Prove that the internal bisector of an angle of a triangle divides the opposite side in the ratio of the lengths of the containing sides.
Can we draw a stetch that includes everything we know? Can we somehow apply the sine or cosine rule to the problem?
In a triangle \(ABC\), \(I\) is the incentre, \(H\) the orthocentre, and \(O\) the circumcentre. Prove that \(AI\) bisects \(\angle OAH\).
![Triangle with incentre, orthocentre and circumcentre and two angles](/thinking-about-geometry/r8917/images/sketch3.png)
Can we show these two angles are equal?
Hence, or otherwise, prove that if \(O\), \(H\) and \(I\) are collinear the triangle must be isosceles.
![a diagram showing O, H and I in a line with the vertices A and B added.](/thinking-about-geometry/r8917/images/img-8917-last.png)
Why are the two lengths labelled \(r\) equal here? Can we use our results from the first and second parts?