Cho tam giac ABC tia phan giac cua goc A cvat BC tai D .Tinh cac goc cua tam giac ABC biet ADH =80 0 va goc B =1.5 goc C
Cho tam giac ABC co goc A khac goc B cac tia phan giac trong va ngoai cua goc C lan luot cat duong thang BA tai D va E tinh goc CED theo goc A va goc B cua tam giac ABC
cho tam giac ABC co A=60 cac tia phan giac cua goc B va C cat nhau tai I,cat cac canh AC,AB lan luot tai D vaE tia phan gia cua goc BIC cat BC o F tinh goc BIC
cho tam giac ABC co goc A=a . Cac tia phan giac cua goc B va C cat nhau tai I . Cac tia phan giac cua cac goc ngoai ding B va C cat nhau tai K. Tia phan giac cua g oc B cat tia phan giac ngoai ding C tai E. Tinh so do cua BKC, B . C theo a
cho tu giac abcd biet goc a: goc b: goc c: goc d =4:3:2:1
a) tinh cac goc cua tu giac abcd
b) cac tia phan giac cua goc c va goc d cat nhau tai e. cac duong phan giac cua goc ngoai tai ca dinh c va d cat nhau tai f. tinh goc ced va goc cfd
a) Ta có: \(\frac{\widehat{A}}{4}=\frac{\widehat{B}}{3}=\frac{\widehat{C}}{2}=\frac{\widehat{D}}{1}\)
Áp dụng t/c dãy tỉ số bằng nhau ta được:
\(\frac{\widehat{A}}{4}=\frac{\widehat{B}}{3}=\frac{\widehat{C}}{2}=\frac{\widehat{D}}{1}=\frac{\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}}{4+3+2+1}=\frac{360}{10}=36\)
\(\Rightarrow\widehat{A}=144^0;\widehat{B}=108^0;\widehat{C}=72^0;\widehat{D}=36^0\)
Cho tam giac ABC, goc A =60 do. Tia phan giac cua goc Bva C cat cac canh doi dien tai D va E, BD va CE cat nhau tai O. Tia phan giac cua goc BOC cat BC tai F. Chung minh a, OD=OE=OF b, Tam giac DEF la tam giac deu.
cho tam giac ABC co B=110 do C =50 do phan giac goc ngoai tai dinh A cat tia doi cua tia BC tai D tinh cac goc cua tam giac ABD
Cho tam giac ABC, goc A bang 60 dọ. Tia phan giac trong cua goc B va goc C cat cac canh doi dien tai D va E, BD va CE cat nhau o O. Tia phan giac cua goc BOC cat BC tai F. CM:
a) OD=OE=OF
b) tam giac DEF deu
bai 4:cho tam giac ABC co goc A=90 do.Goi M la trung diem cua AC,tren tia Bm lay diem N sao cho M la trung diem cua doan BN.CMR:
a)CN vuong goc voi AC va CN=AB
b)AN=BC va AN song song voi BC
bai 4:cho tam giac ABC ke AH vuong goc voi BC(H thuoc BC)goi M la trung diem cua canh BC.Biet goc BAH=goc HAM=goc MAC.Tinh cac goc cua tam giac ABC
bai 6:cho tam giac ABC vuong tai A,phan giac BD.Tren canh BC lay diem H sao cho BH=BA
a)CMR:DH vuong goc voi BC
b)BIET goc ADH=120 do.Tinh goc ABD
Bài 6:
b) Theo câu a) ta có \(\Delta ABD=\Delta HBD.\)
=> \(\widehat{ADB}=\widehat{HDB}\) (2 góc tương ứng).
Ta có: \(\widehat{ADB}+\widehat{HDB}=\widehat{ADH}\left(gt\right)\)
=> \(\widehat{ADB}+\widehat{HDB}=120^0\)
Mà \(\widehat{ADB}=\widehat{HDB}\left(cmt\right)\)
=> \(2.\widehat{ADB}=120^0\)
=> \(\widehat{ADB}=120^0:2\)
=> \(\widehat{ADB}=60^0.\)
=> \(\widehat{ADB}=\widehat{HBD}=60^0\)
Xét \(\Delta ABD\) có:
(định lí tổng ba góc trong một tam giác).
=> \(90^0+\widehat{ABD}+60^0=180^0\)
=> \(150^0+\widehat{ABD}=180^0\)
=> \(\widehat{ABD}=180^0-150^0\)
=> \(\widehat{ABD}=30^0\)
Vậy \(\widehat{ABD}=30^0.\)
Chúc bạn học tốt!
Cho tam giac ABC ,tia phan giac cua goc B cat tia phan giac cua goc C tai diem I .Tinh goc BIC biet :
a) Goc A =80 do
b) Goc A = m do
Trong tam giác BIC có: góc BIC = 180o - (IBC + ICB)
Mà góc IBC = ABC /2 ; góc ICB = ACB / 2 nên góc BIC = 180o - \(\frac{ABC+ACB}{2}\) = 180o - \(\frac{180^o-BAC}{2}\)
a) Góc BAC = 80o thì góc BIC = 180o - \(\frac{180^o-80^o}{2}\) = 130o
b) Góc BAC = m thì góc BIC = 180o - \(\frac{180^o-m}{2}=90^o+\frac{m}{2}\)