Chứng minh hình tam giác có 3 cạnh
K
Khách
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19 tháng 5 2017
A B C D E F G K
Trên tia đối của DG, lấy điểm K sao cho DK=DG. Nối K với B. Ta được \(\Delta\)BGK với 3 cạnh BG,GK,BK.
AD là trung tuyến của \(\Delta\)ABC, G là trọng tâm của tam giác ABC \(\Rightarrow\)AG=2/3AD \(\Rightarrow\)DG=1/3AD.
DK=DG \(\Rightarrow\)DK=1/3 AD \(\Rightarrow\)DG+DK=1/3AD+1/3AD=2/3AD \(\Rightarrow\)GK=2/3 AD (1)
Ta có: BG là 1 cạnh của \(\Delta\)BGK và BG=2/3BE (2)
Xét \(\Delta\)CGD và \(\Delta\)BKD có:
CD=BD
\(\widehat{CDG}\)=\(\widehat{BDK}\) (Đối đỉnh) \(\Rightarrow\)\(\Delta\)CGD=\(\Delta\)BKD (c.g.c)
DG=DK
\(\Rightarrow\)CG=BK (2 cạnh tương ứng). Mà theo tính chất 3 đường trung tuyến của tam giác : CG=2/3 CF \(\Rightarrow\)BK=2/3CF (3)
Từ (1),(2) và (3) \(\Rightarrow\)3 đường trung tuyến AD,BE,CF tỉ lệ với 3 cạnh của \(\Delta\)BGK lần lượt là GK,BG,BK.
\(\Rightarrow\)AD,BE,CF thỏa mãn bất đẳng thức tam giác hay ta có thể nói AD,BE,CF là 3 cạnh của một hình tam giác (đpcm).
16 tháng 12 2017
1/ Phần này đơn giản thôi bạn! Khi chứng minh tâm của đường tròn ngoại tiếp tam giác vuồn là trung điểm cạnh huyền thì ta chứng minh ngược lại là trung điểm của cạnh huyền trong 1 tam giác vuông là tâm của đường tròn ngoại tiếp.
Giả sử ta có tam giác ABC vuông tại A và O là trung điểm của cạnh huyền BC
=> AO là đường trung tuyến ứng với cạnh huyền
=> OA = OB =OC = 1/2 BC
=> O là tâm của đường tròn ngoại tiếp tam giác ABC
Vậy ....
2/ Giả sử ta có tam giác ABC có BC là đường kính của đường tròn ngoại tiếp tam giác.
Gọi O là tâm của đường tròn ngoại tiếp tam giác ABC
=>OA = OB =OC (*)
mà BC là đường kính của đường tròn ngoại tiếp
=> O là trung điểm BC
=> OB = OC = 1/2 BC(**)
từ (*) và (**) => OA = OB = OC = 1/2 BC
=> tam giác ABC vuông tại A
20 tháng 2 2018
@Nhoc_sieu_pham đây là toán lớp 7 mà, sao lại giải cách lớp 9 như vậy được?
15 tháng 6 2022
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3 tháng 9 2018
Bài 2:
kẻ hình thang ABCD
kẻ 2 đường cao AH và BK nối B với H
xét tam giác ABH và tam giác KBH
có ^ABH = ^KBH ( 2gocs so le trong )
HB chung
=> tam giác ABH = tam giác KBH (cạnh huyền +góc nhọn )
=> AB =HK ( 2 cạnh tương ứng )
xét tam giác BKC có BC>KC ( trong tam giác vuông cạnh huyền là cạnh lớn nhất )(1)
xét tam giác AHD có AD>HD (trong tam giác vuông cạnh huyền là cạnh lớn nhất)(2)
từ (1) và (2) => BC+AD >KC+HD
ta lại có DH+DK +HK =DC
mà AB=HK (C/m )
=> DH+DK+AB =dc
ta có DC-AB = DH+DK+AB-AB= DH+DK
mà DH+DK<BC+AD(c/m)
=>DC -AB< BC+AD
vậy tổng hai cạnh bên của hình thang lớn hơn hiệu hai đáy
Giải
Tam là 3
Giác là cạnh
=> Tam + giác = tam giác = 3 cạnh
Study well
Ai gian lận đánh “sai”vào DTTALTTL 12-21 đó? Đúng rồi lại còn “sai”