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電子課本網(wǎng) 第128頁

第128頁

信息發(fā)布者:
?$解: 因?yàn)镕G//AC , HI//AB$?
?$所以∠FPD=∠IEP,∠FDP=∠IPE$?
?$所以△FDP∽△IPE$?
?$因?yàn)椤鱂DP、△IPE的面積分別為4、 9$?
?$所以\frac {S_{△FDP}}{S_{△IPE}}=(\frac {DP}{PE})2=\frac {4}{9}$?
?$所以\frac {DP}{PE}=\frac {2}{3}$?
?$同理可得,\frac {FP}{PG}=\frac {2}{7}$?
?$設(shè)DP=2x ,則PE=3x .$?
?$因?yàn)镈P//BC , HI//AB$?
?$所以∠FPD=∠PGH ,∠DFP=∠HPG , $?
?$所以△FDP∽△PHG$?
?$所以\frac {DP}{HG}=\frac {FP}{PG}=\frac {2}{7}$?
?$因?yàn)镈P=2x, $?
?$所以HG=7x$?
?$因?yàn)镕G//AC , HI//AB , DE//BC$?
?$所以四邊形DPHB和四邊形PECG都為平行四邊形$?
?$所以BH=DP=2x,CG=PE=3x,$?
?$所以BC=BH+HG+CG=12x$?
?$因?yàn)镈E//BC,F(xiàn)G//AC,$?
?$所以∠FDP=∠B ,∠DFP=∠A$?
?$所以△FDP∽△ABC$?
?$所以\frac {S_{△FDP}}{S_{△ABC}}=(\frac {DP}{BC})2=(\frac {2x}{12x})2=\frac {1}{36}$?
?$因?yàn)镾_{△FDP}=4 ,$?
?$所以S_{△ABC}= 144$?

?$解:(1)△ABP∽△CQP∽△DQR,△BPC∽△BRE$?
?$(2)延長(zhǎng)AD、BR_{交于}F$?
?$由平行四邊形A BCD和平行四邊形ACED$?
?$得AD//BC , AB//CD, AC//DE , AD= BC= CE,$?
?$所以\frac {FR}{BR}=\frac {DF}{BE}=\frac {DR}{RE}=1,DF=BE,$?
?$\frac {QF}{BQ}=\frac {DF}{BC}=2$?
?$\frac {PF}{BP}=\frac {AF}{BC}=3$?
?$即BP: (BP+ PQ+ PR): (BP+ 2PQ+ 2PR)=1 : 2: 3$?
?$設(shè)BP=x,則x+PQ+PR=2x,x+2PQ+PR=3x$?
?$得PQ=\frac {1}{3}x,PR=\frac {2}{3}x$?
?$所以BP: PQ : QR=3:1: 2$?