a: \(\widehat{AEB}< 90^0\)
nên \(\widehat{BEC}>90^0\)(đpcm)
b: Đặt \(\widehat{C}=a;\widehat{B}=c\)
Theo đề, ta có:
\(\left\{{}\begin{matrix}a-c=10\\a+c=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=50\\b=40\end{matrix}\right.\)
\(\widehat{AEB}=90^0-20^0=70^0\)
=>\(\widehat{BEC}=110^0\)
Cho ∆ ABC, goc A=90°. Ke AH vuong goc vs BC tai H. Goi AD la tia p/giac cua goc HAC
a, CMrang goc BAD = goc BDA
b, biet goc C=40°. Tinh goc B, goc BDA , goc DAC
Cho tam giac ABC. Cac tia phan giac cua cac goc B va C cat nhau tai I. Chung minh rang AI la phan giac goc A
Cho tam giac ABC biet AB nho hon AC.Tren tia BA lay diem D sao cho BC bang BD. Noi C voi D. Phan giac goc B cat canh AC ,DC tai E, I. Chung minh tam giac BEC bang tam giac BID .Tu A ve duong vuong goc voi DC , H thuoc DC.Chung minh AH song song voi BI
cho tam giac ABC = tam giac MNP biet AC=MN,BC=MP . Khang dinh nao sau day la dung ? a, goc A = goc N b , goc B = goc M c, goc A= goc M d, goc C= goc P
Cho tam giac ABC co goc B= 80° , goc C = 30°.Tia phan giac cua goc A cat BC ow D .Tinh goc ADc,ADB
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.
tam giac ABC co tia phan giac cua goc B cat AC o D.Qua A ke duong thang song song voi BD,duong thang nay cat duong thang BC o E.Chung minh rang goc BEA = goc BAE
Cho tam giac ABC, diem O nam trong tam giac do.
a. CMR : goc BOC = goc A + goc ABO + goc ACO.
b. Biet goc ABO + goc ACO = 90 do - goc A/2 va tia BO la p/g cua goc B. CMR :CO la p/g cua goc C.
Huhu gjup mk vs m.n oj chju nay dj hc r. thanks nha !
cho tam giac abc vuong tai a ( ab<ac) tia phan giac cua goc b cắt ac o d. ke dh vuong bc. tren tia ac lay diem e sao cho ae= ab. dg thanh vuong goc ab cat tia dh o k cmr goc dbk = 45 độ
