Cho tam giác ABC cân tại A vẽ BD vuông goác vơia AC tại D, CE vuông góc với AB tại E . Gọi H là giao điểm của BD và CE . Cmr
a,AH vuông góc BC
b, AD =CE , BD = AE
c, MB mũ 2 + MC mũ 2 = 2 MA mũ 2
b, góc
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cho tam giác abc vuông cân tại a. h là trung điểm cạnh bc. m là trung điểm cạnh bc. m là điểm nằm giữa b và h. vẽ md vuông góc ab tại d, me vuông góc với ac tại e. Cm:
a) ah vuông góc với bc
b) ad= ce, bd= ae
c) mb mũ 2 + mc mũ 2= 2ma mũ 2
18 tháng 2 2022 lúc 17:33
Cho tam giác ABC vuông cân tại A, H là trung điểm của BC, M là điểm nằm giữa B và H. Vẽ MB vuông góc AB tại D, ME vuông góc AC tại E.Chứng minh:
a) AH vuông góc với BC
b) AD=CE; BD=AE
c)MB^2+MC^2=2MA^2
Mn giúp mik vs, lát 7h mik phải nộp bài rồi ạ
cho tam giác ABC cân tại A , vẽ BD vuông góc vs AC tại D, CE vuông góc vs AB tại E. gọi H là giao điểm của BD và CE. CM
a)BD=CE
b)AH vuông góc vs BC
c)góc EAH= góc DAH
a) Xét 2 tam giác vuông \(\Delta EBC\)và \(\Delta DCB\)có:
\(BC:\)cạnh chung
\(\widehat{EBC}=\widehat{DCB}\)
suy ra: \(\Delta EBC=\Delta DCB\) (ch_gn)
\(\Rightarrow\)\(BD=EC\) (cạnh tương ứng)
b) \(\Delta ABC\)có các đường cao \(BD,EC\)cắt nhau tại \(H\)
\(\Rightarrow\)\(H\)là trực tâm của \(\Delta ABC\)
\(\Rightarrow\)\(AH\)là đường cao của \(\Delta ABC\)
\(\Rightarrow\)\(AH\perp BC\)
c) \(\Delta ABC\)cân tại A có AH là đường cao
nên AH đồng thời là đường phân giác
\(\Rightarrow\)\(\widehat{EAH}=\widehat{DAH}\) (đpcm)
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Cho tam giác ABC cân tại A. Vẽ BD vuông với AC tại D, CE vuông với AB tại E . Gọi H là giao điểm của BD và CE. Chứng minh rằng: a) BD = CE b) IH vuông góc BC .giúp mik với ạ 😩🥺❤️❤️
Xét tam giác vuông AEC và tam giác vuông ADB,có:
Góc A: chung
AB=AC ( ABC cân )
Vậy tam giác vuông AEC và tam giác vuông ADB ( ch.gn )
=> BD=CE ( 2 cạnh tương ứng )
b. bạn xem lại đề nhé
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Bài 4. Cho tam giác ABC cân tại A (Â 90o). Vẽ BD vuông góc với AC tại D, CE vuông góc với AB tại E. Gọi H là giao điểm của BD và CE.a)Chứng minh tam giác ABD tâm giác ACE để suy ra CE BDb)Chứng minh AH là phân giác của góc BAC.c)Chứng minh DE // BCd)Trên tia CE lấy điểm M sao cho E là trung điểm của HM. Trên tia BD lấy điểm N sao cho D là trung điểm của HN. Chứng minh AM AH và tam giác AMN cân.e)Tam giác ABC cho trước phải có điều kiện gì để tam giác AMN là tam giác đều.
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Bài 4. Cho tam giác ABC cân tại A (Â < 90o). Vẽ BD vuông góc với AC tại D, CE vuông góc với AB tại E. Gọi H là giao điểm của BD và CE.
a)Chứng minh tam giác ABD = tâm giác ACE để suy ra CE = BD
b)Chứng minh AH là phân giác của góc BAC.
c)Chứng minh DE // BC
d)Trên tia CE lấy điểm M sao cho E là trung điểm của HM. Trên tia BD lấy điểm N sao cho D là trung điểm của HN. Chứng minh AM = AH và tam giác AMN cân.
e)Tam giác ABC cho trước phải có điều kiện gì để tam giác AMN là tam giác đều.
a: Xét ΔABD vuông tại D và ΔACE vuông tại E có
AB=AC
\(\widehat{BAD}\) chung
Do đó: ΔABD=ΔACE
Suy ra; BD=CE
b: Xét ΔAEH vuông tại E và ΔADH vuông tại D có
AH chung
AE=AD
Do đó: ΔAEH=ΔADH
Suy ra: \(\widehat{EAH}=\widehat{DAH}\)
hay AH là tia phân giác của góc BAC
c: Xét ΔABC cso AE/AB=AD/AC
nên DE//BC
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cho tam giác abc cân tại a( ab>ac) kẻ bd vuông góc với ac, d thuộc ac. ce vuông góc với ab, e thuộc ab. gọi h là giao điểm của bd và ce.
c) ah>hc
3 tháng 3 2022 lúc 10:16
Cho tam giác ABC cân tại A (góc A < 90 độ ). Vẽ BD vuông góc AC tại D ; CE vuông góc AB tại E . Gọi I là giao điểm của BD và CE . Chứng minh: a) tam giác BEC= tam giác CDB .
b) AD =AE .
c) AI là tia phân giác của góc BAC .
d) DE / /BC .
e) Gọi M là trung điểm của cạnh BC . Chứng minh ba điểm A ,I ,M thẳng hàng.
a: Xét ΔBEC vuông tại E và ΔCDB vuông tại D có
BC chung
\(\widehat{EBC}=\widehat{DCB}\)
Do đó:ΔBEC=ΔCDB
b: Xét ΔABD vuông tại D và ΔACE vuông tại E có
AB=AC
\(\widehat{BAD}\) chung
Do đó:ΔABD=ΔACE
Suy ra: AD=AE
c: Ta có: ΔBEC=ΔCDB
nên \(\widehat{IBC}=\widehat{ICB}\)
hayΔIBC cân tại I
Xét ΔABI và ΔACI có
AB=AC
AI chung
BI=CI
Do đó:ΔABI=ΔACI
Suy ra: \(\widehat{BAI}=\widehat{CAI}\)
hay AI là tia phân giác của góc BAC
d: Xét ΔABC có AE/AB=AD/AC
nên DE//BC
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Cho tam giác ABC cân tại A (góc A < 90 độ) . Vẽ BD vuông góc AC tại D ; CE vuông góc AB tại E,
a) Chứng minh: tam giác ADB= tam giác AEC.
b) Gọi H là giao điểm của BD & CE. Chứng minh HE= HD
c) Vẽ AM vuông góc với BC tại M. Chứng minh AM đi qua điểm H.
d) Chứng minh AB2 +AC2+BC2= 3EC2+2EA2+EB2.
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 \(AB^2+AC^2+BC^2=2AC^2+BC^2=2\left(EA^2+EC^2\right)+EB^2+EC^2\)
\(=3EC^2+2EA^2+BC^2\).
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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
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