Intersect extended lines by comparing independent coefficients

$AC=kAB$ gives $\mathbf c-\mathbf a=k(\mathbf b-\mathbf a)$, so $\mathbf c=k\mathbf b-k\mathbf a+\mathbf a=(1-k)\mathbf a+k\mathbf b$.

B lies between A,C with AB:BC=1:(k−1). Hence $\mathbf b=[\mathbf c+(k-1)\mathbf a]/k$. Rearranging gives $\mathbf c=-(k-1)\mathbf a+k\mathbf b=(1-k)\mathbf a+k\mathbf b$.

Line OC:$\mathbf r=\lambda\mathbf c$; line DB:$\mathbf r=\frac12\mathbf a+\mu(\mathbf b-\frac12\mathbf a)$, with independent real parameters. For k=3/2, $\mathbf c=-\mathbf a/2+3\mathbf b/2$. At the intersection,

$\begin{aligned}&-\frac\lambda2\mathbf a+\frac{3\lambda}2\mathbf b\\ & =\left(\frac12-\frac\mu2\right)\mathbf a+\mu\mathbf b.\end{aligned}$

Under the nonzero, nonparallel a,b assumption invoked by the source solution, compare coefficients: $-\lambda/2=1/2-\mu/2$ and $3\lambda/2=\mu$. Substitution gives $-\lambda/2=1/2-3\lambda/4$, so λ=2, μ=3. Hence $OE=2\mathbf c=-\mathbf a+3\mathbf b$. Also $DE=OE-OD=-3\mathbf a/2+3\mathbf b=3(\mathbf b-\mathbf a/2)=3DB$, so DB:DE=1:3.