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Sửa đề: AM/MC=1/3
a: Qua N, kẻ NK//BM(K∈MC)
Xét ΔANK có OM//NK
nên \(\frac{OM}{NK}=\frac{AM}{AK}\) (1)
Xét ΔBMC có NK//BM
nên \(\frac{NK}{BM}=\frac{CK}{CM}\) (2)
AM/MC=1/3
=>MC=3AM
AM+MC=AC
=>AC=3AM+AM=4AM
=>\(AM=\frac14AC;CM=\frac34AC\)
\(\frac{OM}{OB}=\frac23\)
=>\(OM=\frac23OB\)
OM+OB=BM
=>\(BM=\frac23OB+OB=\frac53OB\)
=>\(BO=\frac35BM\)
BO+OM=BM
=>\(OM=BM-BO=BM-\frac35BM=\frac25BM\)
Từ (1),(2) suy ra \(\frac{OM}{NK}\cdot\frac{NK}{BM}=\frac{AM}{AK}\cdot\frac{CK}{CM}=\frac{AM}{CM}\cdot\frac{CK}{AK}\)
=>\(\frac{OM}{BM}=\frac{AM}{CM}\cdot\frac{CK}{AK}\)
=>\(\frac25=\frac13\cdot\frac{CK}{AK}\)
=>\(\frac{CK}{AK}=\frac25:\frac13=\frac25\cdot3=\frac65\)
=>\(\frac{CK}{CA}=\frac{6}{11}\)
=>\(CK=\frac{6}{11}CA\)
\(\frac{CK}{CM}=\frac{6}{11}:\frac34=\frac{6}{11}\cdot\frac43=\frac{24}{33}=\frac{8}{11}\)
Xét ΔCMB có NK//BM
nên \(\frac{CN}{CB}=\frac{CK}{CM}=\frac{8}{11}\)
=>\(\frac{CN}{NB}=\frac83\)
CK+KM=CM
=>KM=CM-CK=3/4AC-6/11CA=33/44AC-24/44AC=9/44AC
=>\(\frac{AM}{MK}=\frac13:\frac{9}{44}=\frac13\cdot\frac{44}{9}=\frac{44}{27}\)
Xét ΔANK có OM//NK
nên \(\frac{AM}{MK}=\frac{AO}{ON}\)
=>\(\frac{AO}{ON}=\frac{44}{27}\)
a: Qua M, kẻ MK//BN(K∈NC)
\(\frac{BM}{MC}=\frac23\)
=>\(CM=\frac35CB;BM=\frac25CB\)
\(\frac{CN}{NA}=\frac35\)
=>\(CN=\frac38CA;AN=\frac58CA\)
Xét ΔBNC có MK//BN
nên \(\frac{CM}{CB}=\frac{CK}{CN}\)
=>\(\frac{CK}{CN}=\frac35\)
=>\(CK=\frac35\cdot CN=\frac35\cdot\frac38\cdot CA=\frac{9}{40}\cdot CA\)
AN+NK+KC=AC
=>\(NK=AC-AN-CK=AC-\frac58AC-\frac{9}{40}AC=\frac38AC-\frac{9}{40}AC=\frac{15}{40}AC-\frac{9}{40}AC=\frac{6}{40}AC=\frac{3}{20}AC\)
\(AK=AN+NK=\frac58AC+\frac{3}{20}AC=\frac{25}{40}AC+\frac{6}{40}AC=\frac{31}{40}AC\)
=>\(\frac{AN}{AK}=\frac58:\frac{31}{40}=\frac58\cdot\frac{40}{31}=\frac{5\cdot5}{31}=\frac{25}{31}\)
Xét ΔAMK có ON//MK
nên \(\frac{AO}{AM}=\frac{AN}{AK}=\frac{25}{31}\)
a: Qua M, kẻ MK//BN(K∈NC)
\(\frac{BM}{MC}=\frac23\)
