cho tam giac abc can tai a co goc bac =50do tren tia doi cua tia bc lay diem d tren tia doi cua tia cb lay diem e sao cho bd =ba ce=ca tinh goc dae
cho tam giac abc deu ve ben ngoai tam giac cac tam giac abd vuong can tai b tam giac ace vuong can tai c tinh so goc nhon cua ade
XÉT \(\Delta ABC\)CÂN TẠI A
\(\Rightarrow\hept{\begin{cases}AB=AC\\\widehat{B}=\widehat{C}\end{cases}}\)
TA CÓ \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\left(Đ/L\right)\)
THAY\(50^0+\widehat{B}+\widehat{C}=180^o\)
\(\widehat{B}+\widehat{C}=130^o\)
MÀ\(\widehat{B}=\widehat{C}\)
\(\Rightarrow\widehat{B}=\widehat{C}=\frac{130^o}{2}=65^o\)
TA CÓ \(\widehat{DBA}+\widehat{ABC}=180^o\left(KB\right)\)
\(\Rightarrow\widehat{DBA}=180^o-65^o=115^o\)
TA CÓ\(\widehat{ACE}+\widehat{ACB}=180^o\left(KB\right)\)
\(\Rightarrow\widehat{ACE}=180^o-65^0=115^o\)
XÉT \(\Delta ACE\)CÓ AC=CE (GT) =>\(\Delta ACE\)CÂN TẠI C
\(\Rightarrow\widehat{CAE}=\widehat{AEC}=\frac{180^o-115^0}{2}=32,5^0\)
XÉT \(\Delta ABD\)CÓ AB=BD (GT) =>\(\Delta ABD\)CÂN TẠI B
\(\Rightarrow\widehat{DAB}=\widehat{ADB}=\frac{180^o-115^0}{2}=32,5^0\)
TA CÓ\(\widehat{DAB}+\widehat{BAC}+\widehat{EAC}=\widehat{DAE}\)
THAY\(32,5^o+50^0+32,5^0=\widehat{DAE}\)
\(\Rightarrow\widehat{DAE}=115^0\)
Cho tam giac abc, am la trung tuyen.Tren tia am lay diem Nsao cho MN=AM
C/M:CN//AB
C/M:tam giac abc=tam giac ncb
Dung ngoai tam giac abc cac tam giac abd va tam giac ace vuong can a
C/M:Be=CD va BE vuong goc CD
C/m:AN=DE va an vuong de
ke ah vuong goc bc C/M:AH di qua trung diem cua de
cho tam giac abc vuong tai A ve ra phia ngoai tam giac abd can tai B ve ra phia ngoai tam giac ACE can tai C biet ACE=ABD cho M trung diem BC so sanh MD,ME
cho tam giac ABC nhon co AB<AC , ve ra phia ngoai tam giac ABC cac tam giac ABD , ACE cung vuong can tai A
a) c/m BE = CD
b) C/m BE vuong goc voi CD
c) lay M la trung diem cua BC . Chung minh DE = 2AM
TÍCH MÌNH ĐÚNG NHA MÌNH CẢM ƠN RẤT NHIỀU
cho tam giac ABC can tai A . M la mot diem thay doi tren BC . c/m rang khi M la mot diem bat ki tren BC thi tong khoang canh tu M den 2 canh AB va AC la ko doi
cho tam giac nhon ABC ve ra phia ngoai tam giac vuong can ABD va AEC(vuong can tai B va tai C ). tren tia doi cua tia AH lay diem I sao cho AI=BC(AH vuong goc voiBC(H thuoc BC)cm
a)tam giac ABI=tam giac BDC
b)Bivuong goc voi CD
cho tam giac ABC co : AB= 9cm, AC= 12cm, BC= 15cm
1) tam giac ABC co dang dac biet nao? Vi sao ?
2) ve trung tuyen AM cua tam giac ABC, ke MH vuong goc voi AC
Tren tia doi tia MH lay diem K sao cho MK= MH
a) CM tam giac MHC= tam giac MKB. tu do suy ra BK//AC
b)BH x AM tai G. CM: G la trong tam cua tam giac ABC
a) vuông ; cân hoặc đều hihi!!!!!!!!!!!!!!
Cho tam giac ABC can tai A co AD la duong trung tuyen
a)Chung minh tam giac ABD= tam gaic ACD va AD vuong goc voi BC
b)Cho AB=10cm,BC=16cm. Tinh do dai AD va so sanh cac goc cua tam giac ABC.
c) Ve duong trung tuyen CF cua tam giac ABC cat AD tai M. Tinh do dai AM.
d) Ve DH vuong goc AC tai H, tren canh AC va canh DC lan luot lay hai diem E,K sao cho AE=AD va DK=DH. Chung minh: EK vuong goc voi BC
A,
xét \(\Delta ABD\)và \(\Delta ACD\)
CÓ \(\hept{\begin{cases}AB=AC\\chungAD\\BD=DC\end{cases}}\)
SUY RA \(\Delta ABD\)=\(\Delta ACD\) (C.C.C) (1)
=> \(\widehat{BDA}\)=\(\widehat{CDA}\)
MÀ \(\widehat{BDA}\)+\(\widehat{CDA}\)=180
=> \(\widehat{BDA}\)=\(\widehat{CDA}\)=90
B, (1) => BC=DC=1/2 BC=8
ÁP DỤNG ĐỊNH LÍ PITAGO TA CÓ
\(AB^2=AD^2+BD^2\)
=> AD^2=36
=>AD=6
cho tam giac ABC. Ve phia ngoai tam giac tai dinh A, ke Ax vuong goc AB, lay diem E thuo)c tia Ax sao cho AE=AB ( E,C nam o 2 phia cua diem A). Ke Ay vuong goc voi AC, lay diem F tren Ay sao cho AF=AC (F,B nam o 2 phia cua A). Lay M la trung diem cua BC
a. AM=1/2EF
B.Keo dai AM cat EF tai I. CM tam giac IAF vuong tai I
cho tam giac ABC co AM la trung tuyen. tren AM lay diem d sao cho M la trung diem cua AD.Tren BC lay diem K SAO CHO C la trung diem cua BK.
CMR: a,KM trung tuyen cua tam giac ADK
b, M la trong tam cua tam giac ADK
c, so sanh cac canh cua tam giac ABC voi dg trung tuyen cua tam giac ADK.
