Cho tam giác ABC = tam giác A'B'C'. Trên cạnh BC lấy điểm M, trên cạnh B'C' lấy điểm M'sao cho BM = \(\frac{1}{3}\)BC, C'M' = \(\frac{2}{3}\)B'C'. Chứng minh rằng AM = A'M'.
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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 \(\widehat{A}< \widehat{C}< \widehat{B}\)
a: Xét ΔHBA vuông tại H và ΔABC vuông tại A có
góc B chung
Do đó: ΔHBA∼ΔABC
b: Xét ΔHCA vuôg tại H và ΔACB vuông tại A có
góc C chung
Do đó: ΔHCA∼ΔACB
* 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.
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a: Xét ΔABC vuông tại A và ΔHBA vuông tại H có
góc B chung
Do đó: ΔABC\(\sim\)ΔHBA
b: Xét ΔBAH vuông tại H và ΔACH vuông tại H có
\(\widehat{BAH}=\widehat{ACH}\)
DO đó: ΔBAH\(\sim\)ΔACH
ta có BM=\(\frac{1}{3}\)BC
\(\Rightarrow\)MC=\(\frac{2}{3}\)BC
mà BC=B'C'\(\Rightarrow\)MC=M'C'
Xét 2 tam giác ACM và tam giác A'C'M'
có AC=A'C'(tam giác ABC=tam giác A'B'C')
MC=M'C'
\(\widehat{C}\)=\(\widehat{C'}\)(tam giác ABC=tam giác A'B'C')
\(\Rightarrow\)Tam giác ACM =tam giác A'C'M' (cạnh . góc . cạnh)
\(\Rightarrow\)AM=A'M'(cặp cạnh tương ứng)