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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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Nguyễn Thảo Vy
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Phạm Quý Hưng
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Thành Phan
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Thành Phan
30 tháng 10 2017 lúc 13:37
ΔΔ ADB vuông tại D nên: DBAˆ+DABˆ=900DBA^+DAB^=900 Lại có: EACˆ+DABˆ=1800−BACˆ=1800−900=900EAC^+DAB^=1800−BAC^=1800−900=900 ⇒⇒ DBAˆ=EACˆDBA^=EAC^ (1) ΔΔ ABC cân tại A nên AB = AC Kết hợp với (1) ⇒⇒ ΔADB=ΔCEAΔADB=ΔCEA (cạnh huyền - góc nhọn) ⇒BD=AE,AD=CE⇒BD=AE,AD=CE ⇒BD+CE=AE+AD=DE⇒BD+CE=AE+AD=DE b. ΔΔ AMB và ΔΔ AMC có: AB=ACAB=AC (ΔΔ ABC cân tại A) MB=MCMB=MC (M là trung điểm của BC) AM là cạnh chung ⇒ΔAMB=ΔAMC⇒ΔAMB=ΔAMC (c.c.c) ⇒MABˆ=MACˆ=900:2=450⇒MAB^=MAC^=900:2=450 Mà ΔΔ ABC vuông cân tại A nên: ABMˆ=450⇒MABˆ=ABMˆ=450ABM^=450⇒MAB^=ABM^=450 ⇒⇒ ΔΔ AMB vuông cân tại M ⇒⇒ MA=MBMA=MB Ta lại có: DBAˆ=EACˆ⇒DBAˆ+450=EACˆ+450DBA^=EAC^⇒DBA^+450=EAC^+450 ⇒DBAˆ+MBAˆ=EACˆ+MACˆ⇒MBDˆ=MAEˆ⇒DBA^+MBA^=EAC^+MAC^⇒MBD^=MAE^ Kết hợp với MA=MBMA=MB và BD=AEBD=AE ⇒⇒ ΔBDM=ΔAEMΔBDM=ΔAEM (c.g.c) ⇒BMDˆ=AMEˆ,MD=ME⇒BMD^=AME^,MD=ME (*) Lại có: DMAˆ+BMDˆ=DMAˆ+AMEˆ=900DMA^+BMD^=DMA^+AME^=900 (**) Từ (*) và (**) ta suy ra ΔΔ DME vuông cân tại M.
Thành Phan
30 tháng 10 2017 lúc 13:39

tilado.edu.vn/student/facebook_view_question/code/747142 link đó bạn nào cần

Châu Thị Diễm Huỳnh
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Lynn ;-;
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Thanh Hoàng Thanh
27 tháng 3 2022 lúc 18:21

a) Xét \(\Delta AHB\) vuông tại H và \(\Delta AHC\) vuông tại H:

AB = AC (\(\Delta ABC\) cân tại A).

\(\widehat{B}=\widehat{C}\) (\(\Delta ABC\) cân tại A).

\(\Rightarrow\Delta AHB=\Delta AHC\) (cạnh huyền - góc nhọn).

b) Xét \(\Delta DHC:\)

DI là trung tuyến (I là trung điểm của HC).

DI là đường cao \(\left(DI\perp HC\right).\)

\(\Rightarrow\Delta DHC\) cân tại D.

lê tiến san
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Kirigawa Kazuto
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soyeon_Tiểubàng giải
7 tháng 12 2016 lúc 23:20

Hình thím tự vẽ

a) Có AD // BM (gt), DM // AB (gt) => DA = BM; DM = AB ( tính chất đoạn chắn) (1)

AE // CM (gt); AC // EM (gt) => AE = CM; AC = EM ( tính chất đoạn chắn) (2)

Từ (1) và (2) => AD + AE = BM + CM

=> DE = BC

Xét \(\Delta ABC\)\(\Delta MDE\) có:

AB = DM (cmt)

BC = DE (cmt)

AC = EM (cmt)

Do đó, \(\Delta ABC=\Delta\)MDE (c.c.c)

 

 

Kirigawa Kazuto
7 tháng 12 2016 lúc 20:45

soyeon_Tiểubàng giải , giúp tớ vs !

Em Không Có Tên
14 tháng 10 2018 lúc 11:37

https://diendan.hocmai.vn/threads/toan-7-bai-tap-hinh-hoc.317320/

Lê Ngọc Linh
Xem chi tiết
Nguyễn Tất Đạt
19 tháng 3 2017 lúc 19:59

A B C D E M d

a)  Ta có: \(\widehat{DAB}+\widehat{CAE}=180^0-\widehat{BAC}=90^0\)(1)

               \(\widehat{DAB}+\widehat{DBA}=180^0-\widehat{BDA}=90^0\)(2)

Từ (1) và (2) \(\widehat{DAB}+\widehat{CAE}=\widehat{DAB}+\widehat{DBA}\Rightarrow\widehat{CAE}=\widehat{DBA}\)

Xét\(\Delta DAB\)\(\Delta ECA\)có:\(\hept{\begin{cases}\widehat{BDA}=\widehat{AEC}=90^0\\AB=AC\\\widehat{DBA}=\widehat{CAE}\end{cases}\Rightarrow\Delta DAB=\Delta ECA}\)(cạnh huyền góc nhọn)

\(\Rightarrow\hept{\begin{cases}EC=AD\\BD=AE\end{cases}\Rightarrow BD+EC=AD+AE}=DE\)