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We first show that \(\vect {b}\) lies in the plane \(XOY\). Since \(\vect {b}\) is a linear combination of \(\vect {x}\) and \(\vect {y}\), it must lie in the plane containing \(\vect {x} = \bvect {OX}\) and \(\vect {y} = \bvect {OY}\), which is the plane \(XOY\).
Let \(\alpha \) be the angle between \(\vect {b}\) and \(\vect {x}\), and let \(\beta \) be the angle between \(\vect {b}\) and \(\vect {y}\), where \(0 \leq \alpha , \beta \leq \pi \).
We have \begin {align*} \cos \alpha & = \frac {\vect {b} \cdot \vect {x}}{\abs *{\vect {b}} \abs *{\vect {x}}} \\ & = \frac {1}{\abs *{\vect {b}}} \cdot \frac {\left (\abs *{\vect {x}} \vect {y} + \abs *{\vect {y}} \vect {x}\right ) \cdot \vect {x}}{\abs *{\vect {x}}} \\ & = \frac {1}{\abs *{\vect {b}}} \cdot \frac {\abs *{\vect {x}} \cdot \left (\vect {x} \cdot \vect {y}\right ) + \abs *{\vect {y}} \cdot \abs *{\vect {x}}^2}{\abs *{\vect {x}}} \\ & = \frac {1}{\abs *{\vect {b}}} \cdot \left (\vect {x} \cdot \vect {y} + \abs *{\vect {x}} \cdot \abs *{\vect {y}}\right ). \end {align*}
Similarly, \[ \cos \beta = \frac {1}{\abs *{\vect {b}}} \cdot \left (\vect {x} \cdot \vect {y} + \abs *{\vect {x}} \cdot \abs *{\vect {y}}\right ) = \cos \alpha . \]
Since the \(\cos \) function is one-to-one on \([0, \pi ]\), we must have \(\alpha = \beta \).
Since \(\vect {x} \cdot \vect {y} = \abs *{\vect {x}} \cdot \abs *{\vect {y}} \cdot \cos \theta \) where \(\theta \) is the angle between \(\vect {x}\) and \(\vect {y}\), we have \(\vect {x} \cdot \vect {y} \geq - \abs *{\vect {x}} \abs *{\vect {y}}\), and since \(\theta \neq \pi \) (since \(OXY\) are non-collinear), we have \(\vect {x} \cdot \vect {y} > - \abs *{\vect {x}} \abs *{\vect {y}}\), and hence \(\cos \alpha = \cos \beta > 0\).
This shows that both angles are less than \(\frac {\pi }{2} = 90\degree \).
Hence, the three conditions
- \(\vect {b}\) lies in the plane \(OXY\),
- the angle between \(\vect {b}\) and \(\vect {x}\) is equal to the angle between \(\vect {b}\) and \(\vect {y}\),
- both angles are less than \(\frac {\pi }{2} = 90\degree \)
are all satisfied, and we can conclude that \(\vect {b}\) is a bisecting vector for the plane \(OXY\).
All bisecting vectors must lie on the line containing \(\vect {b}\) (the dashed line on the diagram), and hence a scalar multiple of \(\vect {b}\).
Furthermore, since both angles must be less than \(\frac {\pi }{2}\), it must not on the opposite as where \(\vect {b}\) is situated, and hence it must be a positive multiple of \(\vect {b}\).
