Cho tam giác ABC vuông ở A có AB=1;AC=3. Trên AC lấy D, E sao cho AD=DE=EC.
a) Tính BD
b) C/m: Tam giác BDE đồng dạng tam giác CDB
c) Tính góc DEB + góc DCB
Những câu hỏi liên quan
Cho tam giác ABC vuông ở A và tam giác DEF vuông ở D có AB = DE và góc ABC = góc DEF. Chứng minh tam giác ABC = tam giác DEF.
xét 2 tam giác vuông ABC và tam giác EDF, ta có:
cạnh góc vuông : AB = DE
góc nhọn : ABC = DEF
=> tam giác ABC = tam giác DEF ( cgv - gn )
Lý thuyết : Cạnh góc vuông - góc nhọn: Nếu một cạnh góc vuông và một góc nhọn kề cạnh ấy của tam giác vuông này bằng một cạnh góc vuông và một góc nhọn kề cạnh ấy của tam giác vuông kia thì hai tam giác đó bằng nhau (cgv-gn)
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xét 2 tam giác vuông ABC và tam giác EDF, ta có:
cạnh góc vuông : AB = DE
góc nhọn : ABC = DEF
=> tam giác ABC = tam giác DEF ( cgv - gn )
Lý thuyết : Cạnh góc vuông - góc nhọn: Nếu một cạnh góc vuông và một góc nhọn kề cạnh ấy của tam giác vuông này bằng một cạnh góc vuông
và một góc nhọn kề cạnh ấy của tam giác vuông kia thì hai tam giác đó bằng nhau (cgv-gn)
cho tam giác ABC vuông ở A có cạnh AB = 1/2 BC .tính các góc của tam giác ABC
Lấy điểm M là trung điểm BC => AM = BM = CM. Vậy tam giác ABM đều => góc B = 60 độ.
=> .................................................... tự xử nha :v
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cho tam giác abc vuông ở A có chu vi = 24m có cạch ab = 3/4 ac ,ab= 10m .Tính diện tích tam giác abc.
Tổng độ dài hai cạnh AB và AC là :
24 - 10 = 14 ( cm )
Độ dài cạnh AB là :
14 : ( 3 + 4 ) x 3 = 6 ( cm )
Độ dài cạnh AC là :
14 - 6 = 9 ( cm )
Diện tích hình tam giác ABC là :
6 x 9 : 2 = 27 ( cm2)
Đáp số : 27 cm2
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tổng độ dài hai cạnh là
24-10=14 cm
độ dại cạnh AB là
14:(3+4).3=6 cm
độ dài cạnh AC là
14-6=8 cm
diện tích là
6.7:2=27cm2
đáp số...............
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Thảo Mai bạn tham khảo đây nhé:
Câu hỏi của Tran Quynh Anh - Toán lớp 5 - Học toán với OnlineMath
......
Thảo Mai
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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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Cho tam giác ABC vuông ở A có A B = 10 c m , A C = 24 c m . So sánh các góc của tam giác ABC
A. A < B < C
B. A > B > C
C. B < A < C
D. C < A < B
Do tam giác ABC vuông tại A nên góc A là góc lớn nhất
Có AB < AC ⇒ C < B . Từ đó suy ra ∠C < ∠B < ∠A hay ∠A > ∠B > ∠C . Chọn B
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bài 1/ cho tam giác ABC có góc BC.Tia phân giác của góc B cắt AC ở D.Tia phân giác của góc C cắt AB ở E.So sánh BD và CE.bài 2/Cho tam giác ABC vuông tại A có ABAC. Lấy D thuộc cạnh AB, điểm E thuộc cạnh AC sao cho ADAE. Đường thẳng đi qua D vuông góc với DE cắt CA ở K. Chứng minh: AKAC.bài 3/cho tam giác ABC có góc A90 độ;ABAC.Lấy điểm D thuộc AB,điểm E thuộc AC sao cho ADAE.Đường thẳng qua D và vuông góc với BE cắt đường CA ở K.CMR:AKAC.