=>\(CM=\frac35CB;BM=\frac25CB\)
\(\frac{CN}{NA}=\frac35\)
=>\(CN=\frac38CA;AN=\frac58CA\)
Xét ΔBNC có MK//BN
nên \(\frac{CM}{CB}=\frac{CK}{CN}\)
=>\(\frac{CK}{CN}=\frac35\)
=>\(CK=\frac35\cdot CN=\frac35\cdot\frac38\cdot CA=\frac{9}{40}\cdot CA\)
AN+NK+KC=AC
=>\(NK=AC-AN-CK=AC-\frac58AC-\frac{9}{40}AC=\frac38AC-\frac{9}{40}AC=\frac{15}{40}AC-\frac{9}{40}AC=\frac{6}{40}AC=\frac{3}{20}AC\)
\(AK=AN+NK=\frac58AC+\frac{3}{20}AC=\frac{25}{40}AC+\frac{6}{40}AC=\frac{31}{40}AC\)
=>\(\frac{AN}{AK}=\frac58:\frac{31}{40}=\frac58\cdot\frac{40}{31}=\frac{5\cdot5}{31}=\frac{25}{31}\)
Xét ΔAMK có ON//MK
nên \(\frac{AO}{AM}=\frac{AN}{AK}=\frac{25}{31}\)
a. Tứ giác ABCD là hình bình hành.
⇒AB=CD⇒AB=CD(tính chất hình bình hành)
và AB//CD⇒ˆABD=ˆBDCAB//CD⇒ABD^=BDC^(so le trong)
Xét ΔAMBΔAMBvà ΔCNDΔCNDcó:
AB=CDAB=CD(cmt)
ˆABM=ˆCDNABM^=CDN^(cmt)
BM=DNBM=DN(GT)
⇒ΔAMB=ΔCND(c.
Sửa đề: AM/MC=1/3
a: \(\frac{AM}{MC}=\frac13\)
=>MC=3AM
AM+MC=AC
=>AC=3AM+AM=4AM
=>\(AM=\frac14AC;CM=\frac34AC\)
\(\frac{OM}{OB}=\frac23\)
=>\(OM=\frac23OB\)
OM+OB=BM
=>\(BM=\frac23OB+OB=\frac53OB\)
=>\(BO=\frac35BM;MO=\frac25BM\)
Qua N, kẻ NI//BM(I∈CM)
Xét ΔBMC có NI//BM
nên \(\frac{NI}{BM}=\frac{CI}{CM}=\frac{CN}{CB}\) (2)
Xét ΔANI có OM//NI
nên \(\frac{OM}{NI}=\frac{AM}{AI}=\frac{AO}{AN}\) (1)
Từ (1),(2) suy ra \(\frac{NI}{BM}\cdot\frac{OM}{NI}=\frac{CI}{CM}\cdot\frac{AM}{AI}\)
=>\(\frac{CI}{CM}\cdot\frac{AM}{AI}=\frac{OM}{BM}=\frac25\)
=>\(\frac{AM}{MC}\cdot\frac{CI}{AI}=\frac25\)
=>\(\frac{CI}{AI}=\frac25:\frac13=\frac25\cdot3=\frac65\)
=>\(\frac{CI}{CA}=\frac{6}{11}\)
=>CI=6/11CA
AM+MI+CI=AC
=>MI=AC-CI-AM=AC-6/11AC-1/4AC=5/11AC-1/4AC=20/44AC-11/44AC=9/44AC
\(\frac{CI}{IM}=\frac{6}{11}:\frac{9}{44}=\frac{6}{11}\cdot\frac{44}{9}=\frac23\cdot4=\frac83\)
Xét ΔCMB có NI//BM
nên \(\frac{CN}{NB}=\frac{CI}{IM}=\frac83\)
\(\frac{AM}{MI}=\frac14:\frac{9}{44}=\frac14\cdot\frac{44}{9}=\frac{11}{9}\)
Xét ΔANI có OM//NI
nên \(\frac{AO}{ON}=\frac{AM}{MI}=\frac{11}{9}\)