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第52頁

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$證明：∵?△ABC∽△A'B'C'?$

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

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

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

$解：連接?AP?并延長交?BC?于點?D?$

$∵點?P?為?△ABC?的重心$

$∴?\frac {AP}{AD}=\frac 23$

$?∵?EF//BC?$

$∴?△AEP∽△ABD，??△AEF∽△ABC$

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

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

$解：∵?DG//AB、??QH//BC ?$

$∴?△PKQ∽△DPE?$

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

$∴?\frac {KP}{PE}=\frac 12 ?$

$∴?\frac {KP}{KE}=\frac 13?$

$又∵?△KQP∽△KBE?$

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

$∴?\frac {4}{S_{△KBE}}=\frac 19 ?$

$∴?S_{△KBE}=36?$

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

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

$∴?S_{△ABC}=16+12+24+16+9+4=81?$

$證明：如圖四邊形?ABCD∽?四邊形?A'B'C'D'?$

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

$∴?△ADC∽△A'D'C'?$

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

命題得證

解：它們對應頂點相連形成的直線交于一點

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