In any triangle, (if) one of the sides (is) produced (then) the external angle is equal to the (sum of the) two internal and opposite (angles), and the (sum of the) three internal angles of the triangle is equal to two right angles. * Let $ABC$ be a triangle, and let one of its sides $BC$ have been produced to $D$. * I say that the external angle $ACD$ is equal to the (sum of the) two internal and opposite angles $CAB$ and $ABC$, and the (sum of the) three internal angles of the triangle - $ABC$, $BCA$, and $CAB$ - is equal to two right angles.
In any triangle \(\triangle{ABC}\) in a plane, the sum of its angles equals $\angle{ACB}+\angle{BAC}+\angle{CBA}=180^\circ.$ Moreover, if one of the sides is extended (without loss of generality extend segment \(BC\) to segment \(BD\)), then the exterior angle equals the sum of the its interior and opposite angles: $\angle{DCA}=\angle{BAC}+\angle{CBA}.$
Proofs: 1 Corollaries: 1 2 3 4 5
Explanations: 1
Proofs: 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
Sections: 33