第1个回答 2012-11-12
1) E、F分别是AB、AC的中点,则EF平行于BC且BC=2EF
因为EF平行于BC,则△DPE∽△DBC;
因为EP:BC=(1/3EF):(2EF)=1/6,则S△DPE:S△DBC=(1/6)^2=1/36
2) 延长BQ交EF于G点,由于EF平行于BC,则∠BGE=∠QBC
又因为BQ是角平分线,所以∠PBQ=∠QBC=∠BGP
所以三角形BPG是等腰三角形 BP=PQ=y
因为EF平行于BC,CQ=1/2CE,所以三角形BQC与三角形GQE全等,即BC=EG
而EG=EP+PG=x+y,BC=6,则x+y=6
3)同上,可得BP=PQ=y
因为EF平行于BC,CQ=1/3CE,所以三角形BQC与三角形GQE相似,即EG:BC=EQ:CQ=2
而EG=EP+PG=x+y,BC=6,则x+y=12
第2个回答 2012-12-09
(1)
E、F为△ABC上AB、AC中点
EF∥BC
EF=1/2BC=3
EP=1/3EF=1
△DPE∽△DBC
△DPE以EP为底高:△DBC以BC为底的高=1:6
S△DPE:S△DBC=1:36
(2)
延长BQ交EF延长线于G
EF∥BC
△BQC∽△EQG
Q是EC中点,EQ=QC
△BQC≌△EQG
EG=BC=6
BQ平分∠CBP
∠CBG=∠GBP=∠PGB
△BPG为等腰三角形
BP=PG=y
EG=EP+PG
y=6-x
(3)
①
△BQC∽△EQG
CQ=1/3EC,CQ=1/2EQ
EG=2BC=12
BP=PG=y
EG=EP+PG=12
x+y=12
②
y=6(n-1)-6
第3个回答 2013-04-09
(1)
E、F为△ABC上AB、AC中点
EF∥BC
EF=1/2BC=3
EP=1/3EF=1
△DPE∽△DBC
△DPE以EP为底高:△DBC以BC为底的高=1:6
S△DPE:S△DBC=1:36
(2)
延长BQ交EF延长线于G
EF∥BC
△BQC∽△EQG
Q是EC中点,EQ=QC
△BQC≌△EQG
EG=BC=6
BQ平分∠CBP
∠CBG=∠GBP=∠PGB
△BPG为等腰三角形
BP=PG=y
EG=EP+PG
6=x+y
(3)
①
△BQC∽△EQG
CQ=1/3EC,CQ=1/2EQ
EG=2BC=12
BP=PG=y
EG=EP+PG=12
x+y=12
②
x+y=6(n-1)
第4个回答 2013-01-12
(1) E, F for train on ABC AB, AC neutral EF ∥ BCEF = 1/2 BC = 3 EP = 1/3 EF = 1 train DPE ∽ train DBC train DPE to EP for bottom high: train DBC to BC at the bottom for high = S train DPE: S train DBC = 1:36 (2) extend into the BQ EF extension cord in GEF ∥ BC train BQC ∽ train EQGQ is EC midpoint, EQ = QC train BQC ≌ train EQGEG = BC = 6 BQ divide the < CBP < CBG = < GBP = < PGB train BPG for isosceles triangle BP = PG = y EG = EP + PG6 = x + y (3) (1) the train BQC ∽ train EQGCQ = 1/3 EC, CQ = 1/2 eqeg = 2 BC = 12 BP = PG = yEG = EP + PG = 12 x + y = 12 (2) x + y = 6 (n - 1)