(duì)應(yīng)頂點(diǎn)相連形成的直線交于一點(diǎn)

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