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Vẽ hình bình hành DAFH.
Gọi N là giao điểm của hai đường chéo DF và AH, M là giao điểm của EH và BC
Ta có NA = NH, ND = NF
Ta đặt ^ADH = ^AFH = \(\alpha\)thì ^BDH = ^HFC = \(\alpha\)+ 600
^DAF = 1800 -\(\alpha\)
^BAC = 3600 - ^BAD - ^CAF - ^DAF = 3600 - 600 - 600 - (1800 - \(\alpha\)) = \(\alpha\)+ 600
\(\Delta\)BDH và \(\Delta\)HFC có: BD = HF (= AD); ^BDH = ^HFC (cmt); DH = FC (= AF)
Do đó \(\Delta\)BDH = \(\Delta\)HFC (c.g.c) => HB = HC (1)
Chứng minh tương tự, ta được \(\Delta\)BAC = \(\Delta\)HFC (c.g.c) => BC = HC (2)
Từ (1) và (2) suy ra HB = HC = BC
Tứ giác BHCE có các cặp cạnh đối bằng nhau (cùng bằng BC) nên là hình bình hành => MB = MC và MH = ME
- Xét ∆AEH có AM và AN là hai đường trung tuyến nên giao điểm G của chúng là trọng tâm => EG = 2/3EN và AG = 2/3AM.
- Xét ∆ABC có AM là đường trung tuyến mà AG = 2/3AM nên G là trọng tâm của ∆ABC
- Xét ∆EDF có EN là đường trung tuyến mà EG = 2/3EN nên G là trọng tâm của∆EDF
Vậy ∆ABC và ∆EDF có cùng trọng tâm G
chưa học trả lời làm gì cho mất thời gian mất công bạn Thanh Trang Hoàng phải đọc
Gọi M là giao điểm của AE và CF
ADFE là hình bình hành nên ^ADF = ^AEF (hai góc đối)
Suy ra ^BDF = ^FEC
Xét \(\Delta\)BDF và \(\Delta\)FEC có:
BD = FE (cùng bằng AD)
^BDF = ^FEC (cmt)
DF = EC ( cùng bằng AE)
Do đó \(\Delta\)BDF = \(\Delta\)FEC (c.g.c) suy ra BF = CF (1) và ^BFD = ^FCE
Mặt khác ^AMC = ^DFC (do DF // AE)
^AMC = ^MEC + ^FCE = 600 + ^FCE và ^DFC = ^BFC + ^BFD
Do đó ^BFC = 600 (2)
Từ (1) và 2) suy ra \(\Delta\)FBC đều (đpcm)
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Tự vẽ hình nha.
Vì ADKE là hình bình hành.
=> ^ADK = ^ AEK
=> ^ ADK + 60o = ^ AEK + 60o
=> ^BDK = ^KCE
Xét tam giác BDK = tam giác KEC ( c.g.c )
=> BK = KC ( 1 )
Có ^DAE + ^ BAC + ^ DAB + ^ EAC = 360o
=> ^ DAE + ^BAC + 120o = 360o
=> ^BAC = 240o - ^DAE
mà ^DAE = 180o - ^ADK
=> ^BAC = 60o + ^ADK = ^BDA
=> tam giác BAC = tam giác BDK ( c g.c )
=> BC = BK ( 2 )
Từ ( 1 ), ( 2 )
=> BC = BK = CK
=> tam giác KBC đều
LẤY I LÀ TRUNG ĐIỂM CỦA BC, O LÀ TRUNG ĐIỂM CỦA AC
XÉT TAM GIÁC MAN VÀ TAM GIÁC IOF CÓ
OI = AB/2=AE/2=AM
OF=AN ( CÚNG LÀ ĐƯƠNG CAO CỦA TAM GIÁC ĐỀU)
GÓC FOI = GÓC MAN = 90 + GÓC A
=> TAM GIÁC MAN = TAM GIACC IOF ( C.G.C)
=> FI = DM
=> GÓC OFI = GÓC MNA
=> GÓC MND = GÓC ANC - GÓC MNA - GÓC DNC
= 90 - GÓC OFI - GÓC IFC
= 90 - 30 = 60
LẠI CÓ FI = ND/2
FI = MD
=> MD = ND/2
MÀ GÓC MND = 60
-> TAM GIÁC MND LÀ NỬ TAM GIÁC ĐỀU
=> DM VUÔNG GÓC DN