1, Cho tam giác ABC vuông cân tại A. Từ A kẻ đường thẳng d nằm ngoài tam giác ABC. Kẻ BD vuông góc với d tại D, CE vuông góc với d tại E. Gọi M là trung điểm của BC. Chứng minh rằng:a, BD CE=DE?b, Tam...

DN

Dang Nhan

11 tháng 2 2017

1, Cho tam giác ABC vuông cân tại A. Từ A kẻ đường thẳng d nằm ngoài tam giác ABC. Kẻ BD vuông góc với d tại D, CE vuông góc với d tại E. Gọi M là trung điểm của BC. Chứng minh rằng:a, BD+CE=DE?b, Tam giác MDE vuông cân?2, Cho đoạn thẳng AB, lấy C nằm giữa A và B. Tên cùng NMP bờ AB vẽ 2 tam giác đều ACD và BCE. Gọi M và N lần lượt là trung điểm của AE và BD. Chứng minh tam giác MNC đều.3, Cho góc xOy...

Đọc tiếp

#Toán lớp 7

0

DN

Dang Nhan

4 tháng 3 2017

Cho tam giác ABC vuông cân tại A, từ A kẻ đường thẳng d nằm ngoài tam giác ABC, kẻ BD vuông góc với d, CE cũng vuông góc với d; gọi M là TĐ của BC. CMR tam giác MDE vuông cân.

#Toán lớp 7

0

NM

neko Miru

24 tháng 1 2019

Cho tam giác ABC vuông cân tại A. Qua A kẻ đường thẳng d tuỳ ý. Từ B và C kẻ BH và CK vuông góc với d. Chứng minh rằng BH2 CK2 không phụ thuộc vào vị trí của đường thẳng d.vẽ BD vuông góc d tại D, CE vuông góc d tại E. CMR DE = BD+CE, BD²+CE²=AB²

gọi M là trung điểm của cạnh BC. CMR tam giác DME là tam giác vuông cân

#Toán lớp 7

0

PQ

Phạm Quý Hưng

17 tháng 11 2015

Cho tam giác ABC vuông tại A có AB = AC. Qua A kẻ đường thẳng d ở ngoài tam giác ABC. Vẽ BD và CE vuông góc với d. M là trung điểm BC. Chứng minh rằng:

a. BD + CE = DE?

b. Tam giác MDE vuông cân?

#Toán lớp 7

0

HT

Hồ Thu Giang

15 tháng 8 2015

Cho tam giác ABC vuông cân tại đỉnh A. Một đường thẳng d đi qua A nhưng không đi qua miền trong tam giác. Kẻ BD vuông góc với đường thẳng d tại D; kẻ CE vuông góc với đường thẳng d tại E

CMR:

- BD + CE = DE

- Gọi I là trung điểm của BC. CMR: Tam giác DIE vuông cân

#Toán lớp 7

2

TT

Trần Thị Loan

15 tháng 8 2015

dòng cuối: em sửa lại kết luận: tam giác DIE vuông nhé!

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CT

chu thi thuy van

25 tháng 8 2017

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

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HT

Hoàng Trang

20 tháng 7 2015

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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NT

Nguyễn Thảo Vy

29 tháng 3 2017

Cho tam giác ABC vuông cân ( AB=AC ). Qua A vẽ đường thẳng d ngoài tam giác ABC. Vẽ BD vuông góc với d tại D, CE vuông góc với d tại E. M là trung điểm BC. Chứng minh rằng:

a) BD+CE=DE

b) Tam giác MDE là tam giác vuông cân

#Toán lớp 7

0

PT

Pham Trong Bach

11 tháng 10 2017

Cho tam giác ABC cân tại A A ^ < 90 ° . Kẻ BD vuông góc với AC tại D, kẻ CE vuông góc với AB tại E.

a) Chứng minh tam giác ADE cân.

b) Chứng minh DE// BC.

c) Gọi I là giao điểm của BD và CE. Chứng minh IB = IC

d) Chứng minh. A I ⊥ B C .

#Toán lớp 7

1

CM

Cao Minh Tâm

11 tháng 10 2017

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A

AduduOsad

17 tháng 2 2020

Cho tam giác ABC vuông cân tại A, d là 1 đường thẳng bất kỳ qua A (d không cắt đoạn
BC).Từ B và C kẻ BD và CE cùng vuông góc với d, D và E thuộc đường thẳng d. Chứng minh rằng:
a) BD // CE;
b) Tam giác ADB = Tam giác CEA;
c) BD + CE = DE;
d) Gọi M là trung điểm của BC. CMR: Tam giác DAM = Tam giác ECM và Tam giác DME vuông cân.

