Xét ΔOAB và ΔODC có
OA=OD
\(\widehat{AOB}=\widehat{DOC}\)
OB=OC
Do đó: ΔOAB=ΔODC
Xét ΔOAB và ΔODC có
OA=OD
\(\widehat{AOB}=\widehat{DOC}\)
OB=OC
Do đó: ΔOAB=ΔODC
Tren canh Ax va Ay cua \(\widehat{xAY}\) lan luot lay diem B va C sao cho AB =AC . Goi M la trung diem
cua doan thang BC .so sanh tam giac AMB va tam giac MCA
ve \(\widehat{xOy}\) va tia phan giac Ot .Tren Ox va Oy lan luot lay diem A va
B sao cho OA=OB . Tren Ot lay diem C sao cho OC < OA
1,Chung minh CA=CB
cho tam giac ABC can tai A, ke tia phan giac AD (D thuoc BC) tren tia doi cua tia AB lay diem E sao cho AE=AB.Tren tia phan giac cua AE lay diem F sao choAF=BD
CMR: a) AD vuong goc voi BC
b) AF song song voi BC
c) EF=AD
d) cac diem E,F,C thang hang
cho tam giac co diem M la trung diem cua BC. Keo dai AM
lay MD =MA
1 CM tam giac ABM =tam giac DCM; tam giac ACM=tam giac DBM
roi viet cac cap canh va cap goc tuong ung bang nhau
2, so sanh tam giac ABD va DAC
cho doan thang AB tren 2 nua mat phang doi nhau bo AB ve ax⊥ab , By⊥BA. tren Ax va By lan luot lay 2 diem C,D sao cho AC=BD.goi O la trung diem AB chung minh
a)△AOC=△BOD
B)O la trung diem cua CD
Tren canh Ax va Ay cua \(\widehat{xAy}\) lan luot lay cac diem B va C sao cho AB =AC
tia phan giac At cua \(\widehat{xAy}\) cat BC tai D so sanh tam giac ADB va CDA
va so sanh cac cap canh va goc tuong ung giua chung
Cho tam giac ABC can tai A ( AB>AC) lay M la trung diem BC Chung Minh
a) BAM^=CAM^
b) tia phan giac cua goc ABC cat Am tai H. Chung Minh BH=CH
c) Duong thang di qua A song song voi BC cat BH tai I chung Minh tg ACI can
Cho tam giac ABC co AB < AC co AD la duong phan giac tren canh AC
lay E sao cho AE=AB . So sanh tam giac ADB va tam giacAED
CHO TAM GIAC ABC CAN TAI a GOI d LA TRUNG DIEM CUA CANH bc KE dh VUONG GOC VOI ab .dk VUONG GOC VOI ac
tren tia doi DK lay diem E sao cho DE=DK CHUNG MINH GOC bed=90 do