cho tam giác ABC cân tại A vẽ BD vuông góc với AC : CE vuông góc với AB ,chứng minh rằng
a) BD=CE; AE=AD
b) gọi O là giao điểm của BD và CE chứng minh rằng AO là tia phân giác của góc BAC
c) AO cắt BC tại H . Biết AB=15cm , AH=12 cm tính BC
1.cho tam giác ABC (AB<AC) .Vẽ BD vuông góc với AC và CE vuông góc với AB tại E . Chứng minh rằng AB - AD>BD - CE
2.cho tam giác ABC(AB>AC) , vẽ BD vuông góc với AC tại D và CE vuông góc với AB tại E . Chứng minh rằng : AB - AD > BD -CE
3.cho tam giác ABC cân tại A , trên 2 cạnh AB AB và AC lấy 2 điểm M và N sao cho AM =AN . Chứng minh rằng
a)Các hình chiếu của BM và CN trên BC bằng nhau
b) BN > (BC+MN)/2
bài 3 giải giúp mik câu b thoy
3b)
Ta có tg BNK vuông tại K ->BN>BK
Ta có IK=MN(tính chất đoạn chắn)
Ta có : BC+MN=BK+KC+MN=BK+BI+IK=2BK
Vì BK<BN->2BK<2BN->BN>BK/2->BN>BC+MN/2
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
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
Cho tam giác ABC cân tại A. Kẻ BD vuông góc với AC (D thuộc AC) và
CE vuông góc với AB (E thuộc AB).
a) Chứng minh: BD = CE.
b) Chứng minh: Tam giác AED cân.
c) Gọi I là giao điểm của BD và CE. Chứng minh: AI là phân giác của góc A và
AI vuông góc BC
Các bạn giúp mình với
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 ΔAED có AE=AD
nên ΔAED cân tại A
c: Xét ΔEBI vuông tại E và ΔDCI vuông tại D có
EB=DC
\(\widehat{EBI}=\widehat{DCI}\)
Do đó; ΔEBI=ΔDCI
Suy ra: IB=IC
Xét ΔAIB và ΔAIC có
AI chung
IB=IC
AB=AC
Do đó: ΔAIB=ΔAIC
Suy ra: \(\widehat{BAI}=\widehat{CAI}\)
hay AI là tia phân giác của góc BAC
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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Cho tam giác ABC cân tại A ( AB > BC ) . Vẽ BD vuông góc với AC tại D, CE vuông góc với AB tại E
a) Chứng minh rằng : tam giác DAB = tam giác EAC và tam giác ADE cân
b) Gọi H là giao điểm của BD và CE . Chứng minh rằng : AH là tia phân giác của góc BAC
c) Chứng minh rằng : AH > CH
Bài 1: Cho tam giác ABC cân tại A. Kẻ BD vuông góc với AC, CE vuông góc với AB và BD và CE cắt nhau tại H. Chứng minh rằng:
a) Tam giác ABD = tam giác ACE.
b) Tam giác BHC cân.
c) ED//BC
a: Xét ΔABD vuông tại D và ΔACE vuông tại E có
AB=AC
\(\widehat{A}\) chung
Do đó: ΔABD=ΔACE
b: Xét ΔBDC vuông tại D và ΔCEB vuông tại E có
BD=CE
BC chung
Do đó: ΔBDC=ΔCEB
Suy ra: \(\widehat{HBC}=\widehat{HCB}\)
hay ΔHBC cân tại H
c: Xét ΔABC có
AE/AB=AD/AC
Do đó: DE//BC
Cho tam giác ABC cân tại A, kẻ BD vuông góc với AC (D thuộc AC), CE vuông góc với AB( E thuộc AB)
a) Chứng minh BD=CE
b) Gọi I là giao điểm của BD và CE. Chứng minh tam giác IBC cân
Xét tam giácBCE= tam giác CBD (cạnh huyền -mgóc nhọn)
góc ABC = góc ACB ( cân tại A)
BC chung
==> BD=CE
b) Tam giác BCE=tam giác CBD chứng minh ở câu a nên
góc BCE = góc DBC
--> IBC cân tại I
Bài. Cho tam giác ABC cân tại A. Kẻ BD vuông góc với AC (D thuộc AC). Kẻ CE vuông góc với AB (E thuộc AB). BD và CE cắt nhau tại I. Là Là a) Cho BC = 5cm, DC = 3cm. Tính độ dài BD. b) Chứng minh rằng BD =CE. c) thẳng AI cắt BC tại H. Chứng minh rằng AI vuông góc với BC tại H.
Cho tam giác ABC cân tại A. Kẻ BD vuông góc với AC, kẻ CE vuông góc với AB. Gọi K là giao điểm của BD và CE. Chứng minh rằng AK là tia phân giác của góc A.
Xét ΔADB vuông tại D và ΔAEC vuông tại E, ta có:
AB = AC (giả thiết)
∠(BAC) chung
⇒ ΔADB = ΔAEC (cạnh huyền, góc nhọn)
⇒ AD = AE (hai cạnh tương ứng)
Xét ΔADK vuông tại D và ΔAEK vuông tại E có:
AD = AE (chứng minh trên)
AK cạnh chung
⇒ ΔADK = ΔAEK (cạnh huyền, cạnh góc vuông)
⇒ ∠(DAK) = ∠(EAK) (hai góc tương ứng)
Vậy AK là tia phân giác của góc BAC.