chứng minh tổng 3 góc ngoài của một tam giác bằng 4 vuông
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Giả sử tam giác ABC có các góc ngoài là : A1 ; B1 ;C1
Ta có
A1+A =180
B1+B =180
C1+C =180
=> (A1+B1+C1) + (A+B+C) =3.180
Mà A+B+C = 180
=> A1+B1+C1 = 3.180 -180 =2.180 =4.90
goi 3 goc trong tam giac do la A;B;C
goc ngoai la:A1;B1;C1
ta có:A1+A=180 do;B1+B=180 do;C1+C=180 do
=>....
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: ∠(A1 ) +∠(A2 ) =180o(hai góc kề bù)
∠(B1 ) +∠(B2 ) =180o(hai góc kề bù)
∠(C1 ) +∠(C2 )=180o(hai góc kề bù)
Suy ra: ∠(A1 ) +∠(A2 ) +∠(B1) +∠(B2 ) +∠(C1 ) +∠(C2 ) = 180º + 180º + 180º =540o
⇒∠(A2 ) + ∠( B2 ) +∠(C2 ) =540o-(∠(A1 ) +∠(B1 ) +∠(C1 ) ) (1)
Trong ΔABC, ta có:
∠(A1 ) +∠(B1 ) +∠(C1 ) =180o (tổng ba góc trong tam giác) (2)
Từ (1) và (2) suy ra: ∠(A2 ) +∠(B2 ) +∠(C2 ) =540o-180o=360o
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi 3 góc ngoài ở 3 đỉnh của 1 tam giác lần lượt là A1;B1;C1 còn A2;B2;C2 là góc trong của tam giác.
Ta có:
A1 + A2 = 180o
B1 + B2 = 180o
C1 + C2 = 180o
=> A1+B1+C1+A2+B2+C2 = 360o
Mà A2 + B2 + C2 = 180o (tổng 3 góc trong của tam giác)
=> A1+B1+C1 = 360o-180o=180o.2 = 360o
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A+A_1+B+B_1+C+C_1=3.180\)
Mà A+B+C=180=> \(A_1+B_1+C_1=360\)
Góc ngoài của 1 tam giác bằng tổng 2 gốc trong không kề với nó
=>tổng 3 góc ngoài bằng 2 lần tổng 3 góc trong bằng 3600 bằng 4 lần 900 bằng 4 góc vuông
cho tam giác ABC
góc A có góc ngoài là \(A_1\)
góc B có góc ngoài là \(B_1\)
góc C có góc ngoài là \(C_1\)
góc \(A_1\) = góc B + góc C
Góc \(B_1\) = góc A + góc C
góc \(C_1\) = góc B + góc A
góc \(A_1\) + góc \(B_1\) + góc \(C_1\) = góc B+ góc C+góc A + góc C + góc B + góc A
= 2 ( góc B + góc A + góc C)
= 2 x \(180^o\) ( vì tổng 3 góc của tam giác bằng 180 độ)
= 360 độ
tổng 3 góc ngoài của 1 tam giác bằng số góc vuông là : 360 độ : 90 độ = 4
vậy tổng 3 góc ngoài của 1 tam giác = 4 góc vuông