cho tam giac ABC có AC>AB trung tuyen AM.Trên tia doi cua tia AM lay diem D sao cho AM=MD,noi C voi D.
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chotam giac abc nhon co ab<ac ,trung tuyen am tren tia am lay d sao cho am=md ,e la diem doi xung voi a qua bc ,ha la giao cua ae va bc cm ae vuong goc voi ed
Cho tam giac ABC vuong tai A co AB=3cm,AC=4cm.Goi AM la duong trung tuyen cua tam giac ABC tren tia doi cua tia AM lay diem D sao cho AM=MD
a)Tinh BC
b)C/m AB // CD
c)C/m goc BAM > goc CAM
d)Goi H la trung diem AM,tren duong thang AH lay E sao cho AH=HE , CE cat AD tai F . C/m F la trung diem CE
Cho tam giac ABC vuong tai A co AB=3cm,AC=4cm.Goi AM la duong trung tuyen cua tam giac ABC tren tia doi cua tia AM lay diem D sao cho AM=MD
a)Tinh BC
b)C/m AB // CD
c)C/m goc BAM > goc CAM
d)Goi H la trung diem AM,tren duong thang AH lay E sao cho AH=HE , CE cat AD tai F . C/m F la trung diem CE
cho tam giac ABC vuong tai a trung tuyen AM Tu M ke cac duong vuong goc MD va ME lan luot xuong AB va AC D thuoc AB E thuoc AC a)tu giac ADME la hinh gi vi sao b) goi N la diem doi xung voi M qua D chung minh tu giac AMBN la hinh thoi c)tren tia doi cua tia EM lay diem K sao cho KE=EM chung minh ba diem K,A,N thang hang d) tam giac vuong ABC co them dieu hien j thi AMBN la hinh vuong
cho tam giac ABC co AB=AC ,goi M la trung diem cua canh BC
chung minh tam giac ABM=tam giac ACM
chung minh AM vuong goc voi BC
tren tia doi cua tia MA lay diem D sao cho MD=MA
chung minh AB song song voi CD
*Xét ΔABM và ΔACM có:
\(\left\{{}\begin{matrix}AB=AC\left(gt\right)\\BM=MC\left(M.l\text{à}.trung.\text{đ}i\text{ểm}.c\text{ủa}.BC\right)\\AM.c\text{ạnh}.chung\end{matrix}\right.\)
⇒ ΔABM = ΔACM (c - c - c)
*Vì ΔABM = ΔACM (cmt)
⇒ \(\widehat{AMB}=\widehat{AMC}\) (hai góc tương ứng) Ta có: \(\widehat{AMB}+\widehat{AMC}=180^o\) (kề bù) ⇒ \(\widehat{AMB}=\widehat{AMC}\) = \(\dfrac{180^o}{2}=90^o\) ⇒ AM ⊥ BC *Xét ΔAMB và ΔDMC có: \(\left\{{}\begin{matrix}AM=MD\left(gt\right)\\\widehat{AMB}=\widehat{DMC}\left(\text{đ}\text{ối}.\text{đ}\text{ỉnh}\right)\\BM=MC\left(gt\right)\end{matrix}\right.\) ⇒ ΔAMB = ΔDMC (c - g - c) ⇒ \(\widehat{ABM}=\widehat{DCM}\) (hai góc tương ứng) Mà hai góc này ở vị trí so le trong ⇒ AB // CD
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cho tam giac ABC co AC>AB trung tuyen AM tren tia doi cua tia MA lay diem D sao cho MD=MAnoi C voi D
a) chung minh ADC>DAC tu do suy ra MAB>MAC
b) ke duong cao AH goi E la mot diemm nam giua A va H so sanh HC va HB