In obtuse-angled triangles, the square on the side subtending the obtuse angle is greater than the (sum of the) squares on the sides containing the obtuse angle by twice the (rectangle) contained by one of the sides around the obtuse angle, to which a perpendicular (straight line) falls, and the (straight line) cut off outside (the triangle) by the perpendicular (straight line) towards the obtuse angle. * Let $ABC$ be an obtuse-angled triangle, having the angle $BAC$ obtuse. * And let $BD$ be drawn from point $B$, perpendicular to $CA$ produced [Prop. 1.12]. * I say that the square on $BC$ is greater than the (sum of the) squares on $BA$ and $AC$, by twice the rectangle contained by $CA$ and $AD$.
Algebraically, this proposition states that in an obtuse triangle \[BC^2 = AB^2 + AC^2 + 2\,AD\,AC.\]
Note that this is a special case for the law of cosines: \[BC^2 = AB^2 + AC^2 -2\,AB\,AC\,\cos {BAC},\] when $BAC$ is an obtuse angle, since \[\cos {BAC} = - AD / AB.\]
Proofs: 1
Sections: 1