\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Since \(\theta \) is the angle between \(\vect {a}\) and \(\vect {b}\), we have \[ \cos \theta = \frac {\vect {a} \cdot \vect {b}}{\abs *{\vect {a}} \abs *{\vect {b}}} = \vect {a} \cdot \vect {b}. \]
Let \(\lambda \) be the angle between \(\vect {m}\) and \(\vect {a}\). Hence, \begin {align*} \cos \lambda & = \frac {\vect {a} \cdot \vect {m}}{\abs *{\vect {a}} \abs *{\vect {m}}} \\ & = \frac {\vect {a} \cdot \frac {1}{2} \left (\vect {a} + \vect {b}\right )}{\abs *{\vect {m}}} \\ & = \frac {\vect {a} \cdot \left (\vect {a} + \vect {b}\right )}{\abs *{\vect {a} + \vect {b}}} \\ & = \frac {1 + \vect {a} \cdot \vect {b}}{\abs *{\vect {a} + \vect {b}}} \\ & = \frac {1 + \cos \theta }{\abs *{\vect {a} + \vect {b}}}. \end {align*}
Similarly, let \(\mu \) be the angle between \(\vect {m}\) and \(\vect {b}\), and we must have \[ \cos \lambda = \cos \mu = \frac {1 + \cos \theta }{\abs *{\vect {a} + \vect {b}}}. \]
Since \(0 \leq \lambda , \mu \leq \pi \), and \(\cos \) is one-to-one when restricted to \([0, \pi ]\), we must have \(\lambda = \mu \), which shows that \(\vect {m}\) bisects the angle between \(\vect {a}\) and \(\vect {b}\).
We must have \(\cos \alpha = \vect {a} \cdot \vect {c}\), and \(\cos \beta = \vect {b} \cdot \vect {c}\).
By definition of the projection, we have \begin {align*} \vect {a}_1 & = \vect {a} - \left (\vect {a} \cdot \vect {c}\right ) \vect {c} \\ & = \vect {a} - \cos \alpha \vect {c}, \end {align*}
and hence \begin {align*} \vect {a}_1 \cdot \vect {c} & = \vect {a} \cdot \vect {c} - \cos \alpha \vect {c} \cdot \vect {c} \\ & = \cos \alpha - \cos \alpha \\ & = 0, \end {align*}
as desired.
Notice that \begin {align*} \abs *{\vect {a}_1}^2 & = \vect {a}_1 \cdot \vect {a}_1 \\ & = \left (\vect {a} - \cos \alpha \vect {c}\right ) \cdot \left (\vect {a} - \cos \alpha \vect {c}\right ) \\ & = \vect {a} \cdot \vect {a} - 2 \cos \alpha \vect {a} \cdot \vect {c} + \cos ^2 \alpha \vect {c} \cdot \vect {c} \\ & = 1 - 2 \cos ^2 \alpha + \cos ^2 \alpha \\ & = 1 - \cos ^2 \alpha \\ & = \sin ^2 \alpha . \end {align*}
Since \(\abs *{a_1} \geq 0\), and \(0 < \alpha < \frac {\pi }{2}\), \(\sin \alpha > 0\), we must have \[ \abs *{\vect {a}_1} = \abs *{\sin \alpha } = \sin \alpha . \]
The angle \(\phi \) is given by \begin {align*} \cos \phi & = \frac {\vect {a}_1 \cdot \vect {b}_1}{\abs *{\vect {a}_1} \abs *{\vect {b}_1}} \\ & = \frac {\left (\vect {a} - \cos \alpha \vect {c}\right ) \cdot \left (\vect {b} - \cos \beta \vect {c}\right )}{\sin \alpha \sin \beta } \\ & = \frac {\vect {a} \cdot \vect {b} - \cos \alpha \vect {b} \cdot \vect {c} - \cos \beta \vect {a} \cdot \vect {c} + \cos \alpha \cos \beta \vect {c} \vect {c}}{\sin \alpha \sin \beta } \\ & = \frac {\cos \theta - \cos \alpha \cos \beta - \cos \beta \cos \alpha + \cos \beta \cos \alpha }{\sin \alpha \sin \beta } \\ & = \frac {\cos \theta - \cos \alpha \cos \beta }{\sin \alpha \sin \beta }. \end {align*}
