Mấy bạn giải hẳn ra hộ mink nhé !
cho tam giac ABC vuông cân tại A . có BC =6 qua A vẽ D nằm ngoài tam giác . Từ BC vẽ BH ; CK vuông góc cới D . tính BH mũ 2 + CK mũ 2
Những câu hỏi liên quan
cho tam giác ABC cuông cân tại A .AC=6cm. Qua A vẽ đường thẳng d nằm ngoài tam giác. Từ BC vẽ BH, CK vuông góc với d. Tính BH2+CK2
Cho tam giác ABC có góc A,B,C<90,Vẽ ra phía ngoài tam giác các tam giac vuông cân tại A là tam giác ABD,ACE.Từ D,E kẻ đường thẳng vuông góc với BC cắt BC lấn lượi tại J và K.Chứng minh tam giác AJK vuông cân.
cho tam giác ABC cân tại A trên tia đối của tia BA lấy D trên tia đối của tia CA lấy điểm E sao cho BD=CE vẽ DH và EK cùng vuông gócvới BC tỉ số độ dài giữa hai đoạn thẳng BH và CK
giải hộ nhé giải hẳn ra
Cho tam giác ABC cân tại A và góc A bằng 45 độ. Vẽ BH vuông góc với AC. Vẽ điểm D sao cho H là trung điểm của CD. Từ D vẽ đường thẳng song song với AB, cắt BH ở E.
a) Chứng minh AE = BC và tam giác CDE vuông cân.
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Đây chỉ là một phần mình cắt ra từ đề bài. Các bạn chỉ cần giải AE = BC là được rồi. Cảm ơn!
cho tam giác ABC có góc A nhọn. Về phía ngoài của tam giac ABC vẽ tam giac BAD cân tại A, tam giác CAE vuông cân tại A
a, DC=BE; DC vuông góc với BE
b, BD^2+CE^2=BC^2+DE^2
c, đường thẳng đi qua A vuông góc với DE và cắt BC tại K. Chứng Minh K là trung điểm của BC
cho tam giác ABC vẽ ra phía ngoài tam giác ABC các tam giác vuông cân tại A là ABE và ACF. Chứng minh
a) BF=CE
b) BF vuông góc CE
vẽ hình ra nhé mình sẽ tặng like nếu giải hẳn ra
cho tam giac ABC cân tại A. gọi M là trung điểm của cạnh BC
a,chứng minh: tam giác ABM = tam giác ACM
b, Từ M vẽ MH vuông Góc AB và MK vuông góc AC. chứng minh BH = CK
c,từ B vẽ BP vuông góc AC, BP cắt MH Tại I chứng minh tam giác IBM cân
nhanh lên nhé thứ 6 mình thi học kì rồiiiiiiiiii
Hình tự vẽ
C/m: a, Xét \(\Delta ABM\)và \(\Delta ACM\) có:
AB = AC (do tam giác ABC cân tại A)
BM = CM ( do M là trung điểm của BC)
AM chung
=> \(\Delta ABM=\Delta ACM\)(c.c.c)
b, Xét tam giác BHM vuông tại H và CKM vuông tại K có:
BM = MC (do M là trung điểm của BC)
\(\widehat{ABC}=\widehat{ACB}\)(do tam giác ABC cân tại A)
=> \(\Delta BHM=\Delta CKM\)(cạnh huyền - góc nhọn)
=> BH = CK (2 cạnh tương ứng)
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a) Vì tam giác ABC là tam giác cân có :
AM là đường trung tuyến
nên AM vừa là đường cao vừa là đường phân giác
=> Góc BAM = góc MAC
Xét ΔAMB và ΔMACΔ có
góc BAM = góc CAM ( cmt)
AM chung
AMB = góc AMC ( cùng bằng 90 độ )
Vậy Tam giác ABM = tam giác AMC ( c-g-v-g-n-k)
b) Xét tam giác AHM và tam giác AKM có
AM chung
Góc AHM =AKM ( = 90 độ)
HAM =MAK ( cmt câu a)
nên Tam giác AHM = tam giác AKM (c-h-g-n)
=> HM = MK
và BHM = MKC , góc B= C
Nên tam giác BHM = KMC
=> HB = KC
c) Ta có BP VUÔNG GÓC VỚI AC
và MK vuông góc với AC
Nên BP// MK
=> góc PBM = KMC
Mà KMC = HMB ( vÌ tam giác BHM = KMC )
Suy ra : PBM = góc HMB
Hay tam giác IBM cân tại I
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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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Xem thêm câu trả lời
cho tam giác abc cân tại a gọi d là trung điểm của bc qua a vẽ đường thẳng d song song với bcchứng minha tam giác ABCtam giác ACDb AD là tia phân giác của góc BAC c AD vuông tại dbài 2cho tam giác ABC vuông cân tại A M là trung điểm canh BC điểm e giữa M và C vẽ BH vuông với AE tại H Ck vuông với AE tại K chứng minh rằng a BHAK b tam giác HBM tam giác KAM c tam giác MHK vuông cân giúp mình với các bn vẽ hình và giải giúp mình với chiều nay cô giáo kiểm tra rồi
Đọc tiếp
cho tam giác abc cân tại a gọi d là trung điểm của bc qua a vẽ đường thẳng d song song với bc
chứng minh
a tam giác ABC=tam giác ACD
b AD là tia phân giác của góc BAC
c AD vuông tại d
bài 2cho tam giác ABC vuông cân tại A M là trung điểm canh BC điểm e giữa M và C vẽ BH vuông với AE tại H Ck vuông với AE tại K chứng minh rằng
a BH=AK
b tam giác HBM = tam giác KAM
c tam giác MHK vuông cân
'giúp mình với các bn vẽ hình và giải giúp mình với chiều nay cô giáo kiểm tra rồi
Câu 1 (Bạn tự vẽ hình giùm)
a) Mình xin chỉnh lại đề một chút: \(\Delta ABD=\Delta ACD\)
\(\Delta ABD\)và \(\Delta ACD\)có: AB = AC (\(\Delta ABC\)cân tại A)
BD = DC (D là trung điểm của BC)
Cạnh AD chung
=> \(\Delta ABD=\Delta ACD\) (c. c. c) (đpcm)
b) Ta có \(\Delta ABD=\Delta ACD\)(cm câu a) => \(\widehat{BAD}=\widehat{DAC}\)(hai góc tương ứng) => AD là tia phân giác của \(\widehat{BAC}\)(đpcm)
c) Mình xin chỉnh lại đề một chút: AD \(\perp\)BC tại D
Ta có \(\Delta ABD=\Delta ACD\)(cm câu a) => \(\widehat{BDA}=\widehat{CDA}\)(hai góc tương ứng)
Mà \(\widehat{BDA}+\widehat{CDA}\)= 180o (kề bù)
=> \(\widehat{BDA}=\widehat{CDA}=\frac{180^o}{2}\)= 90o => AD \(\perp\)BC tại D (đpcm)
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