\(% 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 \(\abs *{MQ} = \mu \abs *{QB}\), we must have \(\abs *{MQ} = \frac {\mu }{1 + \mu } \abs *{MB}\), and hence \[ \bvect {MQ} = \frac {\mu }{1 + \mu } \bvect {MB}, \] and hence \[ \vect {q} - \vect {m} = \frac {\mu }{1 + \mu } \left (\vect {b} - \vect {m}\right ). \]
Similarly, \[ \vect {q} - \vect {n} = \frac {\nu }{1 + \nu } \left (\vect {a} - \vect {n}\right ). \]
Since \(\vect {q} = \vect {q}\), we have \[ \frac {\mu }{1 + \mu } \left (\vect {b} - \vect {m}\right ) + \vect {m} = \frac {\nu }{1 + \nu } \left (\vect {a} - \vect {n}\right ) + \vect {n}, \] which rearranges to give \[ \frac {1}{1 + \mu } \vect {m} - \frac {1}{1 + \nu } \vect {n} = \frac {\nu }{1 + \nu } \vect {a} - \frac {\mu }{1 + \mu } \vect {b}. \]
Since \(\vect {m}\) is a scalar multiple of \(\vect {a}\) as \(M\) is on the side \(OA\), and \(\vect {n}\) is a scalar multiple of \(\vect {b}\) similarly, and \(\vect {a}\) and \(\vect {b}\) are linearly independent since \(OAB\) forms a triangle, we can conclude that \[ \vect {m} = \frac {1 + \mu }{1} \cdot \frac {\nu }{1 + \nu } \vect {a}, \] and hence \[ \vect {m} = \frac {(1 + \mu ) \nu }{1 + \nu } \vect {a}. \]
Similarly, \[ \vect {n} = \frac {(1 + \nu ) \mu }{1 + \mu } \vect {b}. \]
Since \(L\) lies on \(OB\) with \(\abs *{OL} = \lambda \abs *{OB}\), then we have \[ \vect {l} = \lambda \vect {b}, \] and hence \[ \bvect {ML} = \vect {l} - \vect {m} = \lambda \vect {b} - \frac {(1 + \mu ) \nu }{1 + \nu } \vect {a}. \]
Since \[ \bvect {AN} = \vect {n} - \vect {a} = \frac {(1 + \nu ) \mu }{1 + \mu } \vect {b} - \vect {a}. \]
\(\bvect {ML}\) is parallel to \(\bvect {AN}\) means that the corresponding scalar vectors for \(\vect {a}\) and \(\vect {b}\) are in ratio (since they are linearly independent), and hence \[ \lambda : \frac {(1 + \nu ) \mu }{1 + \mu } = \left (- \frac {(1 + \mu ) \nu }{1 + \nu }\right ) : (-1), \] and hence \[ \lambda = \frac {(1 + \mu ) \nu }{1 + \nu } \cdot \frac {(1 + \nu ) \mu }{1 + \mu } = \mu \nu . \]
The condition \(\mu \nu < 1\) ensured that \(L\) lies on \(OB\) between \(O\) and \(B\) (i.e. on the side \(OB\)).