By definition of a projection, we have \begin {align*} \vect {m}_1 & = \vect {m} - (\vect {m} \cdot \vect {c}) \vect {c} \\ & = \frac {1}{2} \left (\vect {a} + \vect {b}\right ) - \left (\frac {1}{2} \left (\vect {a} + \vect {b}\right ) \cdot \vect {c}\right ) \vect {c} \\ & = \frac {1}{2} \left (\vect {a} + \vect {b}\right ) - \left (\frac {1}{2} \left (\cos \alpha + \cos \beta \right )\right ) \vect {c} \\ & = \frac {1}{2} \left (\vect {a}_1 + \vect {b}_1\right ). \end {align*}
Let \(\nu \) be the angle between \(\vect {m}_1\) and \(\vect {a}_1\), we have \begin {align*} \cos \nu & = \frac {\vect {m}_1 \cdot \vect {a}_1}{\abs *{\vect {m}_1} \abs *{\vect {a}_1}} \\ & = \frac {\frac {1}{2} \left (\vect {a}_1 + \vect {b}_1\right ) \cdot \vect {a}_1}{\frac {1}{2} \abs *{\vect {a}_1 + \vect {b}_1} \sin \alpha } \\ & = \frac {\vect {a}_1 \cdot \vect {a}_1 + \vect {b}_1 \cdot \vect {a}_1}{\abs *{\vect {a}_1 + \vect {b}_1} \sin \alpha } \\ & = \frac {\sin ^2 \alpha + \cos \phi \sin \alpha \sin \beta }{\abs *{\vect {a}_1 + \vect {b}_1} \sin \alpha } \\ & = \frac {\sin ^2 \alpha + \cos \theta - \cos \alpha \cos \beta }{\abs *{\vect {a}_1 + \vect {b}_1} \sin \alpha }. \end {align*}
Similarly, let \(\tau \) be the angle between \(\vect {m}_1\) and \(\vect {b}_1\), we have \[ \cos \tau = \frac {\sin ^2 \beta + \cos \theta - \cos \alpha \cos \beta }{\abs *{\vect {a}_1 + \vect {b}_1} \sin \beta }. \]
Since \(0 \leq \nu , \tau \leq \pi \), \(\nu = \tau \) if and only if \begin {align*} \cos \nu & = \cos \tau \\ \frac {\sin ^2 \alpha + \cos \theta - \cos \alpha \cos \beta }{\abs *{\vect {a}_1 + \vect {b}_1} \sin \alpha } & = \frac {\sin ^2 \beta + \cos \theta - \cos \alpha \cos \beta }{\abs *{\vect {a}_1 + \vect {b}_1} \sin \beta } \\ \sin \beta \left (\sin ^2 \alpha + \cos \theta - \cos \alpha \cos \beta \right ) & = \sin \alpha \left (\sin ^2 \beta + \cos \theta - \cos \alpha \cos \beta \right ) \\ \sin \alpha \sin \beta (\sin \alpha - \sin \beta ) + \cos \alpha \cos \beta (\sin \alpha - \sin \beta ) & = \cos \theta (\sin \alpha - \sin \beta ) \\ (\sin \alpha \sin \beta + \cos \alpha \cos \beta ) (\sin \alpha - \sin \beta ) & = \cos \theta (\sin \alpha - \sin \beta ) \\ \left (\cos (\alpha - \beta ) - \cos \theta \right ) (\sin \alpha - \sin \beta ) & = 0. \end {align*}
This is if and only if \(\sin \alpha = \sin \beta \), or \(\cos \theta = \cos (\alpha - \beta )\).
Since \(0 < \alpha , \beta < \frac {\pi }{2}\), and \(\sin \) is one-to-one when restricted to \(\left (0, \frac {\pi }{2}\right )\), the first condition is true if and only if \(\alpha = \beta \).
Hence, \(\vect {m}_1\) bisects the angle between \(\vect {a}_1\) and \(\vect {b}_1\) if and only if \(\alpha = \beta \) or \(\cos \theta = \cos (\alpha - \beta )\), as desired.