Cho tam giac ABC co AB = AC . Tia phan giac goc A cat BC tai M
a) Chung minh tam giac AMB = tam giac AMC
b) Tren tia doi tia MA lay diem D sao cho MD = MA . Chung minh AB // CD
c) Qua diem M ve ME vuong goc AB ( E thuoc AB ) , MF vuong goc AC ( F thuoc AC ) . Chung minh EM = DC
d) Chung minh EM vuong goc DC
Cho tam giac ABC co AB = AC . Tia phan giac goc A cat BC tai M
a) Chung minh tam giac AMB = tam giac AMC
b) Tren tia doi tia MA lay diem D sao cho MD = MA . Chung minh AB // CD
c) Qua diem M ve ME vuong goc AB ( E thuoc AB ) , MF vuong goc AC ( F thuoc AC ) . Chung minh EM = DC
d) Chung minh EM vuong goc DC
lam cau c) va cau d) la dc
cho tam giac ABC co goc a nhon M la trung diem cua BC tren tia doi cua tia MA lay diem D sao cho MA=MD chung minh BAM=CDM chung minh AC=AD tren nua mat phang Bo AB ko chua C ve tia Ax vuong goc AB tren nua mat phang bo AC ko chua B ve tia Ay vuong goc AC tren tia Ax lay Diem P sao cho AP=AB tren tia Ay lay diem Q sao cho AQ=AC chung minh tam giac ABQ= tam giac APC goi giao diem cua DA va PQ la K chung minh AK vuong goc PQ
cho tam giac ABC vuong tai A. Diem M la trung diem cua canh BC. Tren tia doi cua tia MA lay diem D sao cho MA=MD. Chung minh rang a) tam giac AMC=tam giac DMB, b) AC=BD, c) AB vuong goc voi BD, d) AM=1/2BC. Cac ban co giup minh nhanh nhat co the nha
cho tam giac abc co goc b = goc c, tia phan giac cua goc a cat bc tai d
1. chung minh tam giac adb = tam giac adc
2. chung minh ad vuong goc voi bc
3. tren tia doi cua tia da lay diem m sao cho da = dm, chung minh ab // cm
cho tam giac ABC tren tia doi cua tia AB lay diem D sao cho AC= AD. tren tia doi cua tia AC lay diem E sao cho AE=AB
a) chung minh tam giac ADE= tam giac ACB
b)goi M la trung diem cua BE chung minh tam giac ABM bang tam giac AEM
c) duong thang AM cat CD tai N. Chung minh AN vuong goc CD
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 // CDCho tam giac ABC , AB < AC . AE la tia phan giac cua goc BAC (E thuoc BC) . Tren canh AC lay diem M sao cho AM=AB
a) Chung minh tam giac ABE = tam giac AME
b) AE cat BM tai diem I . Chung minh I la rung diem cua BM
c) Tren tia doi cua tia EM lay diem N sao cho EN=EC. Chung minh tam giac ENB = tam giac ECM
d) Chung minh K,B,M thang hang
Cho tam giac ABC , AB < AC . AE la tia phan giac cua goc BAC (E thuoc BC) . Tren canh AC lay diem M sao cho AM=AB
a) Chung minh tam giac ABE = tam giac AME
b) AE cat BM tai diem I . Chung minh I la rung diem cua BM
c) Tren tia doi cua tia EM lay diem N sao cho EN=EC. Chung minh tam giac ENB = tam giac ECM
d) Chung minh K,B,M thang hang
Xét tam giác ABE và tam giác AME có:
AM=AB(gt)
BAE=MAE(AE là tia phân giác BAC)
AE là cạnh chung
=>tam giác ABE=tam giác AME(c-g-c)
