cho tam giác abc có góc a<90 độ trên nửa mp bờ ab có c vẽ ad vuông góc ab sao ab=ad,trên nửa mp bờ ac có b vẽ ae vuông góc ac sao cho ac=ae m là trung điểm của bc cmr am vuông góc de help me mai mình phải kiểm tra rùi pls
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bài 2:
ta có: AB<AC<BC(Vì 3cm<4cm<5cm)
=> góc C>góc A> góc B (Các cạnh và góc đồi diện trong tam giác)
Bài 3:
*Xét tam giác ABC, có:
góc A+góc B+góc c= 180 độ( tổng 3 góc 1 tam giác)
hay góc A+60 độ +40 độ=180độ
=> góc A= 180 độ-60 độ-40 độ.
=> góc A=80 độ
Ta có: góc A>góc B>góc C(vì 80 độ>60 độ>40 độ)
=> BC>AC>AB( Các cạnh và góc đối diện trong tam giác)
bài 2:
ta có: AB <AC <BC (Vì 3cm <4cm <5cm)
=> góc C>góc A> góc B (Các cạnh và góc đồi diện trong tam giác)
Bài 3:
*Xét tam giác ABC, có:
góc A+góc B+góc c= 180 độ( tổng 3 góc 1 tam giác)
hay góc A+60 độ +40 độ=180độ
=> góc A= 180 độ-60 độ-40 độ.
=> góc A=80 độ
Ta có: góc A>góc B>góc C(vì 80 độ>60 độ>40 độ)
=> BC>AC>AB( Các cạnh và góc đối diện trong tam giác)
HT mik làm giống bạn Dương Mạnh Quyết
* Theo mình thì phần a) Góc A = 90 độ sẽ hợp lý hơn chứ. Vậy nên mình sẽ làm theo cả hai góc A 90 độ và 80 độ nhé ( Nhưng bài của mình phần b) sẽ theo góc A = 90 độ )
a)
Góc A = 80 độ thì sẽ có thể tam giác ABC là tam giác cân, tam giác ⊥ tại B hoặc C, tam giác ABC là tam giác tù hoặc tam giác nhọn
Góc A = 90 độ thì tam giác ABC là tam giác vuông tại A
b)
Theo phần a), ta có: Tam giác ABC cân tại A
=> Góc B = góc C = ( 180 độ - 70 độ ) : 2 = 55 độ
3:
góc C=90-50=40 độ
Xét ΔABC vuông tại A có sin C=AB/BC
=>4/BC=sin40
=>\(BC\simeq6,22\left(cm\right)\)
\(AC=\sqrt{BC^2-AB^2}\simeq4,76\left(cm\right)\)
1:
góc C=90-60=30 độ
Xét ΔABC vuông tại A có
sin B=AC/BC
=>3/BC=sin60
=>\(BC=\dfrac{3}{sin60}=2\sqrt{3}\left(cm\right)\)
=>\(AB=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\left(cm\right)\)
giải:
Lưu ý: Đề thiếu dữ kiện AD = AB nhé.
tham khảo!
Lấy M là trung điểm BC ta sẽ chứng minh A, H, M thẳng hàng.
Trên tia đối của tia MA lấy điểm F sao cho MA = MF. K là giao điểm của AM và DE, ta sẽ chứng minh K trùng với H.
Ta có: △△BMF = △△CMA (c.g.c) ⇒⇒ BF = CA = AE và ˆFBM=ˆACMFBM^=ACM^
⇒ BF // AC ⇒ˆABF+ˆBAC=1800⇒ABF^+BAC^=1800 (1)
Lại có: ˆBAD=ˆCAE=900BAD^=CAE^=900
⇒ˆDAE+ˆBAC=900+ˆBAE+ˆBAC=900+900=1800⇒DAE^+BAC^=900+BAE^+BAC^=900+900=1800 (2)
Từ (1) và (2) suy ra: ˆABF=ˆDAEABF^=DAE^.
Từ giả thiết cùng với chứng minh trên ta lại có: AB = DA và BF = AE
⇒ △△ABF = △△DAE ⇒ˆBAF=ˆADE⇒BAF^=ADE^
Lại có: ˆBAF+ˆDAF=ˆBAD=900⇒ˆADE+ˆDAF=900BAF^+DAF^=BAD^=900⇒ADE^+DAF^=900
⇒ˆDKA=900⇒⇒DKA^=900⇒ AM ⊥⊥ DE. suy ra A,M, H thẳng hàng
Ta có điều phải chứng minh.