Cho tam giác ABC ; điểm M thuộc cạnh BC,kẻ MN// AB; MP//AC (N thuộc AC, P thuộc AB)
a) Chứng minh \(\frac{BP}{AB}+\frac{CN}{AC}=1\)
b) Tìm vị trí của điểm M trên BC để tứ giác ANMP có diện tích lớn nhất?
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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
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cho tam giác ABC=tam giác DEF và tam giác DEF = tam giác HIK. chứng minh tam giác ABC = tam giác HIK
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: tam giác ABC=tam giác DEF (1)
và tam giác DEF = tam giác HIK (2)
Từ (1) và (2) => tam giác ABC = tam giác HIK
cho tam giác ABC=tam giác DEF và tam giác DEF = tam giác HIK. chứng minh tam giác ABC = tam giác HIK
![](https://rs.olm.vn/images/avt/0.png?1311)
Biết tam giác abc bằng tam giác DEF, tg DEF = tg HIK suy ra tam giác ABC = tam giác HIK
a
Áp dụng định lý Thales ta có:
\(\frac{BP}{AB}=\frac{BM}{BC};\frac{CN}{AC}=\frac{CM}{BC}\Rightarrow\frac{PB}{AB}+\frac{CN}{AC}=\frac{BM}{BC}+\frac{CM}{BC}=1\)
b
Gọi \(S_{BPM}=a^2;S_{CMN}=b^2;S_{ABC}=S^2\)
PM//AC nên \(\Delta\)BPM ~ \(\Delta\)BAC =>\(\frac{S_{BPM}}{S_{ABC}}=\frac{a^2}{S^2}=\frac{BM^2}{BC^2}\Rightarrow\frac{BM}{BC}=\frac{a}{S}\)
MN//AB nên \(\Delta\)CMN ~ \(\Delta\)CBA => \(\frac{S_{CMN}}{S_{ABC}}=\frac{b^2}{S^2}=\frac{CM^2}{BC^2}\Rightarrow\frac{CM}{BC}=\frac{b}{S}\)
\(\Rightarrow\frac{a}{S}+\frac{b}{S}=1\Rightarrow a+b=S\Rightarrow S^2=\left(a+b\right)^2\)
\(\Rightarrow S_{AMNP}=\left(a+b\right)^2-a^2-b^2=2ab\le\frac{\left(a+b\right)^2}{2}=\frac{S^2}{2}\) ( không đổi )
Vậy Max \(S_{AMNP}=\frac{S_{ABC}}{2}\) khi M là trung điểm của BC.
Cảm ơn nha