#Toán lớp 7

8

MN

Minh Nguyen

18 tháng 2 2020

a) Ta có : CE ⊥ d

BD ⊥ d

\(\Rightarrow\)CE // BD (ĐPCM)

b) Xét △CEA và △ADB có :

AC = AB

\(\widehat{EAC}=\widehat{ABD}\)(cùng phụ với \(\widehat{DAB}\))

\(\Rightarrow\) △CEA = △ADB (cạnh huyền-góc nhọn)

c) Có △CEA = △ADB

\(\Rightarrow\hept{\begin{cases}BD=AE\\CE=AD\end{cases}}\)(Cặp cạnh tương ứng)

\(\Rightarrow\)BD + CE = AE + AD = DE (ĐPCM)

d) △ABC vuông tại A có AM là trung tuyến

\(\Rightarrow\)AM = BM = CM

\(\Rightarrow\)△ABM cân tại M

Có : \(\widehat{ECA}=\widehat{BAD}\)(△CEA = △ADB)

\(\widehat{ACB}=\widehat{ABC}\) (△ABC cân tại A)

\(\Rightarrow\widehat{ECA}+\widehat{ACB}=\widehat{BAD}+\widehat{ABC}\)

Mà \(\widehat{ABC}=\widehat{MAB}\)(△MAC cân tại M)

\(\Rightarrow\widehat{ECA}+\widehat{ACB}=\widehat{BAD}+\widehat{MAB}\)

\(\Rightarrow\widehat{ECM}=\widehat{MAD}\)

Xét △ADM và △CEM có :

EC = AD

\(\widehat{ECM}=\widehat{MAD}\)

AM = CM

\(\Rightarrow\)△ADM = △CEM (c-g-c) (ĐPCM)

\(\Rightarrow\)EM = MD (Cặp cạnh tương ứng) (1)

Có : \(\widehat{EMA}+\widehat{EMC}=90^o\)

\(\widehat{EMC}=\widehat{DMA}\)(△ADM = △CEM)

\(\Rightarrow\widehat{EMA}+\widehat{DMA}=90^o\)

\(\Rightarrow\widehat{EMD}=90^o\)(2)

Từ (1) và (2) suy ra △DME vuông cân tại M.

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NT

Nguyễn Thị Khánh Huyền

21 tháng 3 2020

mình không biết

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PT

Pham Trong Bach

10 tháng 10 2018

Cho tam giác ABC cân tại A ( < 90 ° ). Kẻ BD vuông góc với AC tại D, kẻ CE vuông góc vói AB tại E. a) Chứng minh tam giác ADE cân. b) Chứng minh DE / / BC c) Gọi I là giao điểm của BD và CE. Chứng minh IB = IC d) Chứng minh. AI BC

#Toán lớp 7

1

CM

Cao Minh Tâm

10 tháng 10 2018

Đúng(0)

H

Hoadayy

27 tháng 12 2022

này là chép mạng mà bro

https://thuvienhoclieu.com/cac-dang-toan-hinh-hoc-7-hoc-ky-1-co-loi-giai/

câu 9a

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S

Sett

10 tháng 3 2020

Cho tam giác ABC cân tại A, kẻ BD vuông góc với ACh(D thuộc AC). Kẻ vuông góc với AB tại E,gọi I là giao điểm của BD và CE chứng minh

A, BD=CE

B, tam giác BIC cân

C, AI là tia phân giác của góc BAC

D, DE//BC

E, gọi H là trung điểm của BC. Chứng minh A,I,H thẳng hàng

F,chứng minh AI vuông góc với BC

#Toán lớp 7

1

TT

Thám Tử THCS Nguyễn Hiếu

11 tháng 3 2020

a) Xét 2 tg vuông AEC và ADB có: AB = AC (vì tam giác ABC cân tại A)

góc A chung

Do đó tg AEC = tg ADB (ch - gn)

=> BD = CE (đpcm)

b) xét 2 tg vuông CEB và BDC có: góc CBE = góc BCD (tam giác ABC cân tại A)

CE = BD (Cmt)

do đó tg CEB = tg BDC (cgv - gnk)

=> góc ECB = góc DBC

=> tam giác BIC cân tại I (đpcm)

c) xét 2 tg AIC và AIB có: AC = AB (tam giác ABC cân tại A)

AI chung

BI = IC (tam giác BIC cân (Cmt))

DO đó tg AIC = tg AIB (c.c.c)

=> góc IAC = góc IAB => AI là tia pg của góc BAC (Đpcm)

d) Ta có: tg CEB = tg BDC (cmt) => CD = BE mà AB = AC => AE = AD => AED cân tại A

Mà AI là tia pg của góc EAD nên AI vuông với DE(1)

Ta lại có: Tam giác ABC cân tại A mà AI là tia pg của góc BAC nên AI vuông BC (2)

Từ (1) và (2) suy ra DE // BC (cùng vuông vs BC) (đpcm)

e) ko bt

F) cm vuông như câu d nha

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