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hhaidz
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Nguyễn Lê Phước Thịnh
27 tháng 7 2023 lúc 10:48

a: Xét ΔAHB vuông tại H và ΔAHC vuông tại H có

AB=AC

AH chung

=>ΔAHB=ΔAHC

=>HB=HC

b: Xét ΔHDB vuông tại D và ΔHEC vuông tại E có

HB=HC

góc B=góc C

=>ΔHDB=ΔHEC

=>BD=CE

vu phuong linh
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%Hz@
8 tháng 3 2020 lúc 9:37

A B C H D E

 TA CÓ \(\Delta ABC\)CÂN TẠI A

\(\Rightarrow\hept{\begin{cases}AB=AC\\\widehat{B}=\widehat{C}\end{cases}}\)

A) VÌ AH VUÔNG GÓC VỚI BC

=> AH LÀ ĐƯỜNG CAO

MÀ TRONG TAM GIÁC CÂN ĐƯỜNG CAO CŨNG CHÍNH LÀ ĐƯỜNG TRUNG TUYẾN

=> AH LÀ TRUNG TUYẾN CỦA BC

=> BH=CH(ĐPCM)

B) XÉT TAM GIÁC NHA

Khách vãng lai đã xóa
Lê Thị Nhung
8 tháng 3 2020 lúc 9:41

A B H C D E

Vì tam giác ABC cân tại A suy ra AB=AC, góc B=góc C

Xét tam giác ABH và tam giác ACH

có AB=AC(CMT)

góc AHC=góc AHB (=900)

góc B=góc C

suy ra tam giác ABH = tam giác ACH (cạnh huyền-góc nhọn)

suy ra BH=CH (hai cạnh tương ứng)

b) Xét tam giac BHD và tam giác CHE

có BH=CH (CMT)

góc B=góc C

góc HDB = góc HEC = 900

suy ra tam giac BHD = tam giác CHE (cạnh huyền-góc nhọn)

suy ra BD=CE (hai cạnh tương ứng)

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Linh Phạm
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Hoàng Trang
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Cao Linh Chi
13 tháng 2 2016 lúc 11:30

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CUTE vô đối
7 tháng 3 2017 lúc 20:37

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

cô nàng bạch dương
18 tháng 3 2017 lúc 12:04

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Phan Thi Hong Nhung Phan...
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Tiểu Thiên Bình
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Nguyễn Ngọc Minh Hương
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Lê Phương Thảo
23 tháng 3 2016 lúc 18:49

1.

Ta có : AC<AD (vì : D là tia đối của tia BC )

=> HD<HC

3. 

Ta có : AB+AC>AH (vì : tog 2 cah cua tam giác luôn lớn hơn cah con lại)

Mà : 1/2AH<AB+AC

=> AB+AC>2AH

4.

Ta có : ko hiu

Nguyễn Ngọc Minh Hương
23 tháng 3 2016 lúc 18:52

bạn giải bài 3 mik hk hiu, bn viết rõ rak dc hk

Hồ Thị Bích Ngọc
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Hồ Thị Bích Ngọc
8 tháng 4 2018 lúc 20:23

help me

Cô Hoàng Huyền
9 tháng 4 2018 lúc 10:34

a) Xét tam giác vuông ADB và tam giác vuông ACE có:

Góc A chung

AB = AC (gt)

\(\Rightarrow\Delta ABD=\Delta ACE\)   (Cạnh huyền - góc nhọn)

b) Do \(\Delta ABD=\Delta ACE\Rightarrow AD=AE\)

Xét tam giác vuông AEH và tam giác vuông ADH có:

Cạnh AH chung

AE = AD (cmt)

\(\Rightarrow\Delta AEH=\Delta ADH\)   (Cạnh huyền - cạnh góc vuông)

\(\Rightarrow HE=HD\)

c) Xét tam giác ABC có BD, CE là đường cao nên chúng đồng quy tại trực tâm. Vậy H là trực tâm giác giác.

Lại có AM cũng là đường cao nên AM đi qua H.

d) Xét các tam giác vuông EBC và EAC, áp dụng định lý Pi-ta-go ta có:

\(BC^2=EB^2+EA^2;AC^2=EA^2+EC^2\)   

Tam giác ABC cân tại A nên AB = AC hay \(AB^2=AC^2\)

Vậy nên \(AB^2+AC^2+BC^2=2AC^2+BC^2=2\left(EA^2+EC^2\right)+EB^2+EC^2\)

\(=3EC^2+2EA^2+BC^2\).

Cô Hoàng Huyền
9 tháng 4 2018 lúc 10:38

Hình vẽ:

Kon Kon
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