Cho tam giác ABC; M, N lần lượt là trung điểm của các cạnh AB và AC. Các đoạn thẳng BN và CM cắt nhau tại G. Nối A với G kéo dài cắt BC tại P. Chứng tỏ các tam giác GMA, GMB, GNA, GNC, GPB, GPC có diện tích bằng nhau.( giúp mik với )
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![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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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
Hai tg GMA và tg GMB có chung đường cao từ G->AB nên và MA=MB nên
\(S_{GMA}=S_{GMB}\) (1)
Tương tự ta cũng có \(S_{GNA}=S_{GNB}\) (2)
Hai tg AMC và tg ABC có chung đường cao tà C->AB nên
\(\dfrac{S_{AMC}}{S_{ABC}}=\dfrac{AM}{AB}=\dfrac{1}{2}\Rightarrow S_{AMC}=\dfrac{1}{2}xS_{ABC}\)
Hai tg ANB và tg ABC có chung đường cao từ B->AC nên
\(\dfrac{S_{ANB}}{S_{ABC}}=\dfrac{AN}{AC}=\dfrac{1}{2}\Rightarrow S_{ANB}=\dfrac{1}{2}xS_{ABC}\)
\(\Rightarrow S_{AMC}=S_{ANB}\) Hai tg này có phần diện tích chung là \(S_{AMGN}\Rightarrow S_{GMB}=S_{GNC}\) (3)
Từ (1) (2) (3) \(\Rightarrow S_{GMA}=S_{GMB}=S_{GNA}=S_{GNC}\) (4)
\(\Rightarrow S_{GMA}+S_{GMB}=S_{GNA}+S_{GNC}\Rightarrow S_{ABG}=S_{ACG}\)
Hai tg ABG và tg ACG có chung AG nên
đường cao từ B->AP = đường cao từ C->AP
Hai tg GPB và tg GPC có chung GP và đường cao từ B->AP = đường cao từ C->AP nên
\(S_{GPB}=S_{GPC}\) (5)
Hai tg GPB và tg GPC có chung đường cao từ G->BC nên
\(\dfrac{S_{GPB}}{S_{GPC}}=\dfrac{BP}{CP}=1\Rightarrow BP=CP\)
Hai tg APB và tg ABC có chung đường cao từ A->BC nên
\(\dfrac{S_{ABP}}{S_{ABC}}=\dfrac{BP}{BC}=\dfrac{1}{2}\Rightarrow S_{ABP}=\dfrac{1}{2}xS_{ABC}\)
\(S_{BMC}=S_{ABC}-S_{AMC}=S_{ABC}-\dfrac{1}{2}xS_{ABC}=\dfrac{1}{2}xS_{ABC}\)
\(\Rightarrow S_{APB}=S_{BMC}\) Hai tg này có phần diện tích chung là \(S_{BMGP}\Rightarrow S_{GMA}=S_{GPC}\) (6)
Từ (4) (5) (6) \(\Rightarrow S_{GMA}=S_{GMB}=S_{GNA}=S_{GNC}=S_{GPB}=S_{GPC}\)