Giải thích các bước giải:

 a) Ta có:

ΔAOB;C∈AO;D∈BO;CD//AB⇒COAO=DOBO=CDAB(DL:Thales)⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩CO=AO.DOBO=12.39=4cmCD=AB.DOBO=18.39=6cmΔAOB;C∈AO;D∈BO;CD//AB⇒COAO=DOBO=CDAB(DL:Thales)⇒{CO=AO.DOBO=12.39=4cmCD=AB.DOBO=18.39=6cmΔAOB;C∈AO;D∈BO;CD//AB⇒COAO=DOBO=CDAB(DL:Thales)⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩CO=AO.DOBO=12.39=4cmCD=AB.DOBO=18.39=6cmΔAOB;C∈AO;D∈BO;CD//AB⇒COAO=DOBO=CDAB(DL:Thales)⇒{CO=AO.DOBO=12.39=4cmCD=AB.DOBO=18.39=6cm

Vậy OC=4cm;CD=6cmOC=4cm;CD=6cm

b) Ta có:

ΔFAB;C∈FB;D∈FA;CD//AB⇒FAAD=FCCB(DL:Thales)⇒FA.BC=FC.ADΔFAB;C∈FB;D∈FA;CD//AB⇒FAAD=FCCB(DL:Thales)⇒FA.BC=FC.AD

c) Ta có:

ΔABD;O∈BD;M∈AD;OM//AB⇒OMAB=DODB(1)ΔABD;O∈BD;M∈AD;OM//AB⇒OMAB=DODB(1)

Lại có:

ΔABC;O∈AC;N∈BC;ON//CD⇒ONAB=COCA(2)ΔABC;O∈AC;N∈BC;ON//CD⇒ONAB=COCA(2)

Mặt khác:

ΔAOB;C∈AO;D∈BO;CD//AB⇒BODO=AOCO⇒BO+DODO=AO+COCO⇒BDDO=ACCO⇒DODB=COCA(3)ΔAOB;C∈AO;D∈BO;CD//AB⇒BODO=AOCO⇒BO+DODO=AO+COCO⇒BDDO=ACCO⇒DODB=COCA(3)

Từ (1),(2),(3)⇒OMAB=ONAB⇒OM=ON