Đọc tiếp
bài 1/ cho tam giác ABC có góc B=C.Tia phân giác của góc B cắt AC ở D.Tia phân giác của góc C cắt AB ở E.So sánh BD và CE.
bài 2/Cho tam giác ABC vuông tại A có AB=AC. Lấy D thuộc cạnh AB, điểm E thuộc cạnh AC sao cho AD=AE. Đường thẳng đi qua D vuông góc với DE cắt CA ở K. Chứng minh: AK=AC.
bài 3/cho tam giác ABC có góc A=90 độ;AB=AC.Lấy điểm D thuộc AB,điểm E thuộc AC sao cho AD=AE.Đường thẳng qua D và vuông góc với BE cắt đường CA ở K.CMR:AK=AC.
1.cho tam giác ABC có góc A=120 độ. Ở phía ngoài tam giác ABC vẽ tam giác đều BCD. chứng minh rằng : AD= AB+AC.
2.cho hình thang vuông ABCD, AD vuông góc với DC, 2 đường chéo vuông góc với nhau. chứng minh: AD^2 = AB x DC.
1)cho tam giác ABC ,D là trung điểm của AB .đường thẳng qua D song song với BC cắt AC ở E,đường thẳng qua E song với AB cắt BC ở F.CMR:
a)AD=EF
b)tam giác ADE= tam giác EFC
c)AE=EC
2/cho tam giác ABC .vẽ về phía ngoài tam giác ABC các tam giác vuông tại A là tam giác ABD,tam giác ACE có AB=AD,AC=AE.kẻ AH vuông góc BC,DM vuông góc AH,EN vuông góc AH.CMR:
a)DM= AH
b)MN đi qua trung điểm DE
Câu hỏi của Hoàng Trang - Toán lớp 7 - Học toán với OnlineMath
1)cho tam giác ABC ,D là trung điểm của AB .đường thẳng qua D song song với BC cắt AC ở E,đường thẳng qua E song với AB cắt BC ở F.CMR:
a)AD=EF
b)tam giác ADE= tam giác EFC
c)AE=EC
2/cho tam giác ABC .vẽ về phía ngoài tam giác ABC các tam giác vuông tại A là tam giác ABD,tam giác ACE có AB=AD,AC=AE.kẻ AH vuông góc BC,DM vuông góc AH,EN vuông góc AH.CMR:
a)DM= AH
b)MN đi qua trung điểm DE
BÀI 1:A, ta có : AD=DB; DE//CB => ED là đường tbinh của tam giác ABC => AE=EC
Ta lại có: AE = EC ; EF//AB=>EF là đường trung bình của tam giác ACB
áp dụng tc đường tb trong tam giác ta có: EF//=1/2 AD hay EF=AD
B, Xét tam giác ADE và tam giác EFC CÓ:
AE = EC
AD = EF
góc A = góc E (cùng bù với góc EFD)
C,Theo phần a, ta có ED là đường tb của tam giác CAB => AE=EC
CHO MK 1 LIK E NHA
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1.CMR nếu ở miền trong tam giác ABC có điểm D sao cho AD=AB thì AB < AC
2 cho tam giác ABC vuông tại A (AB<AC) .Vẽ AH vuông góc với BC (H thuộc BC). CMR AB+AC<AH+BC
1.
Chọn điểm D như hình vẽ. Gọi E là giao điểm của AB và DC.
Ta có: \(\widehat{ADE}\)là góc ngoài của tam giác ADC => \(\widehat{ADE}>\widehat{ACD}\)(1)
Tương tự \(\widehat{BDE}>\widehat{BCD}\)(2)
(1), (2) => \(\widehat{ADB}>\widehat{ACB}\)
Mà \(\widehat{ADB}=\widehat{ABD}\)
=> \(\widehat{ABC}>\widehat{ABD}=\widehat{ADB}>\widehat{ACB}\)
=> AC>AB
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Xét tam giác ABC vuông tại A
Theo BĐT tam giác: \(AB< AC+BC\)
Và tam giác AHC vuông tại H có: \(AC< AH+CH\) (1)
\(\Rightarrow AB+AC< \left(AH+BC\right)+\left(AC+CH\right)\)
Hay \(AB+AC< \left(AH+CH+BH\right)+\left(AC+CH\right)\)
Hay \(AB+AC< AH+2CH+BH+AC\)
Bớt AC ở cả hai vế: \(AB< AH+2CH+BH\) (2)
Từ (1) và (2) suy ra \(AB+AC< 2AH+2CH+BH+CH\)
Hay \(AB+AC< 2AH+2CH+BC\)
Tới đây bí rồi.
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