Cho tam giác ABC có A' B' C' lần lượt là trđ của BC CA AB
a. Cmr: vecto BC'= Vecto C'A = vecto A'B'
b. Tìm các vecto bằng vecto B'C' , C'A'
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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
Bài 1:
a: Xét ΔABC có \(AC^2=AB^2+BC^2\)
nên ΔABC vuông tại B
b: XétΔABC có BC<AB<AC
nên \(\widehat{A}< \widehat{C}< \widehat{B}\)
* 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
a)
Nhận xét: H là một điểm nằm trong tam giác ABC.
b)
Nhận xét: H trùng với đỉnh A của tam giác ABC.
c)
Nhận xét: H nằm ngoài tam giác ABC.
Lời giải:
a)
Vì \(A,C',B\) theo thứ tự là ba điểm thẳng hàng, nên \(\overrightarrow {BC'},\overrightarrow{C'A}\) là hai vector cùng phương, cùng hướng.
Mặt khác \(BC'=C'A\) do $C'$ là trung điểm nên \(\overrightarrow{BC'}=\overrightarrow{C'A}=\frac{\overrightarrow{BA}}{2}(1)\)
Lại có, do \(\frac{B'C}{B'A}=\frac{CA'}{A'B}=1\Rightarrow A'B'\parallel AB\) và \(A'B'=\frac{1}{2}BA\)
Mà \(\overrightarrow {A'B'}\) cùng hướng với \(\overrightarrow{BA}\) nên \(\overrightarrow{A'B'}=\frac{\overrightarrow{BA}}{2}(2)\)
Từ \((1),(2)\Rightarrow \overrightarrow{BC'}=\overrightarrow{C'A}=\overrightarrow{A'B'}\)
b)
Tương tự cách của phần a, ta có:
\(\overrightarrow{B'C'}=\overrightarrow{CA'}=\overrightarrow{A'B}\)
\(\overrightarrow{C'A'}=\overrightarrow{AB'}=\overrightarrow{B'C}\)