Bài 1: Cho tam giác ABC vuông ở A có đường cao AH. Hãy tìm hai góc cùng phụ với B Bài 2: Cho tam giác vuông ở A. Lấy D thuộc cạnh AC. Vẽ DE vuông góc với BC. Chung minh B = CDE Bài 3: Vẽ tam giác AB...

HH

huỳnh hoàng hưng

15 tháng 9 2019

Bài 1: Cho tam giác ABC vuông ở A có đường cao AH. Hãy tìm hai góc cùng phụ với B Bài 2: Cho tam giác vuông ở A. Lấy D thuộc cạnh AC. Vẽ DE vuông góc với BC. Chung minh B = CDE Bài 3: Vẽ tam giác ABC nhọn có hai đường cao AH. Ve HI vuông góc với AC ở I 1) Chứng minh AHI = C 2) Giả sử B = 75 do, BAC = 65 do. Tinh AHI Bài 4: Cho tam giác ABC nhọn có hai đường cao BD và CE. Hãy tìm hai góc cùng phụ với...

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#Toán lớp 7

1

VM

Vũ Minh Tuấn

15 tháng 9 2019

Bài 1:

Ta có: \(\Delta ABC\) vuông tại \(A\left(gt\right)\)

=> \(\widehat{B}+\widehat{C}=90^0\)

=> \(\widehat{B}\) và \(\widehat{C}\) phụ nhau.

Có: \(\Delta ABH\) vuông tại \(H\left(gt\right)\)

=> \(\widehat{B}+\widehat{BAH}=90^0\)

=> \(\widehat{B}\) và \(\widehat{BAH}\) phụ nhau.

Bài 2:

undefined

Bài 4:

Ta có: \(\Delta ABD\) vuông tại \(D\left(gt\right)\)

=> \(\widehat{A}+\widehat{ABD}=90^0\)

=> \(\widehat{A}\) và \(\widehat{ABD}\) phụ nhau.

Có: \(\Delta ACE\) vuông tại \(E\left(gt\right)\)

=> \(\widehat{A}+\widehat{ACE}=90^0\)

=> \(\widehat{A}\) và \(\widehat{ACE}\) phụ nhau.

Chúc bạn học tốt!

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HH

huỳnh hoàng hưng

15 tháng 9 2019

cảm ơn bạn nhé

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AT

Anh Tài Lê

17 tháng 4 2020 - olm

Bài 1. Cho tam giác ABC vuông tại A có góc B= 53 độa) Tính góc C.b) Trên cạnh BC, lấy một điểm D sao cho BD=BA. Tia phân giác của góc B cắt cạnh AC ở điểm E. Chứng minh tam giác BEA = tam giác BED.Bài 2. Cho tam giác ABC có AB= AC và M là trung điểm của cạnh BC.a) Chứng minh tam giác AMB = tam giác AMC.b) Qua A, vẽ đường thẳng a vuông góc với AM. Chứng minh AM vuông góc với BC và a song song với BC.c) Qua C, vẽ...

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#Toán lớp 7

1

I

IS

17 tháng 4 2020

bài 1

có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0=>\widehat{C}=180^0-\widehat{A}-\widehat{B}=180^0-90^0-53^0=37^0\)

b) xét 2 tam giác của đề bài có

góc ABE = góc DBE

BD=BA

BE chung

=> 2 tam giác = nhau

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TD

Trịnh Dung

7 tháng 6 2016 - olm

GIÚP MÌNH NHÉ MỌI NGƯỜI, BÀI NÀO BIẾT GIÚP MÌNH TRƯỚC CŨNG ĐƯỢC. CẢM ƠN RẤT NHIỀU!!! :"3Bài 1: cho tam giác ABC có góc A tù. Ở miền ngoài tam giác vẽ tam giác vuông cân BAD, CAE, ( đỉnh A). Đường cao AH cắt DE tại M. Chứng minh MD=MEBài 2: cho tam giác ABC, góc BAC = 120độ, đường phân giác trong AD. Từ D hạ DE vuông góc AB, DF vuông góc AC.a) Hãy cho nhận xét về tam giác DEFb) qua C vẽ đường thẳng song...

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#Toán lớp 7

0

NT

Nguyễn Thị Kiểm

29 tháng 9 2016 - olm

Bài 1.Cho tam giác ABC có trọng tâm G. Đường thẳng d đi qua G cắt hai cạnh AB và AC. CMR khoảng cách từ A đến d bằng tổng các khoảng cách từ B và C đến d.Bài 2. Cho tam giác ABC cân tại A và đường cao AD. Từ D dựng DE vuông góc AB và DF vuông góc AC (E thuộc AB, F thuộc AC)a) Chứng minh AD là trung trực của đoạn EF.[B]b) [/B]Trên tia đối của tia DE lấy điểm G sao cho DG=DE. Chứng minh tam giác CEG vuông.Bài...