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If \(B\) lies on \(XY\), then \(\vect {OB} = \mu \vect {x} + (1 - \mu ) \vect {y}\) must be a convex combination of \(\vect {x}\) and \(\vect {y}\), and hence \[ \lambda \left (\abs *{\vect {x}} \vect {y} + \abs *{\vect {y}} \vect {x}\right ) = \mu \vect {x} + (1 - \mu ) \vect {y}. \]
Since \(O\), \(X\) and \(Y\) are not collinear, we must have \(\vect {x}\) and \(\vect {y}\) are linearly independent, and hence \(\lambda \abs *{\vect {y}} = \mu \) and \(\lambda \abs *{\vect {x}} = 1 - \mu \), hence giving \[ \lambda = \frac {1}{\abs *{\vect {x}} + \abs *{\vect {y}}} \]
We therefore have \begin {align*} \frac {XB}{BY} & = \frac {\abs *{\bvect {OB} - \vect {x}}}{\abs *{\vect {y} - \bvect {OB}}} \\ & = \frac {\abs *{\frac {\abs *{\vect {x}}}{\abs *{\vect {x}} + \abs *{\vect {y}}} \vect {y} + \frac {\abs *{\vect {y}}}{\abs *{\vect {x}} + \abs *{\vect {y}}} \vect {y} \vect {x} - \vect {x}}}{\abs *{\frac {\abs *{\vect {x}}}{\abs *{\vect {x}} + \abs *{\vect {y}}} \vect {y} + \frac {\abs *{\vect {y}}}{\abs *{\vect {x}} + \abs *{\vect {y}}} \vect {y} \vect {x} - \vect {y}}} \\ & = \frac {\abs *{\frac {\abs *{\vect {x}}}{\abs *{\vect {x}} + \abs *{\vect {y}}} \left (\vect {y} - \vect {x}\right )}}{\abs *{\frac {\abs *{\vect {y}}}{\abs *{\vect {x}} + \abs *{\vect {y}}} \left (\vect {x} - \vect {y}\right )}} \\ & = \frac {\frac {\abs *{\vect {x}}}{\abs *{\vect {x}} + \abs *{\vect {y}}} \cdot \abs *{\vect {y} - \vect {x}}}{\frac {\abs *{\vect {y}}}{\abs *{\vect {x}} + \abs *{\vect {y}}} \cdot \abs *{\vect {x} - \vect {y}}} \\ & = \frac {\abs *{\vect {x}}}{\abs *{\vect {y}}}, \end {align*}
which means \[ XB : BY = \abs *{\vect {x}} : \abs *{\vect {y}}, \] which is precisely the angle bisector theorem.
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Considering the dot product, \begin {align*} \bvect {OB} \cdot \bvect {XY} & = \lambda \vect {b} \cdot \left (\vect {y} - \vect {x}\right ) \\ & = \lambda \left (\abs *{\vect {x}} \vect {y} + \abs *{\vect {y}} \vect {x}\right ) \cdot \left (\vect {y} - \vect {x}\right ) \\ & = \lambda \left [\abs *{\vect {x}} \cdot \vect {y} \cdot \vect {y} + \abs *{\vect {y}} \cdot \vect {x} \cdot \vect {y} - \abs *{\vect {x}} \cdot \vect {x} \cdot \vect {y} - \abs *{\vect {y}} \cdot \vect {x} \cdot \vect {x}\right ] \\ & = \lambda \left [\abs *{\vect {x}} \cdot \abs *{\vect {y}}^2 + \left [\abs *{\vect {y}} - \abs *{\vect {x}}\right ] \vect {x} \cdot \vect {y} - \abs *{\vect {y}} \cdot \abs *{\vect {x}}^2\right ] \\ & = \lambda \left (\abs *{\vect {y}} - \abs *{\vect {x}}\right ) \left (\abs *{\vect {x}} \abs *{\vect {y}} + \vect {x} \cdot \vect {y}\right ) \\ & = 0. \end {align*}
Since \(O, X, Y\) are not collinear, \(\vect {x} \cdot \vect {y} > - \abs *{\vect {x}} \abs *{\vect {y}}\), and hence \(\abs *{\vect {x}} \abs *{\vect {y}} + \vect {x} \cdot \vect {y} > 0\).
Also, \(\lambda = \frac {1}{\abs *{\vect {x}} + \abs *{\vect {y}}} \neq 0\).
So it must be the case that \(\abs *{\vect {x}} - \abs *{\vect {y}} = 0\), which means \(\abs *{\vect {x}} = \abs *{\vect {y}}\).
Hence, \(OX = OY\), and triangle \(OXY\) is isosceles.
Let \(\vect {u}\), \(\vect {v}\) and \(\vect {w}\) be the bisecting vectors for \(QOR\), \(ROP\) and \(POQ\) respectively, and let \(\vect {p} = \bvect {OP}\), \(\vect {q} = \bvect {OQ}\), \(\vect {r} = \bvect {OR}\).
Let \(i, j, k\) be some arbitrary positive real constant.
