= Solution
A <central force> has the form $\mathbf F=f(r)\widehat{\mathbf r}$, so it always points along the line from a fixed centre to the particle. Its torque about the centre vanishes:
$$
\frac{d\mathbf L}{dt}=\mathbf r\times\mathbf F=0,
\qquad
\mathbf L=m\mathbf r\times\dot{\mathbf r}.
$$
Thus <angular momentum> is conserved. Since $\mathbf r\cdot\mathbf L=0$, the trajectory lies in the fixed plane through the origin perpendicular to $\mathbf L$.
The area swept in time $dt$ is $dA=\frac12|\mathbf r\times d\mathbf r|$, so
$$
\boxed{\frac{dA}{dt}=\frac12|\mathbf r\times\dot{\mathbf r}|=\frac{|\mathbf L|}{2m},}
$$
which is constant. This proves <Kepler's second law>.
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