cho tam giac ABC co goc C =30 duong trung truc AC cat AC tai H va cat BC tai D
a)chung minh tam giac ABD deu
b)ke phan giac goc B cat AD tai K cat DH tai I chung minh IA=ID=IC
c)chung minh AE<\(\frac{AB+BC}{2}\) voi IE vuong goc voi BC tai E
cho tam giac ABC co goc BAC =90 do , tia phan giac goc ACB cat AB tai M . Qua A ke duong thanh vuong goc voi CM tai H , AN cat BC tai H.a) cminh tam giac ACN = tam giac HCN b) Qua N ke HM cat AC tai K . chung minh BK song song AH . c) qua N ke duong thang song song AC cat BC tai E . chung minh NE = 1/2 AC
CHO TAM GIAC ABC CO AD LA PHAN GIAC CUA GOC A (D THUOC BC). KE DEVUONG GOC VOI AB, KE DF VUONG GOC VOI AC
a, CHUNG MINH TAM GIAC AED= TAM GIAC AFD
b, TIA FD CAT TIA AB TAI P, TIA ED CAT TIA AC TAI Q. CHUNG MINH AP=AQ
c, TIA AD CAT PQ TAI M. CHUNG MINH AM LA DUONG TRUNG TRUC CUA DOAN THANG PQ
d, CHO GOC BAC=50 DO. TINH SO DO GOC APQ VA GOC AQP
Cho tam giac ABC co AB=3,AC=4,BC=5
a,Chung minh tam giac ABC vuong
b,Tia phan giac goc B cat AC tai D .Ke DH vuong goc voi BC(H thuoc BC) Chung minh DH=DA)
c,Dthang DH cat AB tai E.Chung minh DE=DC
a) Ta có: \(3^2+4^2=25\)
\(5^2=25\)
suy ra: \(AB^2+AC^2=BC^2\)
\(\Rightarrow\)\(\Delta ABC\)\(\perp\)\(A\)
b) Xét 2 tam giác vuông: \(\Delta BAD\)và \(\Delta BHD\)có:
\(\widehat{ABD}=\widehat{HAD}\) (gt)
\(BD:\)cạnh chung
suy ra: \(\Delta BAD=\Delta BHD\)(ch_gn)
\(\Rightarrow\)\(DA=DH\)(cạnh tương ứng)
c) Xét 2 tam giác vuông: \(\Delta ADE\)và \(\Delta HDC\)có:
\(AD=HD\)(cmt)
\(\widehat{ADE}=\widehat{HDC}\) (đđ)
suy ra: \(\Delta ADE=\Delta HDC\)(cgv_gn)
\(\Rightarrow\)\(DE=DC\)(cạnh tương ứng)
CHUNG MINH CHO MIK Y D NUA
d,Chung minh BE=BC
cho tam giac abc nhon noi tiep (O;R) co ab>ac tia phan giac cua goc a cat bc tai i va cat (O) tai d. ha be va cf vuong goc voi ad tai e va f, ve duong cao ah cua tam giac abc
c, ve im vuong goc ab tai m chung minh f,m,h thang hang
d, bf cat ce tai k chung minh ak la phan giac ngoai tam giac abc
cho tam giac ABC co goc A = 64 do hai tia phan giac cua goc B va goc C cat nhau tai I
a+ tinh goc BIC
b) ke duong phan giac qua I song song voi BC cat AB tai M va AC tai N chung minh rang tam giac BMI va tam giac CNI can
c) chung minh MN=BN+CN
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 vuong tai A, co AB=4, AC=5
a) Hay so sanh so do goc B va goc C cua tam giac ABC
b)tia phan giac cua goc ABC cat canh AC tai D. Ke DM vuong goc voi BC tai M chung minh tam giac ABM=tam giac MBD
c)Hai tia MD va BÂct nhau tai E . tia BD cat EC tai N . Chung minh goc BNC=90o
d) Goi K la trung diem cua DE . Chung Minh CK=3/4 EC
cho tam giac ABC can tai C.Ke tia phan giac voi goc C cat AB tai I.Biet AC=5cm,AB=6cm. a,Chung minh tam giac ACI=tam giac BCI va AI=BI. b,Tinh do dai CI. c, Qua A va B lan luot ke cac duong thang vuong goc voiAC va BC chung cat nhau tai K.Chung minh 3 diem C,I,K thang hang
Hình vẽ:
Giải:
a/ Xét \(\Delta ACI\) và \(\Delta BCI\) có:
AI: chung
\(\widehat{ACI}=\widehat{BCI}\left(gt\right)\)
AC = BC (gt)
=> \(\Delta ACI=\Delta BCI\left(c-g-c\right)\left(đpcm\right)\)
=> AI = BI (c t/ứng)(đpcm)
b/ \(\Delta ACI=\Delta BCI\left(ýa\right)\)
\(\Rightarrow\widehat{AIC}=\widehat{BIC}\) (g t/ứng)
mà \(\widehat{AIC}+\widehat{BIC}=180^o\) (kề bù)
=> \(\widehat{AIC}=\widehat{BIC}=90^o\)
=> CI _l_ AB
Vì AI = BI mà AB = 6
=> AI = BI = 3
Áp dụng định lý Py-ta-go vào \(\Delta ACI\) vuông tại I có: \(CI^2+AI^2=AB^2\)
hay \(CI^2+3^2=5^2\)
\(\Rightarrow CI^2=5^2-3^2=16\)
\(\Rightarrow CI=4\left(cm\right)\)
c/ Xét 2 \(\Delta vuông\): \(\Delta ACK\) và \(\Delta BCK\) có:
AK: chung
AC = BC (gt)
=> \(\Delta ACK=\Delta BCK\left(ch-cgv\right)\)
\(\Rightarrow\widehat{ACK}=\widehat{BCK}\) (g t/ứng)
=> CK là tia p/g của góc ACB (1)
Lại có: CI là tia p/g của góc ACB (gt)
=> CK trùng CI
=> 3 điểm C, I, K thẳng hàng (đpcm)
Cho tam giac ABC co A=75, B=35. AD la tia phan giac cua goc A. Qua A ke duong vuong goc voi AD duong thang nay cat uong thang BC tai E. Tren nua mat phang bo AE cung phia voi C ve tia Ax sao cho AEC=EAx. Tia Ax cat ED tai M.
a) Chung minh: ME=MD
b) Chung minh: BC= chu vi tam giac ABC
c) Tu M; N ta ke cac duong thang lan luot vuong goc voi Ae; AM. Chung cat nhau tai O. Duong thang OC cat AE tai N va AD tai K.Chung minh MN vuong goc voi NK