From the question, we have \[ \left \{ \begin {aligned} \vect {u} & = i \left (\abs *{\vect {q}} \vect {r} + \abs *{\vect {r}} \vect {q}\right ), \\ \vect {v} & = j \left (\abs *{\vect {r}} \vect {p} + \abs *{\vect {p}} \vect {r}\right ), \\ \vect {w} & = k \left (\abs *{\vect {p}} \vect {q} + \abs *{\vect {q}} \vect {p}\right ). \\ \end {aligned} \right . \]
Considering a pair of dot-product, we have \begin {align*} \vect {u} \cdot \vect {v} & = ij \cdot \left (\abs *{\vect {q}} \abs *{\vect {r}} \vect {r} \cdot \vect {p} + \abs *{\vect {p}} \abs *{\vect {q}} \vect {r} \cdot \vect {r} + \abs *{\vect {r}} \abs *{\vect {r}} \vect {p} \cdot \vect {q} + \abs *{\vect {r}} \abs *{\vect {p}} \vect {q} \cdot \vect {r}\right ) \\ & = ij \abs *{\vect {r}} \left (\abs *{\vect {q}} \vect {r} \cdot \vect {q} + \abs *{\vect {p}} \vect {r} \cdot \vect {q} + \abs *{\vect {p}} \abs *{\vect {q}} \abs *{\vect {r}} + \abs *{\vect {r}} \vect {p} \cdot \vect {q}\right ) \\ & = ij \abs *{\vect {r}}^2 \abs *{\vect {p}} \abs *{\vect {q}} \left (\cos \ang *{\vect {p}, \vect {r}} + \cos \ang *{\vect {r}, \vect {q}} + \cos \ang *{\vect {p}, \vect {q}} + 1\right ), \end {align*}
where \(\ang *{\vect {a}, \vect {b}}\) denotes the angle between \(\vect {a}\) and \(\vect {b}\), in \([0, \pi ]\).
Denote \[ t = \cos \ang *{\vect {p}, \vect {r}} + \cos \ang *{\vect {r}, \vect {q}} + \cos \ang *{\vect {q}, \vect {p}} + 1, \] and hence \[ \left \{ \begin {aligned} \vect {u} \cdot \vect {v} & = ij \abs *{\vect {r}}^2 \abs *{\vect {p}} \abs *{\vect {q}} t, \\ \vect {u} \cdot \vect {w} & = ik \abs *{\vect {r}} \abs *{\vect {p}} \abs *{\vect {q}}^2 t, \\ \vect {v} \cdot \vect {w} & = jk \abs *{\vect {r}} \abs *{\vect {p}}^2 \abs *{\vect {q}} t. \end {aligned} \right . \]
Since \(i, j, k > 0\), and \(\abs *{\vect {p}}, \abs *{\vect {q}}, \abs *{\vect {r}} > 0\) since none of \(P, Q, R\) are at \(O\), we must have \[ \sgn (\vect {u} \cdot \vect {v}) = \sgn (\vect {u} \cdot \vect {w}) = \sgn (\vect {v} \cdot \vect {w}) = \sgn t, \] where \(\sgn : \RR \to \{-1, 0, -1\}\) is the sign function defined as \[ \sgn x = \begin {cases} 1, & x > 0, \\ 0, & x = 0, \\ -1, & x < 0. \end {cases} \]
But the sign of a dot product also corresponds to the angle between two non-collinear non-zero vectors, since this resembles the sign of the cosine of the angle between them: \begin {align*} \sgn \vect {a} \cdot \vect {b} & = \sgn \abs *{\vect {a}} \abs *{\vect {b}} \cos \ang *{\vect {a}, \vect {b}} \\ & = \sgn \cos \ang *{\vect {a}, \vect {b}} \\ & = \begin {cases} 1, & \ang {a, b} \text { is acute}, \\ 0, & \ang {a, b} \text { is right-angle}, \\ -1, & \ang {a, b} \text { is obtuse}. \end {cases} \end {align*}
This means the angles between \(\vect {u}\) and \(\vect {v}\), \(\vect {u}\) and \(\vect {w}\), \(\vect {v}\) and \(\vect {w}\) must all be acute, obtuse, or right angles. This is exactly what is desired, and finishes our proof.