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Trình Nguyễn Lê
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Nguyễn Linh Chi
9 tháng 1 2020 lúc 9:50

Câu hỏi của nguyen huyen dieu - Toán lớp 7 - Học toán với OnlineMath

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dsfdsf
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Trịnh Việt Dũng
15 tháng 6 2022 lúc 20:31

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daophanminhtrung
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Nguyễn Lê Phước Thịnh
19 tháng 2 2022 lúc 21:32

1: S

2: S

3: Đ

4: S

5: Đ

6: Đ

Dark_Hole
19 tháng 2 2022 lúc 21:32

TT

Nội dung

Đúng

Sai

1

Nếu hai tam giác có ba góc bằng nhau từng đôi một thì hai tam giác đó bằng nhau.

 

x

2

Nếu ABC và DEF có AB = DE, BC =  EF,  thì ABC = DEF

x

 

3

Trong một tam giác, có ít nhất là hai góc nhọn.

x

 

4

Nếu góc A là góc ở đáy của một tam giác cân thì  > 900.

 

x

5

Nếu hai tam giác có ba cạnh tương ứng bằng nhau thì hai tam giác giác đó bằng nhau

x

 

6

Nếu một tam giác vuông có một góc nhọn bằng 450 thì tam giác đó là tam giác vuông cân Đúng

 

Chúc em học giỏi

phốt đuỹ bẹn tên Công Mi...
19 tháng 2 2022 lúc 21:33

tham khảoundefined

phươngtrinh
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Nguyễn Lê Phước Thịnh
6 tháng 10 2021 lúc 15:30

b: Xét ΔAKB vuông tại K và ΔAKC vuông tại K có 

AB=AC
AK chung

Do đó: ΔAKB=ΔAKC

Suy ra: KB=KC

Xét ΔMBK vuông tại M và ΔNCK vuông tại N có 

KB=KC

\(\widehat{B}=\widehat{C}\)

Do đó: ΔMBK=ΔNCK

Suy ra: KM=KN(1)

Xét ΔAKB vuông tại K có KM là đường cao ứng với cạnh huyền AB

nên \(AM\cdot MB=KM^2\left(2\right)\)

Xét ΔAKC vuông tại K có KN là đường cao ứng với cạnh huyền AC

nên \(AN\cdot NC=KN^2\left(3\right)\)

Từ (1), (2) và (3) suy ra \(AM\cdot MB=AN\cdot NC\)

lilith.
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Nguyễn Lê Phước Thịnh
8 tháng 12 2023 lúc 20:48

a: Xét ΔADB và ΔADE có

AD chung

\(\widehat{BAD}=\widehat{EAD}\)

AB=AE

Do đó: ΔADB=ΔADE

b: Ta có: ΔADB=ΔADE

=>\(\widehat{ABD}=\widehat{AED}\)

=>\(\widehat{ABC}=\widehat{AEF}\)

Xét ΔEAF và ΔBAC có

\(\widehat{AEF}=\widehat{ABC}\)

AE=AB

\(\widehat{EAF}\) chung

Do đó: ΔEAF=ΔBAC

=>AF=AC

c: Ta có: AB+BF=AF

AE+EC=AC

mà AB=AE và AF=AC

nên BF=EC

Ta có: \(\widehat{ABD}+\widehat{FBD}=180^0\)(hai góc kề bù)

\(\widehat{AED}+\widehat{CED}=180^0\)(hai góc kề bù)

mà \(\widehat{ABD}=\widehat{AED}\)

nên \(\widehat{FBD}=\widehat{CED}\)

Ta có: ΔABD=ΔAED

=>DB=DE

Xét ΔDBF và ΔDEC có

DB=DE

\(\widehat{DBF}=\widehat{DEC}\)

BF=EC

Do đó: ΔDBF=ΔDEC

tuấn trần
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Nguyễn Lê Phước Thịnh
20 tháng 1 2022 lúc 8:29

\(\cos A=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{8^2+10^2-13^2}{2\cdot8\cdot10}=-\dfrac{1}{32}< 0\)

nên \(\widehat{A}>90^0\)

=>ΔABC tù

Hồ Thị Khánh Hòa
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Mina
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Nguyễn Lê Phước Thịnh
30 tháng 10 2021 lúc 22:53

a: Xét ΔABC có 

M là trung điểm của AB

N là trung điểm của AC

Do đó: MN là đường trung bình của ΔABC

Suy ra: MN//BC

hay BMNC là hình thang

8A2 Dương Duy Khang
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Ami Mizuno
18 tháng 10 2021 lúc 19:27

Bạn tự vẽ hình giúp mình nhé!

Xét tam giác AHC vuông tại H có:

HM là đường trung tuyến ứng với cạnh huyền AC

\(\Rightarrow HM=AM=MC=MN\)

\(\Rightarrow HN=AC\) (1)

Xét tam giác HMC và tam giác NMA có:

\(\left\{{}\begin{matrix}AM=MC\\\widehat{AMN}=\widehat{CMH}\left(đđ\right)\\HM=MN\end{matrix}\right.\)

\(\Rightarrow\Delta HMC=\Delta NMA\)

\(\Rightarrow\widehat{MHC}=\widehat{MNA}\)

Mà hai góc trên nằm ở vị trí so le

\(\Rightarrow\)AN//HC(2)

Chứng minh tương tự ta được AH//NC(3)

Từ (1),(2),(3) suy ra, tứ giác AHCN là hình chữ nhật

 

Nguyễn Thị Hà Vy
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