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#Toán lớp 9

1

I

IS

22 tháng 2 2020

Ta có: ΔABC đều, D ∈ AB, DE⊥AB, E ∈ BC
=> ΔBDE có các góc với số đo lần lượt là: 300
; 600
; 900
=> BD=1/2BE
Mà BD=1/3BA => BD=1/2AD => AD=BE => AB-AD=BC-BE (Do AB=BC)
=> BD=CE.
Xét ΔBDE và ΔCEF: ^BDE=^CEF=900
; BD=CE; ^DBE=^ECF=600
=> ΔBDE=ΔCEF (g.c.g) => BE=CF => BC-BE=AC-CF => CE=AF=BD
Xét ΔBDE và ΔAFD: BE=AD; ^DBE=^FAD=600
; BD=AF => ΔBDE=ΔAFD (c.g.c)
=> ^BDE=^AFD=900
=>DF⊥AC (đpcm).
b) Ta có: ΔBDE=ΔCEF=ΔAFD (cmt) => DE=EF=FD (các cạnh tương ứng)
=> Δ DEF đều (đpcm).
c) Δ DEF đều (cmt) => DE=EF=FD. Mà DF=FM=EN=DP => DF+FN=FE+EN=DE+DP <=> DM=FN=EP
Lại có: ^DEF=^DFE=^EDF=600=> ^PDM=^MFN=^NEP=1200
(Kề bù)
=> ΔPDM=ΔMFN=ΔNEP (c.g.c) => PM=MN=NP => ΔMNP là tam giác đều.
d) Gọi AH; BI; CK lần lượt là các trung tuyến của ΔABC, chúng cắt nhau tại O.
=> O là trọng tâm ΔABC (1)
Do ΔABC đều nên AH;BI;BK cũng là phân giác trong của tam giác => ^OAF=^OBD=^OCE=300
Đồng thời là tâm đường tròn ngoại tiếp tam giác => OA=OB=OC
Xét 3 tam giác: ΔOAF; ΔOBD và ΔOCE:
AF=BD=CE
^OAF=^OBD=^OCE => ΔOAF=ΔOBD=ΔOCE (c.g.c)
OA=OB=OC
=> OF=OD=OE => O là giao 3 đường trung trực Δ DEF hay O là trọng tâm Δ DEF (2)
(Do tam giác DEF đề )
/

(Do tam giác DEF đều)
Dễ dàng c/m ^OFD=^OEF=^ODE=300
=> ^OFM=^OEN=^ODP (Kề bù)
Xét 3 tam giác: ΔODP; ΔOEN; ΔOFM:
OD=OE=OF
^ODP=^OEN=^OFM => ΔODP=ΔOEN=ΔOFM (c.g.c)
OD=OE=OF (Tự c/m)
=> OP=ON=OM (Các cạnh tương ứng) => O là giao 3 đường trung trực của ΔMNP
hay O là trọng tâm ΔMNP (3)
Từ (1); (2) và (3) => ΔABC; Δ DEF và ΔMNP có chung trọng tâm (đpcm).

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LT

Lê Thùy Ánh

28 tháng 6 2020 - olm

Bài 1 : Cho tam giác ABC có AB =6cm , AC = 8cm , BC = 10cm a) Chứng tỏ tam giác ABC vuông b) Gọi M là trung điểm BC . Kẻ MK vuông AC trên tia đối tia MH lấy K sao cho MK = MH chứng minh BK // AC c) BH cắt AG tại G là trọng tâm tam giác ABC Bài 2 : Cho tam giác ABC ở phía ngoài tam giác đó vẽ các tam giác vuông cân tại A là ACD và ACE a) Chứng minh CD = BE và CD vuông góc với BE b) Kẻ đường thẳng đi qua A vuông với BC tại...

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#Toán lớp 7

2

TT

Trí Tiên亗

28 tháng 6 2020

a) xét \(\Delta ABC\)CÓ

\(BC^2=10^2=100\)

\(AB^2+AC^2=6^2+8^2=36+64=100\)

VÌ \(100=100\)

\(\Rightarrow BC^2=AB^2+AC^2\)

VẬY \(\Delta ABC\) VUÔNG TẠI A

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NP

Nguyễn phương anh

28 tháng 6 2020

trong tam giác ABC ta có :

AB²=6²=36

AC²=8²=64

BC²=10²=100

ta thấy : 100=36+64 => BC²=AC²=AB²( định lý pytago đảo )

=> tam giác ABC vuông tại A

CHÚC BẠN HỌC TỐT !!!

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HT

Hoàng Trang

20 tháng 7 2015 - olm

1. Cho tam giác ABC vuông ở A có AB<AC. AH vuông góc với BC tại H, D là điểm trên cạnh BC sao cho AD=AB. Vẽ DE vuông góc với BC tại E. Chứng mih rằng AH=HE.

2. Cho tam giác ABC vuông cân tại A.. Qua A vẽ đường thẳng d ở ngoài tam giác ABC . Vẽ BD vuông góc với d taị D. CE vuông góc với d tại E. M là trung điểm CB. Chứng minh rằng:

a) BD + CE = DE

b) Tam giác MDE là tam giác vuông cân

#Toán lớp 7

12

CL

Cao Linh Chi

13 tháng 2 2016

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CV

CUTE vô đối

7 tháng 3 2017

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

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IL

I love dễ thương

3 tháng 12 2015 - olm

bài 1/ cho tam giác ABC có góc B=C.Tia phân giác của góc B cắt AC ở D.Tia phân giác của góc C cắt AB ở E.So sánh BD và CE.bài 2/Cho tam giác ABC vuông tại A có AB=AC. Lấy D thuộc cạnh AB, điểm E thuộc cạnh AC sao cho AD=AE. Đường thẳng đi qua D vuông góc với DE cắt CA ở K. Chứng minh: AK=AC.bài 3/cho tam giác ABC có góc A=90 độ;AB=AC.Lấy điểm D thuộc AB,điểm E thuộc AC sao cho AD=AE.Đường thẳng qua D và vuông...

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#Toán lớp 7

0