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 \(\frac{AO}{AM}=\frac{AN}{AK}=\frac{25}{31}\)

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 \(\frac{AO}{ON}=\frac{AM}{MI}=\frac{11}{9}\)

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 \(\frac{AM}{MK}=\frac{AO}{ON}\)

=>\(\frac{AO}{ON}=\frac{44}{27}\)

bài 1 làm sao vậy sao ko thấy mấy câu trả lời vậy bạn giúp mình giải bài tập số 1 với cảm ơn nhiều

không biết tớ trả trước mà

a. Tứ giác ABCD là hình bình hành.

$\Rightarrow AB=CD$(tính chất hình bình hành)

và $AB//CD\Rightarrow \widehat{ABD}=\widehat{BDC}$(so le trong)

Xét $\Delta AMB$và $\Delta CND$có:

$AB=CD$(cmt)

$\widehat{ABM}=\widehat{CDN}$(cmt)

$BM=DN$(GT)

$\Rightarrow \Delta AMB=\Delta CND(c.$